<ul> <li> <a to="/exercises/algebra">Algebra</a> <ul> <li><a to="/exercises/algebra/expression-transformations">Expression transformations</a> <ul> <li><a to="/exercises/algebra/expression-transformations/polynomial-factorization"> Polynomial factorization </a></li> <li><a to="/exercises/algebra/expression-transformations/rational-fractions"> Rational fractions </a></li> <li><a to="/exercises/algebra/expression-transformations/expressions-with-radicals"> Expressions with radicals </a></li> <li><a to="/exercises/algebra/expression-transformations/expressions-with-logarithms"> Expressions with logarithms </a></li> </ul> </li> </ul> </li></ul>

<desc> Simplify an expression in two variables containing a product of roots and logarithms </desc> <task> Simplify <tx>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}</tx></task> <solution> <tex>\sqrt{\log _{a}b+\log _{b}a+2}=</tex> <tex>\sqrt{\log _{a}b+\frac{1}{\log _{a}b}+2}=</tex> <tex>\sqrt{\frac{\left( \log _{a}b \right)^{2}+1+2\log _{a}b}{\log _{a}b}}=</tex> <tex>\frac{\sqrt{\left( \log _{a}b \right)^{2}+1+2\log _{a}b}}{\sqrt{\log _{a}b}}=</tex> <tex>\frac{\sqrt{\left( \log _{a}b+1 \right)^{2}}}{\sqrt{\log _{a}b}}=</tex> <tex>\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}</tex> <hr/> <tex>\log _{ab}a=</tex> <tex>\frac{1}{\log _{a}ab}=</tex> <tex>\frac{1}{\log _{a}a+\log _{a}b}=</tex> <tex>\frac{1}{1+\log _{a}b}</tex> <hr/> <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=</tex> <tex>\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}\cdot \frac{1}{1+\log _{a}b}\cdot \left( \sqrt{\log _{a}b} \right)^{3}=</tex> <tex>\left( \sqrt{\log _{a}b} \right)^{2}=</tex> <tex>\log _{a}b</tex> Finally, <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=\log _{a}b</tex> <tex>\text{ when } 0 < a < 1, 0 1, b > 1</tex></solution> <result> <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=\log _{a}b</tex> <tex>\text{ when } 0 < a < 1, 0 1, b > 1</tex></result> <howto> <understand> <question> What do we have? </question> <answer> We have an expression in two variables <tx>a, b</tx> containing a product of three factors <ul> <li>the first factor is a square root of an expression containing two logarithms with base <tx>a</tx> and <tx>b</tx></li> <li>the second factor is a logarithm with base <tx>ab</tx></li> <li>the third factor is a square root of logarithm with base <tx>a</tx> raised to power 3</li> </ul> </answer> <question>What should we find?</question> <answer> We should find the simplified form of the expression. Simplification is the process of replacing a mathematical expression with a simpler equivalent expression. </answer> <question>What conditions should we consider?</question> <answer> Unless otherwise specified, we should consider the expression in the context of real numbers. We should consider the domain of the initial expression, which we can define as the set of the values for which <ul> <li>the logarithms exist</li> <li>the radicant of square roots are not negative</li> </ul> The domain of the simplified expression should be restricted to the domain of the initial expression. </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li>define the domain of the expression</li> <li>transform the first factor by converting logarithms to the same base and using a factoring formula</li> <li>transform the second factor by changing the base of the logarithm</li> <li>combine the results and simplify the whole expression</li> </ul> </answer> <question> What can we say about the domain of the expression? </question> <answer> We can note that <ul> <li><tx>\log _{a}b</tx> is defined when <tx>a > 0, a\ne 1</tx> and <tx>b > 0</tx></li> <li><tx>\log _{b}a</tx> is defined when <tx>b > 0, b\ne 1</tx> and <tx>a > 0</tx></li> <li><tx>\log _{ab}a</tx> is defined when <tx>ab > 0, ab\ne 1</tx> and <tx>a > 0</tx></li> <li><tx>\sqrt{\log _{a}b+\log _{b}a+2}</tx> is defined when <tx>\log _{a}b+\log _{b}a+2 \ge 0</tx></li> <li> <tx>\sqrt{\left( \log _{a}b \right)^{3}}</tx> is defined when <tx>\log _{a}b \ge 0</tx> <ul> <li> <tx>\log _{a}b \ne 0</tx> since <tx>b\ne 1</tx> </li> <li> <tx>\log _{a}b > 0</tx> when <tx>0 < a < 1, 0 or <tx>a > 1, b > 1</tx> </li> </ul> </li> </ul> Therefore, we should consider that <tx>0 < a < 1, 0 or <tx>a > 1, b > 1</tx> </answer> <question> Consider the first factor <tx>\sqrt{\log _{a}b+\log _{b}a+2}</tx>. How can we convert the logarithms to the same base? How can we simplify this expression then? <hint> How can we use the identities <ul> <li><fl>\log _{A}B=\frac{1}{\log _{B}{A}}</fl> where <tx>B>0, B\ne 1</tx></li> <li><tx>\sqrt{\frac{A}{B}}=\frac{\sqrt{A}}{\sqrt{B}}</tx> when <tx>A\ge0, B>0</tx></li> <li><tx>A^{2}+2AB+B^{2}=\left (A+B \right)^2</tx></li> <li><tx>\sqrt{A^2}=A</tx> when <tx>A\ge0</tx></li> </ul> to transform the first factor into <tx>\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}</tx>? </hint> </question> <answer> Let's transform the first factor <tex>\sqrt{\log _{a}b+\log _{b}a+2}=</tex> <note> We note that <tx>a>0, a\ne 1</tx> and use a special case of the change-of-base formula <fl>\log _{A}B=\frac{1}{\log _{B}{A}}</fl> where <tx>B>0, B\ne 1</tx> to rewrite <tx>\log _{b}a=\frac{1}{\log _{a}b}</tx> </note> <tex>\sqrt{\log _{a}b+\frac{1}{\log _{a}b}+2}=</tex> <note> We put the expression over the common denominator <tex>\sqrt{\frac{\left( \log _{a}b \right)^{2}}{\log _{a}b}+\frac{1}{\log _{a}b}+\frac{2\log _{a}b}{\log _{a}b}}=</tex> and combine the fractions </note> <tex>\sqrt{\frac{\left( \log _{a}b \right)^{2}+1+2\log _{a}b}{\log _{a}b}}=</tex> <note> We note that <ul> <li><tx>\log _{a}b > 0</tx></li> <li><tx>\left( \log _{a}b \right)^{2}+1+2\log _{a}b > 0</tx> (since the three terms are positive)</li> </ul> and apply the property for square roots <tx>\sqrt{\frac{A}{B}}=\frac{\sqrt{A}}{\sqrt{B}}</tx> when <tx>A\ge0, B>0</tx> </note> <tex>\frac{\sqrt{\left( \log _{a}b \right)^{2}+1+2\log _{a}b}}{\sqrt{\log _{a}b}}=</tex> <note> We apply the factoring formula <tx>A^{2}+2AB+B^{2}=\left (A+B \right)^2</tx> where <tx>A=\log _{a}b, B=1</tx> </note> <tex>\frac{\sqrt{\left( \log _{a}b+1 \right)^{2}}}{\sqrt{\log _{a}b}}=</tex> <note> We note that <tx>\log_{a}b+1 > 0</tx> and apply the property of square roots <tx>\sqrt{A^2}=A</tx> when <tx>A \ge 0</tx> to rewrite <tx>\sqrt{\left( \log _{a}b+1 \right)^{2}}=\log _{a}b+1</tx> </note> <tex>\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}</tex> </answer> <question> How can we transform <tx>\log _{ab}a</tx>? <hint> Consider the identities <ul> <li><fl>\log _{A}B=\frac{1}{\log _{B}{A}}</fl> where <tx>B>0, B\ne 1</tx></li> <li><tx>\log _{X}AB=\log _{X}A+\log _{X}B</tx> when <tx>A>0, B>0</tx></li> <li><tx>\log _{A}A=1</tx></li> </ul> How can we use them to transform the second factor into <tx>\frac{1}{1+\log _{a}b}</tx>? </hint> </question> <answer> Let's rewrite the second factor <tex>\log _{ab}a=</tex> <note> We note that <tx>a>0, a\ne 1</tx> and use a special case of the change-of-base formula <fl>\log _{A}B=\frac{1}{\log _{B}{A}}</fl> where <tx>B>0, B\ne 1</tx> where <tx>A=ab, B=a</tx> </note> <tex>\frac{1}{\log _{a}ab}=</tex> <note> We note that <tx>a>0, b>0</tx> and use the logarithmic property <tx>\log _{X}AB=\log _{X}A+\log _{X}B</tx> when <tx>A>0, B>0</tx> </note> <tex>\frac{1}{\log _{a}a+\log _{a}b}=</tex> <note> We note that <tx>\log _{a}a=1</tx> </note> <tex>\frac{1}{1+\log _{a}b}</tex> Let's rewrite the whole expression <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=</tex> <note> We use the previous results <ul> <li><tx>\sqrt{\log _{a}b+\log _{b}a+2}=\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}</tx></li> <li><tx>\log _{ab}a=\frac{1}{1+\log _{a}b}</tx></li> </ul> </note> <tex>\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}\cdot \frac{1}{1+\log _{a}b}\cdot \sqrt{\left(\log _{a}b \right)^{3}}</tex> </answer> <question> How can we simplify the resulting expression? <hint> Consider the identities <ul> <li><tx>\sqrt{A^3}=\left(\sqrt{A}\right)^3</tx></li> <li><tx>\sqrt{A^2}=A</tx> when <tx>A\ge 0</tx></li> </ul> </hint> </question> <answer> Let's simplify the resulting expression <tex>\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}\cdot \frac{1}{1+\log _{a}b}\cdot \sqrt{\left(\log _{a}b \right)^{3}}=</tex> <note> We note that <tx>\log _{a}b+1 \ne 0</tx> and cancel out the common factors <tex>\frac{\cancel{\log _{a}b+1}}{\sqrt{\log _{a}b}}\cdot \frac{1}{\cancel{1+\log _{a}b}}\cdot \sqrt{\left( \log _{a}b \right)^{3}}=</tex> </note> <tex>\frac{1}{\sqrt{\log _{a}b}}\cdot \sqrt{\left( \log _{a}b \right)^{3}}=</tex> <note> We use the root property <tx>\sqrt{A^3}=\left(\sqrt{A}\right)^3</tx> to rewrite <tx>\sqrt{\left( \log _{a}b \right)^{3}}=\left(\sqrt{ \log _{a}b }\right)^{3}</tx> </note> <tex>\frac{1}{\sqrt{\log _{a}b}}\cdot \left(\sqrt{ \log _{a}b }\right)^{3}=</tex> <note> We note that <ul> <li><tx>\sqrt{\log _{a}b} \ne 0</tx></li> <li><tx>\left(\sqrt{ \log _{a}b }\right)^{3}=\sqrt{ \log _{a}b }\left(\sqrt{ \log _{a}b }\right)^{2}</tx></li> </ul> and cancel the common factors <tex>\frac{1}{\cancel{\sqrt{\log _{a}b}}}\cdot \cancel{\sqrt{ \log _{a}b }}\left(\sqrt{ \log _{a}b }\right)^{2}=</tex> </note> <tex>\left( \sqrt{\log _{a}b} \right)^{2}=</tex> <note> We note that <tx>\sqrt{\log _{a}b}>0</tx> and use the property of square root <tx>\sqrt{A^2}=A</tx> when <tx>A\ge 0</tx> </note> <tex>\log _{a}b</tex> Finally, <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=\log _{a}b</tex> <tex>\text{ when } 0 < a < 1, 0 1, b > 1</tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <tex>\sqrt{\log _{a}b+\log _{b}a+2}=</tex> <tex>\sqrt{\log _{a}b+\frac{1}{\log _{a}b}+2}=</tex> <tex>\sqrt{\frac{\left( \log _{a}b \right)^{2}+1+2\log _{a}b}{\log _{a}b}}=</tex> <tex>\frac{\sqrt{\left( \log _{a}b \right)^{2}+1+2\log _{a}b}}{\sqrt{\log _{a}b}}=</tex> <tex>\frac{\sqrt{\left( \log _{a}b+1 \right)^{2}}}{\sqrt{\log _{a}b}}=</tex> <tex>\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}</tex> <hr/> <tex>\log _{ab}a=</tex> <tex>\frac{1}{\log _{a}ab}=</tex> <tex>\frac{1}{\log _{a}a+\log _{a}b}=</tex> <tex>\frac{1}{1+\log _{a}b}</tex> <hr/> <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=</tex> <tex>\frac{\log _{a}b+1}{\sqrt{\log _{a}b}}\cdot \frac{1}{1+\log _{a}b}\cdot \left( \sqrt{\log _{a}b} \right)^{3}=</tex> <tex>\left( \sqrt{\log _{a}b} \right)^{2}=</tex> <tex>\log _{a}b</tex> Finally, <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=\log _{a}b</tex> <tex>\text{ when } 0 < a < 1, 0 1, b > 1</tex> </answer> </carryout> <lookback> <question>How can we check the result?</question> <answer> We can evaluate the initial and simplified expressions for specific values of <tx>a, b</tx> and show that the results are identical. <ul> <li>Let <tx>a=\frac{1}{2}, b=\frac{1}{2}</tx> The initial expression: <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=</tex> <tex>\sqrt{\log _{\frac{1}{2}}\frac{1}{2}+\log _{\frac{1}{2}}\frac{1}{2}+2}\; \cdot \; \log _{\frac{1}{4}}\frac{1}{2}\; \cdot \; \sqrt{\left( \log _{\frac{1}{2}}\frac{1}{2} \right)^{3}}=</tex> <tex>\sqrt{1+1+2}\; \cdot \; \frac{1}{2}\; \cdot \; 1=1</tex> The simplified expression: <tex>\log _{a}b=\log _{\frac{1}{2}}\frac{1}{2}=1</tex> </li> <li>Let <tx>a=2, b=4</tx> The initial expression: <tex>\sqrt{\log _{a}b+\log _{b}a+2}\; \cdot \; \log _{ab}a\; \cdot \; \sqrt{\left( \log _{a}b \right)^{3}}=</tex> <tex>\sqrt{\log _{2}4+\log _{4}2+2}\; \cdot \; \log _{8}2\; \cdot \; \sqrt{\left( \log _{2}4 \right)^{3}}=</tex> <tex>\sqrt{2+\frac{1}{2}+2}\; \cdot \; \frac{1}{3}\; \cdot \; \sqrt{8}=</tex> <tex>\sqrt{\frac{9}{2}}\; \cdot \; \frac{1}{3}\; \cdot \; \sqrt{8}=\frac{3}{\sqrt{2}}\; \cdot \; \frac{1}{3}\; \cdot \; 2\sqrt{2}=2</tex> The simplified expression: <tex>\log _{a}b=\log _{2}4=2</tex> </li> </ul> </answer> </lookback></howto>

<desc> Simplify an irrational expression containing exponents and logarithms. </desc> <task> Simplify <tx>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1</tx></task> <solution> <tex>2^{ \dfrac{\log a}{\log \sqrt{2}}}=</tex> <tex>\left(2^{\textstyle \log a}\right)^{\dfrac{1}{\log 2^\frac{1}{2}}}=</tex> <tex>\left(a^{\textstyle \log 2}\right)^{\dfrac{1}{\frac{1}{2}\log 2}}=</tex> <tex>a^{\textstyle \dfrac{2\log 2}{\log 2}}=</tex> <tex>a^2</tex> <hr/> <tex>a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}=</tex> <tex>a\cdot a^{\dfrac{1}{\log _{4}a^{2}}}=</tex> <tex>a\cdot a^{\dfrac{1}{\log _{2}a}}=</tex> <tex>a\cdot a^{\textstyle \log _{a}2}=</tex> <tex>2a</tex> <hr/> <tex>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1=</tex> <tex>\sqrt{1+a^{2}-2a}-1=</tex> <tex>\sqrt{\left( a-1 \right)^{2}}-1</tex> <tex>\left| a-1 \right|-1=</tex> <tex>\begin{cases} -\left(a-1\right)-1 \text{ when } 0 < a < 1 \\ a-1-1 \text{ when } a > 1 \end{cases} </tex> Finally, <tex>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1=\begin{cases} -a \text{ when } 0 < a < 1 \\ a-2 \text{ when } a > 1 \end{cases} </tex></solution> <result> <tex>\begin{cases} -a \text{ when } 0 < a < 1 \\ a-2 \text{ when } a > 1 \end{cases} </tex></result> <howto> <understand> <question> What do we have? </question> <answer> We have an expression in one variable <tx>a</tx> containing square roots, exponents and logarithms. Note: <tx>\log a = \log _{10}a</tx> You can read more about common logarithms <a target="_blank" href="https://en.wikipedia.org/wiki/Common_logarithm">here</a> </answer> <question>What should we find?</question> <answer> We should find the simplified form of the expression. Simplification is the process of replacing a mathematical expression with a simpler equivalent expression. </answer> <question>What conditions should we consider?</question> <answer> Unless otherwise specified, we should consider the expression in the context of real numbers. We should consider the domain of the initial expression, which we can define as the set of the values for which <ul> <li>the logarithms exist</li> <li>the denominators are not 0</li> <li>the radicant of square roots are not negative</li> </ul> The domain of the simplified expression should be restricted to the domain of the initial expression. </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li>define the domain of the expression</li> <li>simplify the terms <tx>2^{ \dfrac{\log a}{\log \sqrt{2}}}</tx> and <tx>a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}</tx></li> <li>transform the resulting expression to eliminate the radical</li> </ul> </answer> <question> What can we say about the domain of the expression? </question> <answer> We can note that <ul> <li><tx>\log a</tx> is defined when <tx>a > 0</tx></li> <li><tx>\log _{4}a^2 \ne0</tx> to prevent division by 0 in <tx>\dfrac{1}{\log _{4}a^{2}}</tx></li> <li><tx>1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}} \ge 0</tx> to prevent a negative radicant</li> </ul> Therefore, we should consider that <tx>a>0, a\ne 1</tx> </answer> <question> How can we rewrite <tx>2^{ \dfrac{\log a}{\log \sqrt{2}}}</tx> to cancel out its logarithms? <hint> How can we use the identities <ul> <li><tx>A^{nm}=\left(A^n\right)^m</tx></li> <li><tx>A^{\textstyle \log _{X}B}=B^{\textstyle \log _{X}A}</tx> were <tx>A>0</tx></li> <li><tx>\log _{A}B^n=n\log _{A}B</tx> where <tx>B>0</tx></li> </ul> to get <tx>a^2</tx>? </hint> </question> <answer> Let's simplify the first exponential expression <tex>2^{ \dfrac{\log a}{\log \sqrt{2}}}=</tex> <note> We rewrite the exponent <tex>2^{\textstyle \log a \cdot \dfrac{1}{\log \sqrt{2}}}=</tex> and use the exponentiation property <tx>A^{nm}=\left(A^n\right)^m</tx> where <tx>A=2, n=\log a, m=\dfrac{1}{\log \sqrt{2}}</tx> </note> <tex>\left(2^{\textstyle \log a}\right)^{\dfrac{1}{\log \sqrt{2}}}=</tex> <note> We apply the identity <tx>A^{\textstyle \log _{X}B}=B^{\textstyle \log _{X}A}</tx> when <tx>A>0</tx> to rewrite <tx>a^{\textstyle \log 2}=2^{\textstyle \log a}</tx> You can try to prove this identity <a target="_blank" href="https://www.dzdx.xyz/exercise/77">here</a> </note> <tex>\left(a^{\textstyle \log 2}\right)^{\dfrac{1}{\log \sqrt{2}}}=</tex> <note> We note that <tx>\sqrt{2}=2^{\frac{1}{2}}</tx> <tex>\left(a^{\textstyle \log 2}\right)^{\dfrac{1}{2^{\frac{1}{2}}}}=</tex> and apply the logarithmic property <tx>\log _{A}B^n=n\log _{A}B</tx> where <tx>B>0</tx> </note> <tex>\left(a^{\textstyle \log 2}\right)^{\dfrac{1}{\frac{1}{2}\log 2}}=</tex> <note> We reduce the complex fraction <tx>\dfrac{1}{\frac{1}{2}\log 2}=\dfrac{2}{\log 2}</tx> <tex>\left(a^{\textstyle \log 2}\right)^{\dfrac{1}{\log 2^{\frac{1}{2}}}}=</tex> and apply the exponential property <tx>\left(A^n\right)^m=A^{nm}</tx> where <tx>A=2, n=\log 2, m=\frac{2}{\log 2}</tx> </note> <tex>a^{\textstyle \dfrac{2\log 2}{\log 2}}=</tex> <note> We note that <tx>\log 2\ne 0</tx> and cancel the common factor <tex>a^{\textstyle \dfrac{2\cancel{\log 2}}{\cancel{\log 2}}}=</tex> </note> <tex>a^2</tex> </answer> <question> How can we simplify <tx>a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}</tx>? <hint> How can we use the identities <ul> <li><tx>A^{n+m}=A^{n}A^{m}</tx></li> <li><tx>\log _{A^n}B^{n}=\log _{A}{B}</tx> when <tx>A>0, B>0</li> <li><tx>\frac{1}{\log _{A}{B}}=\log _{B}A</tx></li> <li><tx>A^{\textstyle \log_{A}B}=B</tx></li> </ul> to get <tx>2a</tx>? </hint> </question> <answer> Let's simplify the second exponential expression <tex>a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}=</tex> <note> We use the exponential property <tx>A^{n+m}=A^{n}A^{m}</tx> where <tx>A=a, n=1, m=\frac{1}{\log _{4}a^{2}}</tx> </note> <tex>a\cdot a^{\dfrac{1}{\log _{4}a^{2}}}=</tex> <note> We note that <tx>a>0</tx> and use the logarithmic property <tx>\log _{A^n}B^{n}=\log _{A}{B}</tx> when <tx>A>0, B>0</tx> to rewrite <tx>\log _{4}a^{2}=\log _{2}a</tx> </note> <tex>a\cdot a^{\dfrac{1}{\log _{2}a}}=</tex> <note> We use the identity <tx>\frac{1}{\log _{A}{B}}=\log _{B}A</tx> where <tx>A=2, B=a</tx> </note> <tex>a\cdot a^{\textstyle \log _{a}2}=</tex> <note> We apply the identity <tx>A^{\textstyle \log_{A}B}=B</tx> where <tx>A=a, B=2</tx> </note> <tex>2a</tex> Let's rewrite the whole expression <tex>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1=</tex> <note> We use the obtained results <ul> <li><tx>2^{ \dfrac{\log a}{\log \sqrt{2}}}=a^2</tx></li> <li><tx>a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}=2a</tx></li> </ul> </note> <tex>\sqrt{1+a^{2}-2a}-1</tex> </answer> <question> How can we transform the resulting expression to eliminate the radical? <hint> How can we use the formulas <ul> <li><tx>A^2-2AB+B^2=\left(A-B\right)^2</tx></li> <li><tx>\sqrt{A^2}=\left|A\right|</tx></li> </ul> to get <tx>\left| a-1 \right|-1</tx>? </hint> </question> <answer> Let's remove the radical from the resulting expression <tex>\sqrt{1+a^{2}-2a}-1=</tex> <note> We apply the formula <tx>A^2-2AB+B^2=\left(A-B\right)^2</tx> where <tx>A=a, B=1</tx> </note> <tex>\sqrt{\left( a-1 \right)^{2}}-1=</tex> <note> We use the identity <tx>\sqrt{A^2}=\left|A\right|</tx> where <tx>A=a-1</tx> </note> <tex>\left| a-1 \right|-1</tex> </answer> <question> Consider the definition of absolute value of a real number <tex> \left|A\right|= \begin{cases} A\phantom{....}\text{if}\phantom{....} A\ge0 \\ -A\phantom{....}\text{if}\phantom{....} A<0 \end{cases} </tex> How can we rewrite the resulting expression considering the different values of <tx>a</tx>? </question> <answer> Let's rewrite the resulting expression <tex>\left| a-1 \right|-1=</tex> <note> We use the definition of absolute value of a real number <tex> \left|A\right|= \begin{cases} A\phantom{....}\text{if}\phantom{....} A\ge0 \\ -A\phantom{....}\text{if}\phantom{....} A<0 \end{cases} </tex> where <tx>A=a-1</tx> and note that <ul> <li><tx>a-1 > 0 </tx> when <tx>a > 1</tx> (<tx>a-1 \ne 0 </tx> since <tx>a\ne 1</tx>)</li> <li><tx>a-1 < 0 </tx> when <tx>0 < a < 1</tx></li> </ul> </note> <tex>\begin{cases} -\left(a-1\right)-1 \text{ when } 0 < a < 1 \\ a-1-1 \text{ when } a > 1 \end{cases} </tex> Finally, <tex>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1=\begin{cases} -a \text{ when } 0 < a < 1 \\ a-2 \text{ when } a > 1 \end{cases} </tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <tex>2^{ \dfrac{\log a}{\log \sqrt{2}}}=</tex> <tex>\left(2^{\textstyle \log a}\right)^{\dfrac{1}{\log 2^\frac{1}{2}}}=</tex> <tex>\left(a^{\textstyle \log 2}\right)^{\dfrac{1}{\frac{1}{2}\log 2}}=</tex> <tex>a^{\textstyle \dfrac{2\log 2}{\log 2}}=</tex> <tex>a^2</tex> <hr/> <tex>a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}=</tex> <tex>a\cdot a^{\dfrac{1}{\log _{4}a^{2}}}=</tex> <tex>a\cdot a^{\dfrac{1}{\log _{2}a}}=</tex> <tex>a\cdot a^{\textstyle \log _{a}2}=</tex> <tex>2a</tex> <hr/> <tex>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1=</tex> <tex>\sqrt{1+a^{2}-2a}-1=</tex> <tex>\sqrt{\left( a-1 \right)^{2}}-1=</tex> <tex>\left| a-1 \right|-1=</tex> <tex>\begin{cases} -\left(a-1\right)-1 \text{ when } 0 < a < 1 \\ a-1-1 \text{ when } a > 1 \end{cases} </tex> Finally, <tex>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1=\begin{cases} -a \text{ when } 0 < a < 1 \\ a-2 \text{ when } a > 1 \end{cases} </tex> </answer> </carryout> <lookback> <question>How can we check the result?</question> <answer> We can evaluate the initial and simplified expressions for specific values of <tx>a</tx> and show that the results are identical. <ul> <li>Let <tx>a=\sqrt{2}</tx> The initial expression: <tex>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1=</tex> <tex>\sqrt{1+2^{ \dfrac{\log \sqrt{2}}{\log \sqrt{2}}}-\left(\sqrt{2}\right)^{\textstyle 1+\dfrac{1}{\log _{4}2}}}-1=</tex> <tex>\sqrt{1+2-\left(\sqrt{2}\right)^{3}}-1=</tex> <tex>\sqrt{\sqrt{2}^2-2\sqrt{2}+1}-1=</tex> <tex>\sqrt{\left(\sqrt{2}-1\right)^2}-1=\sqrt{2}-1-1=\sqrt{2}-2</tex> The simplified expression (<tx>\sqrt{2}>1</tx>): <tex>a-2=\sqrt{2}-2</tex> </li> </ul> <ul> <li>Let <tx>a=\frac{1}{\sqrt{2}}</tx> The initial expression: <tex>\sqrt{1+2^{ \dfrac{\log a}{\log \sqrt{2}}}-a^{\textstyle 1+\dfrac{1}{\log _{4}a^{2}}}}-1=</tex> <tex>\sqrt{1+2^{ \dfrac{\log \frac{1}{\sqrt{2}}}{\log \sqrt{2}}}-\left(\frac{1}{\sqrt{2}}\right)^{\textstyle 1+\dfrac{1}{\log _{4}\frac{1}{2}}}}-1=</tex> <tex>\sqrt{1+2^{ \dfrac{\log \sqrt{2}^{-1}}{\log \sqrt{2}}}-\left(\frac{1}{\sqrt{2}}\right)^{\textstyle 1+\dfrac{1}{\log _{4}2^{-1}}}}-1=</tex> <tex>\sqrt{1+2^{-1}-\left(\frac{1}{\sqrt{2}}\right)^{1-2}}-1=</tex> <tex>\sqrt{1+\frac{1}{2}-\sqrt{2}}-1=</tex> <tex>\sqrt{1-2\cdot\frac{1}{\sqrt{2}}+\left(\frac{1}{\sqrt{2}}\right)^2}-1=</tex> <tex>\sqrt{\left(1-\frac{1}{\sqrt{2}}\right)^2}-1=1-\frac{1}{\sqrt{2}}-1=-\frac{1}{\sqrt{2}}</tex> The simplified expression (<tx>\frac{1}{\sqrt{2}} < 1</tx>): <tex>-a=-\frac{1}{\sqrt{2}}</tex> </li> </ul> </answer> </lookback></howto>

<desc> Simplify an exponential expression containing logarithms </desc> <task> Simplify <tx>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}</tx></task> <solution> <center> <tx>\frac{\log _{100}a}{\log a}=</tx> <tx>\frac{\log _{100}a}{\log _{10}a}=</tx> <tx>\frac{\log _{100}a}{\log _{10^{2}}a^{2}}=</tx> <tx>\frac{\log _{100}a}{2\log _{100}a}=</tx> <tx>\frac{1}{2}</tx> </center> <hr/> <tex>\frac{\log _{100}b}{\log b}=\frac{1}{2}</tex> <hr/> <tex>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=</tex> <tex>\left( b^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=</tex> <tex>\left( ba \right)^{\textstyle \dfrac{1}{2} \cdot 2\log _{ab}\left( a+b \right)}=</tex> <tex>\left( ba \right)^{\textstyle \log _{ab}\left( a+b \right)}=</tex> <tex>a+b</tex> Finally, <tex>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=a+b</tex> <tex>\text{ when } a>0, a\ne1 , b>0, b\ne1, ab\ne 1</tex></solution> <result> <center> <tx>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=a+b</tx> <tx>\text{ when } a>0, a\ne1 , b>0, b\ne1, ab\ne 1</tx> </center></result> <howto> <understand> <question> What do we have? </question> <answer> We have an exponential expression in two variables <tx>a, b</tx> that contains quotients of logarithms. Note: <tx>\log a = \log _{10}a</tx> You can read more about common logarithms <a target="_blank" href="https://en.wikipedia.org/wiki/Common_logarithm">here</a> </answer> <question>What should we find?</question> <answer> We should find the simplified form of the expression. Simplification is the process of replacing a mathematical expression with a simpler equivalent expression. </answer> <question>What conditions should we consider?</question> <answer> Unless otherwise specified, we should consider the expression in the context of real numbers. We should consider the domain of the initial expression, which we can define as the set of the values for which <ul> <li>logarithms exist</li> <li>denominators are not 0</li> </ul> The domain of the simplified expression should be restricted to the domain of the initial expression. </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li>define the domain of the expression</li> <li>simplify <tx>\frac{\log _{100}a}{\log a}</tx> and <tx>\frac{\log _{100}b}{\log b}</tx> by expressing the quotients in terms of logarithms with base 100</li> <li>use the obtained results to simplify the whole expression</li> </ul> </answer> <question> What can we say about the domain of the expression? </question> <answer> We can note that <ul> <li><tx>\log _{100}a</tx> and <tx>\log a</tx> are defined when <tx>a > 0</tx></li> <li><tx>\log _{100}b</tx> and <tx>\log b</tx> are defined when <tx>b > 0</tx></li> <li><tx>\log _{ab}\left(a+b\right)</tx> is defined when <tx>ab > 0, ab\ne1</tx> and <tx>a+b>0</tx></li> <li><tx>\log a \ne0</tx> to prevent division by 0 in <tx>\dfrac{\log _{100}a}{\log a}</tx></li> <li><tx>\log b \ne0</tx> to prevent division by 0 in <tx>\dfrac{\log _{100}b}{\log b}</tx></li> </ul> Therefore, we should consider that <tx>a>0, a\ne 1</tx>, <tx>b>0, b\ne1</tx> and <tx>ab\ne1</tx> </answer> <question> How can we represent <tx>\frac{\log _{100}a}{\log a}</tx> and <tx>\frac{\log _{100}b}{\log b}</tx> as quotients of logarithms with base 100? How can we simplify the expressions then? <hint> Consider the identities <ul> <li><fl>\log _{A}B=\log _{A^n}B^n</fl></li> <li><fl>\log _{A}B^n=n\log _{A}B</fl> when <tx>B>0</tx></li> </ul> How can we use them to transform the initial expression to <tx>\left( b^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)} \text{?}</tx> </hint> </question> <answer> Let's the quotients of logarithms <tex>\frac{\log _{100}a}{\log a}=</tex> <note> We apply the logarithmic property <fl>\log _{A}B=\log _{A^n}B^n</fl> to rewrite <tx>\log a=\log _{10}a=\log _{10^{2}}a^{2}</tx> </note> <tex>\frac{\log _{100}a}{\log _{10^{2}}a^{2}}=</tex> <note> We note that <tx>10^2=100</tx> and apply the property <fl>\log _{A}B^n=n\log _{A}B</fl> when <tx>B>0</tx> </note> <tex>\frac{\log _{100}a}{2\log _{100}a}=</tex> <note> We note that <tx>\log _{100}a\ne0</tx> (since <tx>a\ne 1</tx>) and cancel the common factor <tex>\frac{\cancel{\log _{100}a}}{2\cancel{\log _{100}a}}=</tex> </note> <tex>\frac{1}{2}</tex> In the same way, we can show that <tex>\frac{\log _{100}b}{\log b}=\frac{1}{2}</tex> Let's rewrite the whole expression <tex>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=</tex> <note> We use the obtained results <ul> <li><tx>\frac{\log _{100}a}{\log a}=\frac{1}{2}</tx></li> <li><tx>\frac{\log _{100}b}{\log b}=\frac{1}{2}</tx></li> </ul> </note> <tex>\left( b^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}</tex> </answer> <question> How can we simplify the resulting expression further? What transformations should we perform to apply the identity <fl>A^{\textstyle \log _{A}B}=B</fl>? <hint> Consider the properties <ul> <li><fl>A^nB^n=\left( AB \right)^n</fl></li> <li><fl>\left( A^n \right)^m=A^{nm}</fl></li> </ul> </hint> </question> <answer> Let's simplify the resulting expression <tex>\left( b^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=</tex> <note> We use the exponentiation property <fl>A^nB^n=\left( AB \right)^n</fl> </note> <tex>\left( \left(ab\right)^{\dfrac{1}{2}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=</tex> <note> We use the exponentiation property <fl>\left( A^n \right)^m=A^{nm}</fl> where <tx>A=ab, n=\frac{1}{2}, m=2\log _{ab}\left( a+b \right)</tx> </note> <tex>\left( ab \right)^{\textstyle \dfrac{1}{2} \cdot 2\log _{ab}\left( a+b \right)}=</tex> <note> We note that <tx>\dfrac{1}{2} \cdot 2=1</tx> </note> <tex>\left( ab \right)^{\textstyle \log _{ab}\left( a+b \right)}=</tex> <note> We apply the identity <fl>A^{\textstyle \log _{A}B}=B</fl> </note> <tex>a+b</tex> Finally, <tex>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=a+b</tex> <tex>\text{ when } a>0, a\ne1 , b>0, b\ne1, ab\ne 1</tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <center> <tx>\frac{\log _{100}a}{\log a}=</tx> <tx>\frac{\log _{100}a}{\log _{10}a}=</tx> <tx>\frac{\log _{100}a}{\log _{10^{2}}a^{2}}=</tx> <tx>\frac{\log _{100}a}{2\log _{100}a}=</tx> <tx>\frac{1}{2}</tx> </center> <hr/> <tex>\frac{\log _{100}b}{\log b}=\frac{1}{2}</tex> <hr/> <tex>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=</tex> <tex>\left( b^{\dfrac{1}{2}}\cdot a^{\dfrac{1}{2}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=</tex> <tex>\left( ba \right)^{\textstyle \dfrac{1}{2} \cdot 2\log _{ab}\left( a+b \right)}=</tex> <tex>\left( ba \right)^{\textstyle \log _{ab}\left( a+b \right)}=</tex> <tex>a+b</tex> Finally, <tex>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=a+b</tex> <tex>\text{ when } a>0, a\ne1 , b>0, b\ne1, ab\ne 1</tex> </answer> </carryout> <lookback> <question>How can we check the result?</question> <answer> We can evaluate the initial and simplified expressions for specific values of <tx>a, b</tx> and show that the results are identical. <ul> <li>Let <tx>a=10, b=10</tx> The initial expression: <tex>\left( b^{\dfrac{\log _{100}a}{\log a}}\cdot a^{\dfrac{\log _{100}b}{\log b}} \right)^{\textstyle 2\log _{ab}\left( a+b \right)}=</tex> <tex>\left( 10^{\dfrac{\log _{100}10}{\log 10}}\cdot 10^{\dfrac{\log _{100}10}{\log 10}} \right)^{\textstyle 2\log _{100} 20}=</tex> <tex>\left( 10^{\dfrac{1}{2}} \cdot 10^{\dfrac{1}{2}} \right)^{\textstyle 2\log _{100}20}=</tex> <tex>10^{\textstyle 2\log _{100}20}=100^{\textstyle \log _{100}20}=20</tex> The simplified expression: <tex>a+b=10+10=20</tex> </li> </ul> </answer> </lookback></howto>

<desc> Simplify a logarithmic expression with two variables a, b </desc> <task> Simplify <tx>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1</tx></task> <solution> <tex>\log _{a}b+\log _{b}a+2=</tex> <tex>\frac{1}{\log _{b}a}+\log _{b}a+2=</tex> <tex>\frac{1+\left( \log _{b}a \right)^{2}+2\log _{b}a}{\log _{b}a}=</tex> <tex>\frac{\left( \log _{b}a+1 \right)^{2}}{\log _{b}a}</tex> <hr/> <tex>\log _{a}b-\log _{ab}b=</tex> <tex>\frac{1}{\log _{b}a}-\frac{1}{\log _{b}ab}=</tex> <tex>\frac{\log _{b}ab-\log _{b}a}{\log _{b}a\; \cdot \; \log _{b}ab}=</tex> <tex>\frac{\log _{b}a+\log _{b}b-\log _{b}a}{\log _{b}a\; \cdot \; \left( \log _{b}a+\log _{b}b \right)}=</tex> <tex>\frac{1}{\log _{b}a\; \cdot \; \left( \log _{b}a+1 \right)}</tex> <hr/> <tex>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1=</tex> <tex>\frac{\left( \log _{b}a+1 \right)^{2}}{\log _{b}a}\cdot \frac{1}{\log _{b}a\; \cdot \; \left( \log _{b}a+1 \right)}\cdot \log _{b}a-1=</tex> <tex>\frac{\log _{b}a+1}{\log _{b}a}-1=</tex> <tex>\frac{\log _{b}a+1-\log _{b}a}{\log _{b}a}=</tex> <tex>\frac{1}{\log _{b}a}=\log _{a}b</tex> Finally, <center> <tex>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1=\log _{a}b</tex> when <tx>a>0, a\ne 1, b>0, b\ne 1</tx> and <tx>ab \ne 1</tx> </center></solution> <result> <center> <tex>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1=\log _{a}b</tex> when <tx>a>0, a\ne 1, b>0, b\ne 1</tx> and <tx>ab \ne 1</tx> </center></result> <howto> <understand> <question> What do we have? </question> <answer> We have an expression in two variables <tx>a, b</tx> which consists of two terms. The first term is a product of three logarithmic expressions. The second term is a constant term. </answer> <question>What should we find?</question> <answer> We should find the simplified form of the expression. Simplification is the process of replacing a mathematical expression with a simpler equivalent expression. </answer> <question>What conditions should we consider?</question> <answer> Unless otherwise specified, we should consider the expression in the context of real numbers. We should consider the domain of the initial expression, which we can define as the set of the values for which logarithms exist. The domain of the simplified expression should be restricted to the domain of the initial expression. </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li>define the domain of the expression</li> <li>rewrite the factors of the first term as expressions with logarithms with base <tx>b</tx> <li>simplify the obtained expressions</li> <li>combine the intermediate results and simplify the whole expression</li> </ul> </answer> <question> What can we say about the domain of the expression? </question> <answer> We can note that <ul> <li><tx>\log _{a}b</tx> is defined when <tx>a > 0, a\ne1 </tx> and <tx>b > 0</tx></li> <li><tx>\log _{b}a</tx> is defined when <tx>b > 0, b\ne1 </tx> and <tx>a > 0</tx></li> <li><tx>\log _{ab}b</tx> is defined when <tx>ab > 0, ab\ne1 </tx> and <tx>b > 0</tx></li> </ul> Therefore, we should consider that <tx>a>0, a\ne 1, b>0, b\ne 1</tx> and <tx>ab \ne 1</tx> </answer> <question> How can we rewrite <tx>\log _{a}b+\log _{b}a+2</tx> in terms of logarithm with base <tx>b</tx>? How can we simplify the expression then? <hint> How can we use the identities <ul> <li><fl>\log _{B}A=\frac{1}{\log _{A}B}</fl> where <tx>A>0, A\ne 1</tx></li> <li><fl>A^{2}+2AB+B^{2}=\left( A+B \right)^{2}</fl></li> </ul> to get <tx>\frac{\left( \log _{b}a+1 \right)^{2}}{\log _{b}a} \text{?}</tx> </hint> </question> <answer> Let's simplify the first factor <tex>\log _{a}b+\log _{b}a+2=</tex> <note> We note that <tx>b>0, b\ne 1</tx> and apply the logarithmic property <fl>\log _{B}A=\frac{1}{\log _{A}B}</fl> where <tx>A>0, A\ne 1</tx> to rewrite <tx>\log _{a}b=\frac{1}{\log _{b}a}</tx> </note> <tex>\frac{1}{\log _{b}a}+\log _{b}a+2=</tex> <note> We put the terms over a common denominator <tex>\frac{1}{\log _{b}a}+\frac{\left( \log _{b}a \right)^{2}}{\log _{b}a}+\frac{2\log _{b}a}{\log _{b}a}=</tex> and combine the fractions </note> <tex>\frac{1+\left( \log _{b}a \right)^{2}+2\log _{b}a}{\log _{b}a}=</tex> <note> We use the factoring formula <fl>A^{2}+2AB+B^{2}=\left( A+B \right)^{2}</fl> where <tx>A=\log _{b}a, B=1</tx> </note> <tex>\frac{\left( \log _{b}a+1 \right)^{2}}{\log _{b}a}</tex> </answer> <question> How can we rewrite <tx>\log _{a}b-\log _{ab}b</tx> in terms of logarithm with base <tx>b</tx>? How can we simplify the expression then? <hint> Consider the logarithmic properties <ul> <li><fl>\log _{B}A=\frac{1}{\log _{A}B}</fl> when <tx>A>0, A\ne 1</tx></li> <li><fl>\log _{C}AB=\log _{C}A + \log _{C}B</fl> when <tx>A>0, B>0</tx></li> <li><fl>\log _{A}A=1</fl></li> </ul> How can we use them to get <tx>\frac{1}{\log _{b}a\; \cdot \; \left( \log _{b}a+1 \right)} \text{?}</tx> </hint> </question> <answer> Let's simplify the second factor <tex>\log _{a}b-\log _{ab}b=</tex> <note> We note that <tx>b>0, b\ne 1</tx> and apply the logarithmic property <fl>\log _{B}A=\frac{1}{\log _{A}B}</fl> where <tx>A>0, A\ne 1</tx> to rewrite <ul> <li><tx>\log _{a}b=\frac{1}{\log _{b}a}</tx></li> <li><tx>\log _{ab}b=\frac{1}{\log _{b}ab}</tx></li> </ul> </note> <tex>\frac{1}{\log _{b}a}-\frac{1}{\log _{b}ab}=</tex> <note> We put the terms over a common denominator <tex>\frac{\log _{b}ab}{\log _{b}ab \; \cdot \; \log _{b}a}-\frac{\log _{b}a}{\log _{b}ab \; \cdot \; \log _{b}a}=</tex> and combine the fractions </note> <tex>\frac{\log _{b}ab-\log _{b}a}{\log _{b}a\; \cdot \; \log _{b}ab}=</tex> <note> We note that <tx>a>0, b>0</tx> and apply the logarithmic property <fl>\log _{C}AB=\log _{C}A + \log _{C}B</fl> when <tx>A>0, B>0</tx> to rewrite <tx>\log _{b}ab=\log _{b}a+\log _{b}b</tx> </note> <tex>\frac{\log _{b}a+\log _{b}b-\log _{b}a}{\log _{b}a\; \cdot \; \left( \log _{b}a+\log _{b}b \right)}=</tex> <note> We note that <tx>\log _{b}b=1</tx> and add like terms <tex>\frac{\cancel{\log _{b}a}+1- \cancel{\log _{b}a}}{\log _{b}a\; \cdot \; \left( \log _{b}a+1\right)}=</tex> </note> <tex>\frac{1}{\log _{b}a\; \cdot \; \left( \log _{b}a+1 \right)}</tex> We can rewrite the whole expression <tex>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1=</tex> <note> We use the obtained results <ul> <li><tx>\log _{a}b+\log _{b}a+2=\frac{\left( \log _{b}a+1 \right)^{2}}{\log _{b}a}</tx></li> <li><tx>\log _{a}b-\log _{ab}b=\frac{1}{\log _{b}a\; \cdot \; \left( \log _{b}a+1 \right)}</tx></li> </ul> </note> <tex>\frac{\left( \log _{b}a+1 \right)^{2}}{\log _{b}a}\cdot \frac{1}{\log _{b}a\; \cdot \; \left( \log _{b}a+1 \right)}\cdot \log _{b}a-1=</tex> <note> We note that <ul> <li><tx>\log _{b}a\ne 0</tx> since <tx>a\ne 1</tx></li> <li><tx>\log _{b}a+1\ne0</tx> since <tx>\log _{b}a\ne -1</tx> as <tx>ab\ne1</tx></li> </ul> and cancel the common factors <tex>\frac{\left( \log _{b}a+1 \right)^{\cancel{2}}}{\log _{b}a}\cdot \frac{1}{\cancel{\log _{b}a}\; \cdot \; \cancel{\left( \log _{b}a+1 \right)}}\cdot \cancel{\log _{b}a}-1=</tex> </note> <tex>\frac{\log _{b}a+1}{\log _{b}a}-1</tex> </answer> <question> How can we simplify the resulting expression? </question> <answer> Let's simplify the expression <tex>\frac{\log _{b}a+1}{\log _{b}a}-1=</tex> <note> We put the terms over a common denominator <tex>\frac{\log _{b}a+1}{\log _{b}a}-\frac{\log _{b}a}{\log _{b}a}=</tex> and combine the fractions </note> <tex>\frac{\log _{b}a+1-\log _{b}a}{\log _{b}a}=</tex> <note> We cancel out the terms in the numerator <tex>\frac{\cancel{\log _{b}a}+1-\cancel{\log _{b}a}}{\log _{b}a}=</tex> </note> <tex>\frac{1}{\log _{b}a}=\log _{a}b</tex> Therefore, <center> <tex>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1=\log _{a}b</tex> when <tx>a>0, a\ne 1, b>0, b\ne 1</tx> and <tx>ab \ne 1</tx> </center> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <tex>\log _{a}b+\log _{b}a+2=</tex> <tex>\frac{1}{\log _{b}a}+\log _{b}a+2=</tex> <tex>\frac{1+\left( \log _{b}a \right)^{2}+2\log _{b}a}{\log _{b}a}=</tex> <tex>\frac{\left( \log _{b}a+1 \right)^{2}}{\log _{b}a}</tex> <hr/> <tex>\log _{a}b-\log _{ab}b=</tex> <tex>\frac{1}{\log _{b}a}-\frac{1}{\log _{b}ab}=</tex> <tex>\frac{\log _{b}ab-\log _{b}a}{\log _{b}a\; \cdot \; \log _{b}ab}=</tex> <tex>\frac{\log _{b}a+\log _{b}b-\log _{b}a}{\log _{b}a\; \cdot \; \left( \log _{b}a+\log _{b}b \right)}=</tex> <tex>\frac{1}{\log _{b}a\; \cdot \; \left( \log _{b}a+1 \right)}</tex> <hr/> <tex>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1=</tex> <tex>\frac{\left( \log _{b}a+1 \right)^{2}}{\log _{b}a}\cdot \frac{1}{\log _{b}a\; \cdot \; \left( \log _{b}a+1 \right)}\cdot \log _{b}a-1=</tex> <tex>\frac{\log _{b}a+1}{\log _{b}a}-1=</tex> <tex>\frac{\log _{b}a+1-\log _{b}a}{\log _{b}a}=</tex> <tex>\frac{1}{\log _{b}a}=\log _{a}b</tex> Finally, <center> <tex>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1=\log _{a}b</tex> when <tx>a>0, a\ne 1, b>0, b\ne 1</tx> and <tx>ab \ne 1</tx> </center> </answer> </carryout> <lookback> <question>How can we check the result?</question> <answer> We can evaluate the initial and simplified expressions for specific values of <tx>a, b</tx> and show that the results are identical. <ul> <li>Let <tx>a=2, b=2</tx> The initial expression: <tex>\left( \log _{a}b+\log _{b}a+2 \right)\left( \log _{a}b-\log _{ab}b \right)\log _{b}a-1=</tex> <tex>\left( \log _{2}2+\log _{2}2+2 \right)\left( \log _{2}2-\log _{4}2 \right)\log _{2}2-1=</tex> <tex>\left( 1+1+2 \right)\left( 1-\frac{1}{2} \right) \cdot 1-1=4\cdot\frac{1}{2}-1=1</tex> The simplified expression: <tex>\log _{a}b = \log _{2}2= 1 </tex> </li> </ul> </answer> </lookback> </howto>

<desc> Prove that a logarithmic equation holds for all values satisfying a^2+b^2=c^2 </desc> <task> Prove that <tx>\log _{c+b}a+\log _{c-b}a=2\log _{c+b}a\; \cdot \; \log _{c-b}a</tx> given that <tx>a^{2}+b^{2}=c^{2}</tx> where <tx>a>0, b>0, c>0</tx></task> <solution> Let <tx>a\ne 1</tx> <tex>\log _{c+b}a+\log _{c-b}a\stackrel{a\ne1}{=}</tex> <tex>\frac{1}{\log _{a}\left( c+b \right)}+\frac{1}{\log _{a}\left( c-b \right)}=</tex> <tex>\frac{\log _{a}\left( c-b \right)+\log _{a}\left( c+b \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <tex>\frac{\log _{a}\left( c^{2}-b^{2} \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <tex>\frac{\log _{a}a^{2}}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <tex>2\log _{a}a\; \cdot \; \log _{c+b}a\; \cdot \; \log _{c-b}a=</tex> <tex>2\log _{c+b}a\; \cdot \; \log _{c-b}a</tex> Let's check that the equation holds for <tx>a=1</tx> The left-hand side <tex>\log _{c+b}a+\log _{c-b}a=\log _{c+b}1+\log _{c-b}1=0+0=0</tex> The right-hand side <tex>2\log _{c+b}a\; \cdot \; \log _{c-b}a=2\log _{c+b}1\; \cdot \; \log _{c-b}1=2\cdot0\cdot0=0</tex></solution> <howto> <understand> <question> What do we have? </question> <answer> We have two equations in three variables <tx>a, b, c</tx>. The first equation is a logarithmic equation. The second equation is a polynomial equation of degree 2. </answer> <question> What should we prove? </question> <answer> We should prove that the equation holds for all values <tx>a, b, c</tx> satisfying the second equation. </answer> <question>What conditions should we consider?</question> <answer> Unless otherwise specified, we should consider the equations in the context of real numbers. We should consider the domain of the first equation, which we can define as the set of the values for which logarithms exist. </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li>define the domain of the first equation</li> <li>rewrite the logarithms on the left-hand side of the first equation as logarithms with the base <tx>a</tx></li> <li>transform this expression into the expression of the right-hand side using the second equation</li> </ul> </answer> <question> What can we say about the domain of the first equation? </question> <answer> We can note that <ul> <li><tx>\log _{c+b}a</tx> is defined when <tx>a > 0 </tx> and <tx>c+b > 0</tx>, <tx>c+b \ne 1</tx></li> <li><tx>\log _{c-b}a</tx> is defined when <tx>a > 0 </tx> and <tx>c-b > 0</tx>, <tx>c-b \ne 1</tx></li> </ul> Therefore, we should consider that <tx>a>0, b>0, c>0</tx> (the initial conditions) and <tx>c > b</tx>, <tx>c+b \ne 1</tx>, <tx>c-b \ne 1</tx>. </answer> <question> Consider the left-hand side of the first equation <tx>\log _{c+b}a+\log _{c-b}a</tx>. How can we transform the two logarithms into logarithms with base <tx>a</tx>? How can we simplify the expression then? <hint> Consider the properties of logarithm <ul> <li><fl>\log _{B}A=\frac{1}{\log _{A}B}</fl> where <tx>A>0, A\ne 1</tx></li> <li><fl>\log _{C}A + \log _{C}B=\log _{C}AB</fl></li> </ul> How can we use them to get <tx>\frac{\log _{a}\left( c^{2}-b^{2} \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)} \text{?}</tx> </hint> </question> <answer> We can transform the left-hand side of the first equation. Let's consider that <tx>a\ne1</tx> <tex>\log _{c+b}a+\log _{c-b}a=</tex> <note> We note that <tx>a\ne 1</tx> and apply the formula <fl>\log _{B}A=\frac{1}{\log _{A}B}</fl> where <tx>A>0, A\ne 1</tx> to rewrite <ul> <li><tx>\log _{c+b}a=\frac{1}{\log _{a}\left( c+b \right)}</tx></li> <li><tx>\log _{c-b}a=\frac{1}{\log _{a}\left( c-b \right)}</tx></li> </ul> </note> <tex>\frac{1}{\log _{a}\left( c+b \right)}+\frac{1}{\log _{a}\left( c-b \right)}=</tex> <note> We find a common denominator and combine the fractions <tex>\frac{\log _{a}\left( c-b \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}+\frac{\log _{a}\left( c+b \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> </note> <tex>\frac{\log _{a}\left( c-b \right)+\log _{a}\left( c+b \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <note> We use the property <fl>\log _{C}A + \log _{C}B=\log _{C}AB</fl> where <tx>A=c-b, B=c+b, C=a</tx> </note> <tex>\frac{\log _{a}\left(\left( c-b \right)\left( c+b \right)\right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <note> We apply the factoring formula <fl>\left(A-B\right)\left(A+B\right)=A^{2}-B^{2}</fl> where <tx>A=c, B=b</tx> </note> <tex>\frac{\log _{a}\left( c^{2}-b^{2} \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}</tex> </answer> <question> How can we transform the resulting expression to get <tx>2\log _{c+b}a\; \cdot \; \log _{c-b}a \text{ ?}</tx> How can we use the second equation to rewrite the numerator <tx>\log _{a}\left( c^{2}-b^{2} \right)</tx>? How can we rewrite the logarithms in the denominator? <hint> Consider the logarithmic properties <ul> <li><fl>\log _{B}A=\frac{1}{\log _{A}B}</fl> where <tx>A>0, A\ne 1</tx></li> <li><fl>\log _{A}B^n=n\log _{A}B</fl> when <tx>B>0</tx></li> <li><fl>\log _{A}A=1</fl></li> </ul> </hint> </question> <answer> Let's simplify the resulting expression <tex>\frac{\log _{a}\left( c^{2}-b^{2} \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <note> We note that <tx>c^{2}-b^{2}=a^{2}</tx> since <tx>a^{2}+b^{2}=c^{2}</tx> </note> <tex>\frac{\log _{a}a^{2}}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <note> We use the logarithmic property <fl>\log _{A}B^n=n\log _{A}B</fl> when <tx>B>0</tx> where <tx>A=a, B=a, n=2</tx> <tex>\frac{2\log _{a}a}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> and note that <tx>\log _{a}a=1</tx> </note> <tex>\frac{2}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <note> We use the property <fl>\log _{B}A=\frac{1}{\log _{A}B}</fl> when <tx>A>0, A\ne 1</tx> to rewrite <ul> <li><tx>\log _{a}\left( c+b \right)=\frac{1}{\log _{c+b}a}</tx></li> <li><tx>\log _{a}\left( c-b \right)=\frac{1}{\log _{c-b}a}</tx></li> </ul> </note> <tex>\frac{2}{\dfrac{1}{\log _{c+b}a} \cdot \dfrac{1}{\log _{c-b}a}}=</tex> <note> We simplify the complex fraction by inverting the denominator </note> <tex>2\log _{c+b}a\; \cdot \; \log _{c-b}a</tex> We showed that <tex>\log _{c+b}a+\log _{c-b}a=2\log _{c+b}a\; \cdot \; \log _{c-b}a \text{ when } a\ne 1</tex> </answer> <question> How can show that the first equation also holds for <tx>a=1</tx>? </question> <answer> Let's check that the values of both sides of the equation are equal when <tx>a=1</tx> <ul> <li>The left-hand side <tex>\log _{c+b}a+\log _{c-b}a=\log _{c+b}1+\log _{c-b}1=0+0=0</tex> </li> <li>The right-hand side <tex>2\log _{c+b}a\; \cdot \; \log _{c-b}a=2\log _{c+b}1\; \cdot \; \log _{c-b}1=2\cdot0\cdot0=0</tex> </li> </ul> Therefore, <tex>\log _{c+b}a+\log _{c-b}a=2\log _{c+b}a\; \cdot \; \log _{c-b}a</tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> Let <tx>a\ne 1</tx> <tex>\log _{c+b}a+\log _{c-b}a\stackrel{a\ne1}{=}</tex> <tex>\frac{1}{\log _{a}\left( c+b \right)}+\frac{1}{\log _{a}\left( c-b \right)}=</tex> <tex>\frac{\log _{a}\left( c-b \right)+\log _{a}\left( c+b \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <tex>\frac{\log _{a}\left( c^{2}-b^{2} \right)}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <tex>\frac{\log _{a}a^{2}}{\log _{a}\left( c+b \right)\log _{a}\left( c-b \right)}=</tex> <tex>2\log _{a}a\; \cdot \; \log _{c+b}a\; \cdot \; \log _{c-b}a=</tex> <tex>2\log _{c+b}a\; \cdot \; \log _{c-b}a</tex> Let's check that the equation holds for <tx>a=1</tx> <ul> <li>The left-hand side <tex>\log _{c+b}a+\log _{c-b}a=\log _{c+b}1+\log _{c-b}1=0+0=0</tex> </li> <li>The right-hand side <tex>2\log _{c+b}a\; \cdot \; \log _{c-b}a=2\log _{c+b}1\; \cdot \; \log _{c-b}1=2\cdot0\cdot0=0</tex> </li> </ul> </answer> </carryout> <lookback> <question>How can we check the result?</question> <answer> We can evaluate <tx>\log _{c+b}a+\log _{c-b}a</tx> and <tx>2\log _{c+b}a\; \cdot \; \log _{c-b}a</tx> for specific values of <tx>a, b, c</tx> and show that the results are identical. <ul> <li>Let <tx>a=4, b=3, c=5</tx> We can note that <tx>a^{2}+b^{2}=c^{2}</tx> holds (<tx>4^2+3^2=5^2</tx>). The left-hand side of the first equation: <tex>\log _{c+b}a+\log _{c-b}a=\log _{8}4+\log _{2}4=\frac{2}{3}+2=\frac{8}{3}</tex> The right-hand side of the first equation: <tex>2\log _{c+b}a\; \cdot \; \log _{c-b}a=2\log _{8}4\; \cdot \; \log _{2}4 = 2 \cdot \frac{2}{3} \cdot 2 = \frac{8}{3}</tex> </li> </ul> </answer> <question> Can we prove the statement differently? How can we transform the right-hand side of the first equation to get the expression of the left-hand side? </question> <answer> Let's transform the right-hand side of the first equation <tex>2\log _{c+b}a\; \cdot \; \log _{c-b}a=</tex> <note> We use the property <fl>n\cdot \log _{A}B=\log _{A}B^{n}</fl> where <tx>A=c+b, B=a, n=2</tx> </note> <tex>\log _{c+b}a^{2}\; \cdot \; \log _{c-b}a=</tex> <note> We use the second equation to rewrite <tx>a^2=c^2-b^2</tx> </note> <tex>\log _{c+b}\left( c^{2}-b^{2} \right)\; \cdot \; \log _{c-b}a=</tex> <note> We apply the factoring formula <fl>A^{2}-B^{2}=\left( A+B \right)\left( A-B \right)</fl> where <tx>A=c, B=b</tx> </note> <tex>\log _{c+b}\left( \left( c-b \right)\left( c+b \right) \right)\; \cdot \; \log _{c-b}a=</tex> <note> We note that <tx>c-b>0</tx> and <tx>c+b> 0</tx> and use the logarithmic property <fl>\log _{C}AB=\log _{C}A + \log _{C}B</fl> when <tx>A>0, B>0</tx> </note> <tex>\left( \log _{c+b}\left( c-b \right)+\log _{c+b}\left( c+b \right) \right)\; \cdot \; \log _{c-b}a=</tex> <note> We note that <tx>\log _{c+b}\left( c+b \right)=1</tx> </note> <tex>\left( \log _{c+b}\left( c-b \right)+1 \right)\; \cdot \; \log _{c-b}a=</tex> <note> We distribute <tx>\log _{c-b}a</tx> over <tx>\log _{c+b}\left( c-b \right)+1</tx> </note> <tex>\log _{c+b}\left( c-b \right)\; \cdot \; \log _{c-b}a\; +\log _{c-b}a=</tex> <note> We use the change-of-base formula <fl>\log _{B}A=\frac{\log _{X}A}{\log _{X}B}</fl> when <tx>X>0, X\ne 1</tx> to rewrite <tx>\log _{c-b}a=\frac{\log _{c+b}a}{\log _{c+b}\left( c-b \right)}</tx> </note> <tex>\log _{c+b}\left( c-b \right)\; \cdot \; \frac{\log _{c+b}a}{\log _{c+b}\left( c-b \right)}\; +\log _{c-b}a=</tex> <note> We note that <tx>\log _{c+b}\left( c-b \right)\ne 0</tx> (since <tx>c-b \ne 1</tx>) and cancel out the denominator <tex>\cancel{\log _{c+b}\left( c-b \right)}\; \cdot \; \frac{\log _{c+b}a}{\cancel{\log _{c+b}\left( c-b \right)}}\; +\log _{c-b}a=</tex> </note> <tex>\log _{c+b}a\; +\log _{c-b}a</tex> </answer> </lookback></howto>

<desc> Prove that b^loga(c)=c^loga(b) </desc> <task> Prove that <tx>b^{\textstyle \log _{a}c}=c^{\textstyle \log _{a}b}</tx></task> <solution> <tex>b^{\textstyle \log _{a}c}=</tex> <tex>\left( a^{\textstyle \log _{a}b} \right)^{\textstyle \log _{a}c}=</tex> <tex>a^{\textstyle \log _{a}b \cdot \log _{a}c }=</tex> <tex>\left( a^{\textstyle \log _{a}c} \right)^{\textstyle \log _{a}b}=</tex> <tex>c^{\textstyle \log _{a}b}</tex></solution> <howto> <understand> <question> What do we have? </question> <answer> We have an equation in three variables <tx>a, b, c</tx>. Each side is an exponential expression containing a logarithm. </answer> <question> What should we prove? </question> <answer> We should prove that the equation is an identity. In other words, both sides are equal for all valid values of <tx>a, b, c</tx>. </answer> <question>What conditions should we consider?</question> <answer> Unless otherwise specified, we should consider the equation in the context of real numbers. We should consider the domain of the equation, which we can define as the set of the values for which logarithms exist. </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li>define the domain of the equation</li> <li>transform the base of the exponential expression on the left-hand side</li> <li>simplify the expression on the left-hand side to obtain the expression of the right-hand side</li> </ul> </answer> <question> What can we say about the domain of the equation? </question> <answer> We can note that <ul> <li><tx>\log _{a}c</tx> is defined when <tx>a > 0 </tx> and <tx>a \ne 1</tx>, <tx>c > 0</tx></li> <li><tx>\log _{a}b</tx> is defined when <tx>a > 0 </tx> and <tx>a \ne 1</tx>, <tx>b > 0</tx></li> </ul> Therefore, we should consider that <tx>a>0, a\ne1, b > 0, c > 0</tx> </answer> <question> Consider the left-hand side of the equation. How can we transform the base of the exponential expression to get an expression with the base <tx>a</tx>? <hint> How can we use the properties <ul> <li><fl>A=B^{\textstyle \log_B{A}}</fl> when <tx>A > 0</tx></li> <li><fl>\left( A^n \right)^m = A^{nm}</fl></li> </ul> to transform the left-hand side into <tx>a^{\textstyle \log _{a}b \cdot \log _{a}c } \text{?}</tx> </hint> </question> <answer> Let's change the base of the exponential expression on the left-hand side <tex>b^{\textstyle \log _{a}c}=</tex> <note> We note that <tx>b>0</tx> and use the property <fl>A=B^{\textstyle \log_B{A}}</fl> when <tx>A > 0</tx> where <tx>A=b, B=a</tx> </note> <tex>\left( a^{\textstyle \log _{a}b} \right)^{\textstyle \log _{a}c}=</tex> <note> We use the exponentiation property <fl>\left( A^n \right)^m = A^{nm}</fl> where <tx>A=a, n=\log _{a}b, m=\log _{a}c</tx> </note> <tex>a^{\textstyle \log _{a}b \cdot \log _{a}c }</tex> </answer> <question> How can we transform the resulting expression to the expression on the right-hand side? <hint> Consider the properties <ul> <li><fl>A^{nm}=\left( A^n \right)^m</fl></li> <li><fl>B^{\textstyle log_B{A}}=A</fl></li> </ul> </hint> </question> <answer> Let's transform the resulting expression to the expression on the right-hand side <tex>a^{\textstyle \log _{a}b \cdot \log _{a}c }=</tex> <note> We use the exponentiation property <fl>A^{nm}=\left( A^n \right)^m</fl> where <tx>A=a, n=\log _{a}c, m=\log _{a}b</tx> </note> <tex>\left( a^{\textstyle \log _{a}c} \right)^{\textstyle \log _{a}b}=</tex> <note> We use the logarithmic property <fl>B^{\textstyle log_B{A}}=A</fl> where <tx>B=a, A=c </tx> </note> <tex>c^{\textstyle \log _{a}b}</tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <tex>b^{\textstyle \log _{a}c}=</tex> <tex>\left( a^{\textstyle \log _{a}b} \right)^{\textstyle \log _{a}c}=</tex> <tex>a^{\textstyle \log _{a}b \cdot \log _{a}c }=</tex> <tex>\left( a^{\textstyle \log _{a}c} \right)^{\textstyle \log _{a}b}=</tex> <tex>c^{\textstyle \log _{a}b}</tex> </answer> </carryout> <lookback> <question>How can we check the result?</question> <answer> We can evaluate <tx>b^{\textstyle \log _{a}c}</tx> and <tx>c^{\textstyle \log _{a}b}</tx> for specific values of <tx>a, b, c</tx> and show that the results are identical <ul> <li>Let <tx>a=2, b=2, c=4 </tx> The left-hand side: <tex>b^{\textstyle \log _{a}c}=2^{\textstyle \log _{2}4}=2^{2}=4</tex> The right-hand side: <tex>c^{\textstyle \log _{a}b}=4^{\textstyle \log _{2}2}=4^{1}=4</tex> </li> <li>Let <tx>b=1</tx> The left-hand side: <tex>b^{\textstyle \log _{a}c}=1^{\textstyle \log _{a}c}=1</tex> The right-hand side: <tex>c^{\textstyle \log _{a}b}=c^{\textstyle \log _{a}1}=c^{0}=1</tex> </li> </ul> </answer> <question> Can we prove the statement differently? How can we use the logarithmic property <fl>\log _{A}B=\log _{A^{\textstyle x}}B^{\textstyle x}</fl> ? </question> <answer> Let's transform the logarithm on the left-hand side <tex>\log _{\textstyle a}c=</tex> <note> We apply the property <tx>\log _{A}B=\log _{A^{\textstyle x}}B^{\textstyle x}</tx> where <tx>A=a, B=c, x=\log _{a}b</tx> </note> <tex>\log _{\textstyle a^{\textstyle \log _{a}b}}\left(c^{\textstyle\log _{a}b}\right)=</tex> <note> We use the property <fl>A^{\log_{A}B}=B</fl> to rewrite <tx>a^{\textstyle \log _{a}b}=b</tx> </note> <tex>\log _{\textstyle b}\left(c^{\textstyle \log _{a}b}\right)</tex> We can transform the left-hand side of the equation <tex>b^{\textstyle \log _{a}c}=</tex> <note> We note that <tx>\log _{\textstyle a}c=\log _{\textstyle b}\left(c^{\textstyle \log _{a}b}\right)</tx> </note> <tex>b^{\textstyle \log _{b}\left(c^{\textstyle \log _{a}b}\right)}=</tex> <note> We use the property <fl>A^{\log_{A}B}=B</fl> where <tx>A=b, B=c^{\textstyle \log _{a}b}</tx> </note> <tex>c^{\textstyle \log _{a}b}</tex> </answer> </lookback></howto>

<desc> Express a logarithm in terms of the value of another logarithm </desc> <task> Express <tx>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}</tx> in terms of <tx>k</tx> given that <tx>\log _{ab}a=k</tx> where <tx>a > 0</tx>, <tx>b > 0</tx> and <tx>ab\ne 1</tx></task> <solution> <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=</tex> <tex>\log _{ab}\sqrt[3]{a}\; -\; \log _{ab}\sqrt{b}=</tex> <tex>\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b</tex> <hr/> <tex>\log _{ab}a=</tex> <tex>\log _{ab}\frac{ab}{b}=</tex> <tex>1-\log _{ab}b</tex> <tex>1-\log _{ab}b=k</tex> <tex>\log _{ab}b=1-k</tex> <hr/> <tex>\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b=</tex> <tex>\frac{1}{3}k\; -\; \frac{1}{2}\left(1-k\right)=</tex> <tex>\frac{2k-3+3k}{6}=</tex> <tex>\frac{5k-3}{6}</tex> Finally, <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=\frac{5k-3}{6}</tex></solution> <result> <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=\frac{5k-3}{6}</tex></result> <howto> <understand> <question> What do we have? </question> <answer> We have two logarithms with the same base <tx>ab</tx>. The value of the second logarithm is <tx>k</tx> </answer> <question> What should we find? </question> <answer> We should rewrite the first logarithm as an algebraic expression using only the constant <tx>k</tx>. </answer> <question> What conditions should we consider? </question> <answer> Unless otherwise specified, we should consider the expressions in the context of real numbers. We should consider the domain of the expressions, which we can define as the set of the values for which logarithms exist. We should use algebraic transformations based on logarithmic and exponentiation properties. You can read more about logarithmic properties <a href="https://en.wikipedia.org/wiki/List_of_logarithmic_identities" target="_blank">here</a> You can read more about exponentiation properties <a href="https://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties" target="_blank">here</a> </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li>define the domain of the expressions</li> <li>express the first logarithm in terms of <tx>\log _{ab}a</tx> and <tx>\log _{ab}b</tx></li> <li>transform the second logarithm to express <tx>\log _{ab}b</tx> in terms of <tx>k</tx></li> <li>use the obtained expressions to rewrite the first logarithm in terms of <tx>k</tx></li> </ul> </answer> <question> What can we say about the domain of the expressions? </question> <answer> We can note that <ul> <li><tx>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}</tx> is defined when <tx>\frac{\sqrt[3]{a}}{\sqrt{b}} > 0 </tx> and <tx>ab > 0</tx>, <tx>ab \ne 1</tx></li> <li><tx>\log _{ab}a</tx> is defined when <tx>ab>0</tx>, <tx>ab \ne1</tx> and <tx>a > 0</tx></li> </ul> Therefore, we should consider that <tx>a>0, b>0</tx> and <tx>ab \ne1</tx> </answer> <question> How can we expand <tx>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}} \text{?}</tx> <hint> Consider the properties <ul> <li> <fl>\log _{C}\frac{A}{B}=\log _{C}A-\log _{C}B</fl> when <tx>A>0, B>0</tx> </li> <li> <fl>\log _{B}A^{n}=n\log _{B}A</fl> when <tx>A>0</tx> </li> <li> <fl>\sqrt[n]{A}=A^{\textstyle \frac{1}{n}}</fl> </li> </ul> How can we use them to get <tx>\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b \text{?}</tx> </hint> </question> <answer> Let's expand the first logarithm <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=</tex> <note> We note that <tx>a>0, b>0</tx> and apply the property <fl>\log _{C}\frac{A}{B}=\log _{C}A-\log _{C}B</fl> when <tx>A>0, B>0</tx> where <tx>A=\sqrt[3]{a}, B=\sqrt{b}, C=ab</tx> </note> <tex>\log _{ab}\sqrt[3]{a}\; -\; \log _{ab}\sqrt{b}=</tex> <note> We use the property <fl>\sqrt[n]{A}=A^{\textstyle \frac{1}{n}}</fl> to rewrite <ul> <li><tx>\sqrt[3]{a}=a^{\textstyle \frac{1}{3}}</tx></li> <li><tx>\sqrt{b}=b^{\textstyle \frac{1}{2}}</tx></li> </ul> </note> <tex>\log _{ab}a^{\textstyle \frac{1}{3}}\; -\; \log _{ab}b^{\textstyle \frac{1}{2}}=</tex> <note> We note that <tx>a>0, b>0</tx> and apply the property <fl>\log _{B}A^{n}=n\log _{B}A</fl> when <tx>A>0</tx> </note> <tex>\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b</tex> </answer> <question> How can we transform the second logarithm to express <tx>\log _{ab}b</tx> in terms of <tx>k</tx> ? <hint> How can we use the logarithmic property <fl>\log _{C}\frac{A}{B}=\log _{C}A-\log _{C}B</fl> when <tx>A>0, B>0</tx> ? </hint> </question> <answer> Let's transform the second logarithm <tex>\log _{ab}a=</tex> <note> We note that <tx>b\ne0</tx> and rewrite <tx>a=\frac{ab}{b}</tx> </note> <tex>\log _{ab}\frac{ab}{b}=</tex> <note> We note that <tx>a>0, b>0</tx> and use the property <fl>\log _{C}\frac{A}{B}=\log _{C}A-\log _{C}B</fl> when <tx>A>0, B>0</tx> where <tx>A=ab, B=b, C=ab</tx> </note> <tex>\log _{ab}ab-\log _{ab}{b}=</tex> <note> We note that <tx>\log _{ab}ab=1</tx> </note> <tex>1-\log _{ab}b</tex> Now, we can express <tx>\log _{ab}b</tx> in terms of <tx>k</tx> <tex>1-\log _{ab}b=k</tex> <tex>\log _{ab}b=1-k</tex> Let's rewrite the obtained results <ul> <li><tx>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b</tx></li> <li><tx>\log _{ab}a=k</tx></li> <li><tx>\log _{ab}b=1-k</tx></li> </ul> </answer> <question> How can we use these results to express <tx>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}</tx> in terms of <tx>k</tx>? </question> <answer> Let's express the first logarithm in terms <tx>k</tx> <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=</tex> <note> We note that <tx>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b</tx> </note> <tex>\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b=</tex> <note> We note that <ul> <li><tx>\log _{ab}a=k</tx></li> <li><tx>\log _{ab}b=1-k</tx></li> </ul> </note> <tex>\frac{k}{3}\; -\; \frac{1-k}{2}=</tex> <note> We find a common denominator and combine the fractions <tex>\frac{2k}{6}\; -\; \frac{3\left(1-k\right)}{6}=</tex> </note> <tex>\frac{2k-3+3k}{6}=</tex> <note> We add like terms </note> <tex>\frac{5k-3}{6}</tex> Thus, <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=\frac{5k-3}{6}</tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=</tex> <tex>\log _{ab}\sqrt[3]{a}\; -\; \log _{ab}\sqrt{b}=</tex> <tex>\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b</tex> <hr/> <tex>\log _{ab}a=</tex> <tex>\log _{ab}\frac{ab}{b}=</tex> <tex>1-\log _{ab}b</tex> <tex>1-\log _{ab}b=k</tex> <tex>\log _{ab}b=1-k</tex> <hr/> <tex>\frac{1}{3}\log _{ab}a\; -\; \frac{1}{2}\log _{ab}b=</tex> <tex>\frac{1}{3}k\; -\; \frac{1}{2}\left(1-k\right)=</tex> <tex>\frac{2k-3+3k}{6}=</tex> <tex>\frac{5k-3}{6}</tex> Finally, <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=\frac{5k-3}{6}</tex> </answer> </carryout> <lookback> <question>How can we check the result?</question> <answer> We can evaluate <tx>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}</tx> and <tx>\frac{5k-3}{6}</tx> for specific values of <tx>a</tx>, <tx>b</tx> and show that the results are identical <ul> <li>Let <tx>b=1</tx> The initial expression: <tex>\log _{ab}\frac{\sqrt[3]{a}}{\sqrt{b}}=\log _{a}\sqrt[3]{a}=\frac{1}{3}</tex> The resulting expression: <tex>k=\log _{ab}a=\log _{a}a=1</tex> <tex>\frac{5k-3}{6}=\frac{5\cdot1-3}{6}=\frac{1}{3}</tex> </li> </ul> </answer> </lookback> </howto>

<desc> Express logc(ab) in terms of p, q, r given that loga(k)=p, logb(k)=q, logc(k)=r </desc> <task> Express <tx>\log _{c}ab</tx> in terms of <tx>p</tx>, <tx>q</tx> and <tx>r</tx> given that <tx>\log _{a}k=p</tx>, <tx>\log _{b}k=q</tx> and <tx>\log _{c}k=r</tx> where <tx>k\ne 1</tx></task> <solution> <tex>\log _{k}a=\frac{1}{p}</tex> <tex>\log _{k}b=\frac{1}{q}</tex> <tex>\log _{k}c=\frac{1}{r}</tex> <hr/> <tex>\log _{c}ab=</tex> <tex>\frac{\log _{k}ab}{\log _{k}c}=</tex> <tex>\frac{\log _{k}a+\log _{k}b}{\log _{k}c}=</tex> <tex>\frac{\dfrac{1}{p}+\dfrac{1}{q}}{\dfrac{1}{r}}=</tex> <tex>\frac{r\left( p+q \right)}{pq}</tex></solution> <result> <tex>\log _{c}ab=\frac{r\left( p+q \right)}{pq}</tex></result> <howto> <understand> <question> What do we have? </question> <answer> We have four logarithms, three of which have the same argument <tx>k</tx>. The value of <ul> <li>the second logarithm is <tx>p</tx></li> <li>the third logarithm is <tx>q</tx></li> <li>the fourth logarithm is <tx>r</tx></li> </ul> </answer> <question> What should we find? </question> <answer> We should rewrite the first logarithm as an algebraic expression using only the constants <tx>p</tx>, <tx>q</tx> and <tx>r</tx>. </answer> <question> What conditions should we consider? </question> <answer> Unless otherwise specified, we should consider the expressions in the context of real numbers. We should consider the domain of the expressions, which we can define as the set of the values for which logarithms exist. We should use algebraic transformations based on logarithmic and exponentiation properties. You can read more about logarithmic properties <a href="https://en.wikipedia.org/wiki/List_of_logarithmic_identities" target="_blank">here</a> You can read more about exponentiation properties <a href="https://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties" target="_blank">here</a> </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li>define the domain of the expressions</li> <li>express each logarithm in terms of logarithm with base <tx>k</tx></li> <li>use the obtained expressions to rewrite the first logarithm in terms of <tx>p</tx>, <tx>q</tx> and <tx>r</tx></li> <li>simplify the resulting expression</li> </ul> </answer> <question> What can we say about the domain of the expressions? </question> <answer> We can note that <ul> <li><tx>\log _{c}ab</tx> is defined when <tx>c>0, c\ne1</tx> and <tx>ab > 0</tx></li> <li><tx>\log _{a}k=p</tx> is defined when <tx>a>0, a\ne1</tx> and <tx>k > 0</tx></li> <li><tx>\log _{b}k=q</tx> is defined when <tx>b>0, b\ne1</tx> and <tx>k > 0</tx></li> <li><tx>\log _{c}k=r</tx> is defined when <tx>c>0, c\ne1</tx> and <tx>k > 0</tx></li> </ul> Therefore, we should consider that <tx>a>0, a\ne1</tx>, <tx>b>0, b\ne1</tx>, <tx>c>0, c\ne1</tx>, <tx>k>0, k\ne1</tx> </answer> <question> How can we rewrite the last three logarithms as logarithms with base <tx>k</tx>? What values do we obtain for these logarithms then? <hint> Consider a special case of the change-of-base formula <fl>\log _{B}A=\frac{\log _{X}A}{\log _{X}B}</fl> when <tx>X=A</tx>: <formula>\log _{B}A=\frac{\log _{A}A}{\log _{A}B}= \frac{1}{\log _{A}B}</formula> How can we apply this identity? </hint> </question> <answer> Let's rewrite the last three logarithms as logarithms with base <tx>k</tx> <ul> <li><tx>\log _{a}k</tx> <tex>\log _{a}k=p</tex> <note> We use the formula <tx>\log _{B}A=\frac{1}{\log _{A}{B}}</tx> where <tx>A=k, B=a</tx> </note> <tex>\frac{1}{\log _{k}a}=p</tex> <note> We multiply both sides by <tx>\frac{\log _{k}a}{p}</tx> </note> <tex>\log _{k}a=\frac{1}{p}</tex> </li> <li><tx>\log _{b}k</tx> <tex>\log _{b}k=q</tex> <tex>\frac{1}{\log _{k}b}=q</tex> <tex>\log _{k}b=\frac{1}{q}</tex> </li> <li><tx>\log _{c}k</tx> <tex>\log _{c}k=r</tex> <tex>\frac{1}{\log _{k}c}=r</tex> <tex>\log _{k}c=\frac{1}{r}</tex> </li> </ul> </answer> <question> How can we use the previous results to express <tx>\log _{c}ab</tx> in terms of <tx>p</tx>, <tx>q</tx> and <tx>r</tx> ? How should we rewrite the first logarithm? <hint> Consider the logarithmic properties <ul> <li> <fl>\log _{B}A=\frac{\log _{C}{A}}{\log _{C}{B}}</fl> when <tx>C>0, C\ne 1</tx> </li> <li> <fl>\log _{C}AB=\log _{C}A + \log _{C}B</fl> when <tx>A>0, B>0</tx> </li> </ul> How can we use them to rewrite the first logarithm and get <tx>\frac{\dfrac{1}{p}+\dfrac{1}{q}}{\dfrac{1}{r}} \text{ ?}</tx> </hint> </question> <answer> Let's express the first logarithm as a quotient of logarithms with base <tx>k</tx> <tex>\log _{c}ab=</tex> <note> We note that <tx>k\ne1</tx> and apply the change-of-base formula <fl>\log _{B}A=\frac{\log _{C}{B}}{\log _{C}{B}}</fl> when <tx>C>0, C\ne 1</tx> where <tx>A=ab, B=c, C=k</tx> </note> <tex>\frac{\log _{k}ab}{\log _{k}c}=</tex> <note> We use the logarithmic property <fl>\log _{C}AB=\log _{C}A + \log _{C}B</fl> when <tx>A>0, B>0</tx> where <tx>A=a, B=b, C=k</tx> </note> <tex>\frac{\log _{k}a+\log _{k}b}{\log _{k}c}</tex> <note> We note that <ul> <li><tx>\log _{k}a=\frac{1}{p}</tx></li> <li><tx>\log _{k}b=\frac{1}{q}</tx></li> <li><tx>\log _{k}c=\frac{1}{r}</tx></li> </ul> </note> <tex>\frac{\dfrac{1}{p}+\dfrac{1}{q}}{\dfrac{1}{r}}</tex> </answer> <question> How can we simplify the resulting expression? </question> <answer> Let's simplify the resulting expression <tex>\frac{\dfrac{1}{p}+\dfrac{1}{q}}{\dfrac{1}{r}}=</tex> <note> We note that <tex>\frac{1}{p}+\frac{1}{q}=\frac{q}{pq}+\frac{p}{pq}</tex> and add the fractions </note> <tex>\frac{\dfrac{q+p}{pq}}{\dfrac{1}{r}}=</tex> <note> We invert the denominator and multiply the fractions <tex>\dfrac{q+p}{pq} \cdot r=</tex> </note> <tex>\frac{r\left( p+q \right)}{pq}</tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <tex>\log _{k}a=\frac{1}{p}</tex> <tex>\log _{k}b=\frac{1}{q}</tex> <tex>\log _{k}c=\frac{1}{r}</tex> <hr/> <tex>\log _{c}ab=</tex> <tex>\frac{\log _{k}ab}{\log _{k}c}=</tex> <tex>\frac{\log _{k}a+\log _{k}b}{\log _{k}c}=</tex> <tex>\frac{\dfrac{1}{p}+\dfrac{1}{q}}{\dfrac{1}{r}}=</tex> <tex>\frac{r\left( p+q \right)}{pq}</tex> </answer> </carryout> <lookback> <question>How can we check the result?</question> <answer> We can evaluate <tx>\log _{c}ab</tx> and <tx>\frac{r\left( p+q \right)}{pq}</tx> for specific values of <tx>a</tx>, <tx>b</tx> and <tx>c</tx> and show that the results are identical <ul> <li>Let <tx>a=b=c=k</tx> The initial expression: <tex>\log _{c}ab=\log _{c}c^2=2\log _{c}c=2</tex> The resulting expression: <tex>\log _{a}k=\log _{b}k=\log _{c}k=1</tex> <tex>p=q=r=1</tex> <tex>\frac{r\left( p+q \right)}{pq}=\frac{1\left( 1+1 \right)}{1\cdot 1}=2</tex> </li> </ul> </answer> <question> Can we derive the result differently? </question> <answer> <tex>p=\log _{a}k=</tex> <note> We use the change-of-base formula <fl>\log _{B}A=\frac{\log _{X}A}{\log _{X}B}</fl> when <tx>X>0, X\ne 1</tx> where <tx>B=a, A=k, X=c</tx> </note> <tex>\frac{\log _{c}k}{\log _{c}a}=</tex> <note> We note that <tx>\log _{c}k=r</tx> </note> <tex>\frac{r}{\log _{c}a}</tex> Therefore, <tex>\log _{c}a=\frac{r}{p}</tex> <hr/> <tex>q=\log _{b}k=</tex> <note> We use the change-of-base formula <fl>\log _{B}A=\frac{\log _{X}A}{\log _{X}B}</fl> when <tx>X>0, X\ne 1</tx> where <tx>B=b, A=k, X=c</tx> </note> <tex>\frac{\log _{c}k}{\log _{c}b}=</tex> <note> We note that <tx>\log _{c}k=r</tx> </note> <tex>\frac{r}{\log _{c}b}</tex> Therefore, <tex>\log _{c}b=\frac{r}{q}</tex> <hr/> <tex>\log _{c}ab=</tex> <note> We use the logarithmic property <tx>\log _{C}AB=\log _{C}A+\log _{C}B</tx> when <tx>A>0, B>0</tx> where <tx>A=a, B=b, C=c</tx> </note> <tex>\log _{c}a+\log _{c}b=</tex> <note> We note that <ul> <li><tx>\log _{c}a=\frac{r}{p}</tx></li> <li><tx>\log _{c}b=\frac{r}{q}</tx></li> </ul> </note> <tex>\frac{r}{p}+\frac{r}{q}=</tex> <note> We find a common denominator and combine the fractions </note> <tex>\frac{r\left( p+q \right)}{pq}</tex> </answer> </lookback></howto>

<desc> Express log2(360) in terms of a and b given that log3(20)=a and log3(15)=b </desc> <task> Express <tx>\log _{2}360</tx> in terms of <tx>a</tx> and <tx>b</tx> given that <tx>\log _{3}20=a</tx> and <tx>\log _{3}15=b</tx></task> <solution> <tex>\log _{2}360=\frac{\log _{3}360}{\log _{3}2}</tex> <hr/> <tex>\frac{\log _{3}360}{\log _{3}2}=</tex> <tex>\frac{\log _{3}3^{2}\cdot 20\cdot 2}{\log _{3}2}=</tex> <tex>\frac{2\log _{3}3+\log _{3}20+\log _{3}2}{\log _{3}2}=</tex> <tex>\frac{2+a+\log _{3}2}{\log _{3}2}</tex> <hr/> <tex>\frac{\log _{3}360}{\log _{3}2}=</tex> <tex>\frac{\log _{3}6\cdot 60}{\log _{3}2}=</tex> <tex>\frac{\log _{3}2\cdot 3\cdot 15\cdot 4}{\log _{3}2}=</tex> <tex>\frac{\log _{3}3\cdot 15\cdot 2^{3}}{\log _{3}2}=</tex> <tex>\frac{\log _{3}3+\log _{3}15+3\log _{3}2}{\log _{3}2}=</tex> <tex>\frac{1+b+3\log _{3}2}{\log _{3}2}</tex> <hr/> <tex> \begin{cases} \log _{2}360=\dfrac{2+a+\log _{3}2}{\log _{3}2} \\ \\ \log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2} \end{cases} </tex> <tex>\log _{2}360-\log _{2}360=\dfrac{2+a+\log _{3}2}{\log _{3}2}-\dfrac{1+b+3\log _{3}2}{\log _{3}2}</tex> <tex>0=\dfrac{2+a+\log _{3}2 - 1-b-3\log _{3}2}{\log _{3}2}</tex> <tex>0=1+a-b-2\log _{3}2</tex> <tex>\log _{3}2=\dfrac{1+a-b}{2}</tex> <hr/> <tex> \begin{cases} \log _{3}2=\dfrac{1+a-b}{2} \\ \\ \log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2} \end{cases} </tex> <tex>\log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2}</tex> <tex>\log _{2}360=\dfrac{1+b+\dfrac{3\left(1+a-b\right)}{2}}{\dfrac{1+a-b}{2}}</tex> <tex>\log _{2}360=\dfrac{2+2b+3+3a-3b}{1+a-b}</tex> <tex>\log _{2}360=\dfrac{3a-b+5}{a-b+1}</tex></solution> <result> <tex>\log _{2}360=\dfrac{3a-b+5}{a-b+1}</tex></result> <howto> <understand> <question> What do we have? </question> <answer> We have three logarithms, two of which have the same base 3. The value of the second logarithm is <tx>a</tx> and that of the third is <tx>b</tx>. </answer> <question> What should we find? </question> <answer> We should rewrite the first logarithm as an algebraic expression using only the constants <tx>a</tx> and <tx>b</tx>. </answer> <question> What conditions should we consider? </question> <answer> We should use algebraic transformations based on logarithmic and exponentiation properties. You can read more about logarithmic properties <a href="https://en.wikipedia.org/wiki/List_of_logarithmic_identities" target="_blank">here</a> You can read more about exponentiation properties <a href="https://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties" target="_blank">here</a> </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li> rewrite the first logarithm using the change-of-base formula to get a division of logarithms with base 3 </li> <li> transform this quotient of logarithms into an expression containing <tx>\log _{3}20</tx> and <tx>\log _{3}15</tx> </li> <li> replace <tx>\log _{3}20</tx> with <tx>a</tx> and <tx>\log _{3}15</tx> with <tx>b</tx> </li> </ul> </answer> <question> Consider the base of the three logarithms. How can we change the base of <tx>\log _{2}360</tx>? <hint> How can we use the change-of-base formula <fl>\log _{B}A=\frac{\log _{X}A}{\log _{X}B}</fl> where <tx>X>0, X\ne 1</tx>? </hint> </question> <answer> Let's change the base of the first logarithm <tex>\log _{2}360=</tex> <note> We use the change-of-base formula <fl>\log _{B}A=\frac{\log _{X}A}{\log _{X}B}</fl> where <tx>A=360, B=2, X=3</tx> </note> <tex>\frac{\log _{3}360}{\log _{3}2}</tex> </answer> <question> How can we transform the resulting expression so that it contains <tx>\log _{3}20</tx>? <hint> Consider the properties of logarithms <ul> <li><fl>\log _{C}AB=\log _{C}A+\log _{C}B</fl> when <tx>A>0, B>0</tx></li> <li><fl>\log _{B}A^{n}=n\log _{B}A</fl> when <tx>A>0</tx></li> </ul> How can we use them to get <tx>\frac{2+a+\log _{3}2}{\log _{3}2} \text{ ?}</tx> </hint> </question> <answer> We can rewrite the resulting expression <tex>\frac{\log _{3}360}{\log _{3}2}=</tex> <note> We note that <tx>360=9\cdot40</tx>, <tx>9=3^2</tx>, and <tx>40=20\cdot 2</tx> so, <tx>360=3^2\cdot 20\cdot 2</tx> </note> <tex>\frac{\log _{3}3^{2}\cdot 20\cdot 2}{\log _{3}2}=</tex> <note> We apply the logarithmic identities <ul> <li><fl>\log _{C}AB=\log _{C}A+\log _{C}B</fl> when <tx>A>0, B>0</tx></li> <li><fl>\log _{B}A^{n}=n\log _{B}A</fl> when <tx>A>0</tx></li> </ul> </note> <tex>\frac{2\log _{3}3+\log _{3}20+\log _{3}2}{\log _{3}2}=</tex> <note> We note that <tx>\log _{3}3=1</tx> and <tx>\log _{3}20=a</tx> </note> <tex>\frac{2+a+\log _{3}2}{\log _{3}2}</tex> So, <tex>\log _{2}360=\frac{2+a+\log _{3}2}{\log _{3}2}</tex> </answer> <question> How can we transform differently <tx>\frac{\log _{3}360}{\log _{3}2}</tx> to get an expression containing <tx>\log _{3}15</tx>? <hint> How can we get <tx>\frac{1+b+3\log _{3}2}{\log _{3}2} \text{ ?}</tx> </hint> </question> <answer> Let's find another expression for the first logarithm <tex>\frac{\log _{3}360}{\log _{3}2}=</tex> <note> We note that <tx>360=6\cdot60</tx>, <tx>6=2\cdot 3</tx>, and <tx>60=15\cdot 4</tx> so, <tx>360=3\cdot 15\cdot 8=3\cdot 15\cdot 2^3</tx> </note> <tex>\frac{\log _{3}3\cdot 15\cdot 2^{3}}{\log _{3}2}=</tex> <note> We apply the logarithmic identities <ul> <li><fl>\log _{C}AB=\log _{C}A+\log _{C}B</fl> when <tx>A>0, B>0</tx></li> <li><fl>\log _{B}A^{n}=n\log _{B}A</fl> when <tx>A>0</tx></li> </ul> </note> <tex>\frac{\log _{3}3+\log _{3}15+3\log _{3}2}{\log _{3}2}=</tex> <note> We note that <tx>\log _{3}3=1</tx> and <tx>\log _{3}15=b</tx> </note> <tex>\frac{1+b+3\log _{3}2}{\log _{3}2}</tex> So, <tex>\frac{\log _{3}360}{\log _{3}2}=\frac{1+b+3\log _{3}2}{\log _{3}2}</tex> </answer> <question> We have expressed <tx>\log _{2}360</tx> in two different ways. How can we use these results to express <tx>\log _{2}360</tx> in terms of <tx>a</tx> and <tx>b</tx> and eliminate <tx>\log _3{2}</tx>? <hint> What system of equations can we set up? How can we express <tx>\log _3{2}</tx> in terms of <tx>a</tx> and <tx>b</tx> ? </hint> </question> <answer> Let's start by expressing <tx>\log _3{2}</tx> in terms of <tx>a</tx> and <tx>b</tx> <tex> \begin{cases} \log _{2}360=\dfrac{2+a+\log _{3}2}{\log _{3}2} \\ \\ \log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2} \end{cases} </tex> We can subtract the second equation from the first <tex>\log _{2}360-\log _{2}360=\dfrac{2+a+\log _{3}2}{\log _{3}2}-\dfrac{1+b+3\log _{3}2}{\log _{3}2}</tex> <note> We combine the fractions </note> <tex>0=\dfrac{2+a+\log _{3}2 - 1-b-3\log _{3}2}{\log _{3}2}</tex> <note> We note that <tx>\log _{3}2 \ne 0</tx> so we can multiply both sides by <tx>\log _{3}2</tx> <tex>0\cdot \log _{3}2=\dfrac{2+a+\log _{3}2 - 1-b-3\log _{3}2}{\cancel{\log _{3}2}}\cdot \cancel{\log _{3}2}</tex> then we add like terms </note> <tex>0=1+a-b-2\log _{3}2</tex> <note> We isolate <tx>\log _{3}2</tx> by adding <tx>2\log _{3}2</tx> to both sides of the equation <tex>2\log _{3}2=1+a-b-\cancel{2\log _{3}2}+\cancel{2\log _{3}2}</tex> and dividing both sides by 2 </note> <tex>\log _{3}2=\dfrac{1+a-b}{2}</tex> Let's rewrite the system <tex> \begin{cases} \log _{3}2=\dfrac{1+a-b}{2} \\ \\ \log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2} \end{cases} </tex> We can rewrite the second equation to express <tx>\log _{2}360</tx> in terms of <tx>a, b</tx> <tex>\log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2}</tex> <note> We note that <tx>\log _{3}2=\dfrac{1+a-b}{2}</tx> </note> <tex>\log _{2}360=\dfrac{1+b+\dfrac{3\left(1+a-b\right)}{2}}{\dfrac{1+a-b}{2}}</tex> <note> We put <tx>1+b+\dfrac{3\left(1+a-b\right)}{2}</tx> over the common denominator </note> <tex>\log _{2}360=\dfrac{\dfrac{2\left(1+b\right) + 3\left(1+a-b\right)}{2}}{\dfrac{1+a-b}{2}}</tex> <note> We invert the denominators and multiply the fractions <tex>\log _{2}360=\frac{2\left(1+b\right) + 3\left(1+a-b\right)}{\cancel{2}} \cdot \frac{\cancel{2}}{1+a-b}</tex> and expand the numerator </note> <tex>\log _{2}360=\dfrac{2+2b+3+3a-3b}{1+a-b}</tex> <note> We add like terms </note> <tex>\log _{2}360=\dfrac{3a-b+5}{a-b+1}</tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <tex>\log _{2}360=\frac{\log _{3}360}{\log _{3}2}</tex> <hr/> <tex>\frac{\log _{3}360}{\log _{3}2}=</tex> <tex>\frac{\log _{3}3^{2}\cdot 20\cdot 2}{\log _{3}2}=</tex> <tex>\frac{2\log _{3}3+\log _{3}20+\log _{3}2}{\log _{3}2}=</tex> <tex>\frac{2+a+\log _{3}2}{\log _{3}2}</tex> <hr/> <tex>\frac{\log _{3}360}{\log _{3}2}=</tex> <tex>\frac{\log _{3}6\cdot 60}{\log _{3}2}=</tex> <tex>\frac{\log _{3}2\cdot 3\cdot 15\cdot 4}{\log _{3}2}=</tex> <tex>\frac{\log _{3}3\cdot 15\cdot 2^{3}}{\log _{3}2}=</tex> <tex>\frac{\log _{3}3+\log _{3}15+3\log _{3}2}{\log _{3}2}=</tex> <tex>\frac{1+b+3\log _{3}2}{\log _{3}2}</tex> <hr/> <tex> \begin{cases} \log _{2}360=\dfrac{2+a+\log _{3}2}{\log _{3}2} \\ \\ \log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2} \end{cases} </tex> <tex>\log _{2}360-\log _{2}360=\dfrac{2+a+\log _{3}2}{\log _{3}2}-\dfrac{1+b+3\log _{3}2}{\log _{3}2}</tex> <tex>0=\dfrac{2+a+\log _{3}2 - 1-b-3\log _{3}2}{\log _{3}2}</tex> <tex>0=1+a-b-2\log _{3}2</tex> <tex>\log _{3}2=\dfrac{1+a-b}{2}</tex> <hr/> <tex> \begin{cases} \log _{3}2=\dfrac{1+a-b}{2} \\ \\ \log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2} \end{cases} </tex> <tex>\log _{2}360=\dfrac{1+b+3\log _{3}2}{\log _{3}2}</tex> <tex>\log _{2}360=\dfrac{1+b+\dfrac{3\left(1+a-b\right)}{2}}{\dfrac{1+a-b}{2}}</tex> <tex>\log _{2}360=\dfrac{2+2b+3+3a-3b}{1+a-b}</tex> <tex>\log _{2}360=\dfrac{3a-b+5}{a-b+1}</tex> </answer> </carryout> <lookback> <question> Can we derive the result differently? <hint> How can we decompose each of the three logarithms? </hint> </question> <answer> We can rewrite <tx>\log _{2}360</tx> as the quotient of logarithms with base 3 <tex>\log _{2}360=\frac{\log _{3}360}{\log _{3}2}</tex> <tex>\log _{3}360=\log _{3}6^{2}+\log _{3}10=</tex> <tex>2\log _{3}6+\log _{3}\frac{20}{2}=</tex> <tex>2\log _{3}2+2\log _{3}3+\log _{3}20-\log _{3}2=</tex> <tex>\log _{3}2+2+\log _{3}4+\log _{3}5=</tex> <tex>3\log _{3}2+\log _{3}5+2</tex> So, <tex>\log _{2}360=\frac{3\log _{3}2+\log _{3}5+2}{\log _{3}2}</tex> Let's rewrite <tx>\log _{3}20</tx> <tex>a=\log _{3}20=2\log _{3}2+\log _{3}5</tex> <tex>\log _{3}2=\frac{a-\log _{3}5}{2}</tex> Let's rewrite <tx>\log _{3}15</tx> <tex>b=\log _{3}15=\log _{3}3+\log _{3}5=1+\log _{3}5</tex> <tex>\log _{3}5=b-1</tex> We can combine the three results <tex>\log _{2}360=\dfrac{3\log _{3}2+2+\log _{3}5}{\log _{3}2}=</tex> <tex>\frac{3\left( \dfrac{a-\left( b-1 \right)}{2} \right)+2+b-1}{\dfrac{a-\left( b-1 \right)}{2}}=</tex> <tex>\frac{\dfrac{3a-3b+3+2+2b}{2}}{\dfrac{a-b+1}{2}}=</tex> <tex>\frac{3a-b+5}{a-b+1}</tex> </answer> <question> How can we check the result? </question> <answer> We can use a calculator or we can estimate the values of <tx>\log _{2}360</tx> and <tx>\frac{3a-b+5}{a-b+1}</tx> and show that they are close. Let's use Thales's theorem to estimate these values. <center> <img style="height:320px" height="320" alt="approximation using thales theorem" src="https://cdn.dzdx.xyz/thales.png"> </center> Let's estimate <tx>\log _{2}360</tx>. We can note that <tx>2^8=256</tx> and <tx>2^9=512</tx>, so <tex>8 < \log _{2}360 < 9</tex> Let <tx>x'</tx> be an approximate value of <tx>\log _{2}360</tx> <tex>\frac{x'-8}{9-8}=\frac{360-256}{512-256}</tex> <tex>x'=8+\frac{104}{256}\approx 8.4</tex> Therefore, <tex>\log _{2}360\approx 8.4</tex> Let's estimate the value of <tx>a</tx>. We can note that <tx>3^2=9</tx> and <tx>3^3=27</tx>, so <tex>2 < \log _{3}20 < 3</tex> <note> Let <tx>a'</tx> be an approximate value of <tx>\log _{3}20</tx>. <tex>\frac{a'-2}{3-2}=\frac{20-9}{27-9}</tex> <tex>a'=2+\frac{11}{18}\approx 2.6</tex> </note> <tex>a=\log _{3}20\approx 2.6</tex> Let's estimate the value of <tx>b</tx>. <tex>2 < \log _{3}15 < 3</tex> <note> Let <tx>b'</tx> be an approximate value of <tx>\log _{3}15</tx>. <tex>\frac{b'-2}{3-2}=\frac{15-9}{27-9}</tex> <tex>b'=2+\frac{6}{18}=2+\frac{1}{3}\approx 2.3</tex> </note> <tex>b=\log _{3}15\approx 2.3</tex> Now, we can evaluate <tx>\frac{3a-b+5}{a-b+1}</tx> where <tx>a=2.6</tx> and <tx>b=2.3</tx> <tex>\frac{3a-b+5}{a-b+1}=</tex> <tex>\frac{3\cdot 2.6-2.3+5}{2.6-2.3+1}=\frac{10.5}{1.3}\approx 8.1</tex> We see that the estimated value of <tx>\frac{3a-b+5}{a-b+1}\approx 8.1</tx> is close to that of the <tx>\log _{2}360\approx 8.4</tx> You can read more about Thales's theorem <a target="_blank" href="https://en.wikipedia.org/wiki/Intercept_theorem">here</a> </answer> </lookback></howto>

<desc> Express log35(28) in terms of a and b given that log14(7)=a and log14(5)=b </desc> <task> Express <tx>\log _{35}28</tx> in terms of <tx>a</tx> and <tx>b</tx> given that <tx>\log _{14}7=a</tx> and <tx>\log _{14}5=b</tx></task> <solution> <tex>\log _{35}28=</tex> <tex>\frac{\log _{14}28}{\log _{14}35}</tex> <hr/> <tex>\frac{\log _{14}28}{\log _{14}35}=</tex> <tex>\frac{\log _{14}7\cdot 4}{\log _{14}7\cdot 5}=</tex> <tex>\frac{\log _{14}7+\log _{14}4}{\log _{14}7+\log _{14}5}=</tex> <tex>\frac{a+2\log _{14}2}{a+b}</tex> <hr/> <tex>\frac{\log _{14}28}{\log _{14}35}=</tex> <tex>\frac{\log _{14}14\cdot 2}{\log _{14}7\cdot 5}=</tex> <tex>\frac{\log _{14}14+\log _{14}2}{\log _{14}7+\log _{14}5}=</tex> <tex>\frac{1+\log _{14}2}{a+b}</tex> <hr/> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \log _{35}28=\dfrac{1+\log _{14}2}{a+b} \end{cases} </tex> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ 2\log _{35}28=\dfrac{2\left(1+\log _{14}2\right)}{a+b} \end{cases} </tex> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ 2\log _{35}28-\log _{35}28=\dfrac{2\left(1+\log _{14}2\right)}{a+b}-\dfrac{a+2\log _{14}2}{a+b} \end{cases} </tex> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \log _{35}28=\dfrac{2-a}{a+b} \end{cases} </tex> Finally, <tex>\log _{35}28=\frac{2-a}{a+b}</tex></solution> <result> <tex>\frac{2-a}{a+b}</tex></result> <howto> <understand> <question> What do we have? </question> <answer> We have three logarithms, two of which have the same base 14. The value of the second logarithm is <tx>a</tx> and that of the third is <tx>b</tx>. </answer> <question> What should we find? </question> <answer> We should rewrite the first logarithm as an algebraic expression using only the constants <tx>a</tx> and <tx>b</tx>. </answer> <question> What conditions should we consider? </question> <answer> We should use algebraic transformations based on logarithmic and exponentiation properties. You can read more about logarithmic properties <a href="https://en.wikipedia.org/wiki/List_of_logarithmic_identities" target="_blank">here</a> You can read more about exponentiation properties <a href="https://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties" target="_blank">here</a> </answer> </understand> <devise> <question> What can be our plan to solve it? </question> <answer> We can <ul> <li> rewrite the first logarithm using the change-of-base formula to get a division of logarithms with base 14 </li> <li> transform this quotient of logarithms into an expression containing <tx>\log _{14}7</tx> and <tx>\log _{14}5</tx> </li> <li> replace <tx>\log _{14}7</tx> with <tx>a</tx> and <tx>\log _{14}5</tx> with <tx>b</tx> </li> </ul> </answer> <question> Consider the base of the three logarithms. How can we change the base of the first logarithm? <hint> How can we use the change-of-base formula <fl>\log _{B}A=\frac{\log _{X}A}{\log _{X}B}</fl> where <tx>X>0, X\ne 1</tx>? </hint> </question> <answer> Let's change the base of the first logarithm <tex>\log _{35}28=</tex> <note> We use the change-of-base formula <fl>\log _{B}A=\frac{\log _{X}A}{\log _{X}B}</fl> where <tx>A=28, B=35, X=14</tx> </note> <tex>\frac{\log _{14}28}{\log _{14}35}</tex> </answer> <question> How can we transform the resulting expression so that it contains <tx>\log _{14}7</tx> and <tx>\log _{14}5</tx> ? <hint> Consider the properties of logarithms <ul> <li><fl>\log _{C}AB=\log _{C}A+\log _{C}B</fl> when <tx>A>0, B>0</tx></li> <li><fl>\log _{B}A^{n}=n\log _{B}A</fl> when <tx>A>0</tx></li> </ul> How can we use them to get <tx>\frac{a+2\log _{14}2}{a+b} \text{ ?}</tx> </hint> </question> <answer> We can rewrite the resulting expression <tex>\frac{\log _{14}28}{\log _{14}35}=</tex> <note> We note that <tx>28=7\cdot4</tx> and <tx>35=7\cdot 5</tx> <tex>\frac{\log _{14}7\cdot 4}{\log _{14}7\cdot 5}=</tex> and apply the logarithmic identity <tx>\log _{C}AB=\log _{C}A+\log _{C}B</tx> when <tx>A>0, B>0</tx> </note> <tex>\frac{\log _{14}7+\log _{14}4}{\log _{14}7+\log _{14}5}=</tex> <note> We note that <tx>4=2^2</tx> <tex>\frac{\log _{14}7+\log _{14}2^2}{\log _{14}7+\log _{14}5}=</tex> and use the property <fl>\log _{B}A^{n}=n\log _{B}A</fl> to rewrite <tx>\log _{14}2^2=2\log _{14}2</tx> </note> <tex>\frac{\log _{14}7+2\log _{14}2}{\log _{14}7+\log _{14}5}=</tex> <note> We note that <tx>\log _{14}7=a</tx> and <tx>\log _{14}5=b</tx> </note> <tex>\frac{a+2\log _{14}2}{a+b}</tex> So, <tex>\log _{35}28=\frac{a+2\log _{14}2}{a+b}</tex> </answer> <question> Let's analyze the resulting expression. How can we eliminate the logarithm <tx>\log _{14}2</tx>? Can we transform <tx>\frac{\log _{14}28}{\log _{14}35}</tx> differently? <hint> How can we rewrite the first logarithm to get <tx>\frac{1+\log _{14}2}{a+b} \text{?}</tx> </hint> </question> <answer> Let's find another way to rewrite the first logarithm <tex>\frac{\log _{14}28}{\log _{14}35}=</tex> <note> We note that <tx>28=14\cdot2</tx> and <tx>35=7\cdot 5</tx> <tex>\frac{\log _{14}14\cdot 2}{\log _{14}7\cdot 5}=</tex> and apply the logarithmic identity <tx>\log _{C}AB=\log _{C}A+\log _{C}B</tx> when <tx>A>0, B>0</tx> </note> <tex>\frac{\log _{14}14+\log _{14}2}{\log _{14}7+\log _{14}5}=</tex> <note> We note that <tx>\log _{14}14=1</tx>, <tx>\log _{14}7=a</tx>, <tx>\log _{14}5=b</tx> </note> <tex>\frac{1+\log _{14}2}{a+b}</tex> So, <tex>\log _{35}28=\frac{1+\log _{14}2}{a+b}</tex> </answer> <question> We have expressed <tx>\log _{35}28</tx> in two different ways. How can we use these results to eliminate <tx>\log _{14}2</tx>? <hint> What system of equations can we set up? </hint> </question> <answer> Let's eliminate <tx>\log _{14}2</tx> from previous results by solving the system <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \\ \log _{35}28=\dfrac{1+\log _{14}2}{a+b} \end{cases} </tex> <note> We multiply both sides of the second equation by <tx>2</tx> </note> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \\ 2\log _{35}28=\dfrac{2\left(1+\log _{14}2\right)}{a+b} \end{cases} </tex> <note> We subtract the first equation from the second </note> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \\ 2\log _{35}28-\log _{35}28=\dfrac{2\left(1+\log _{14}2\right)}{a+b}-\dfrac{a+2\log _{14}2}{a+b} \end{cases} </tex> <note> We combine the fractions, expand the numerators and cancel the terms in the second equation <tex>\log _{35}28=\frac{2+\cancel{2\log _{14}2}-a-\cancel{2\log _{14}2}}{a+b}</tex> </note> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \\ \log _{35}28=\dfrac{2-a}{a+b} \end{cases} </tex> Finally, <tex>\log _{35}28=\frac{2-a}{a+b}</tex> </answer> </devise> <carryout> <question> We can write and check each step of the solution now. </question> <answer> <tex>\log _{35}28=</tex> <tex>\frac{\log _{14}28}{\log _{14}35}</tex> <hr/> <tex>\frac{\log _{14}28}{\log _{14}35}=</tex> <tex>\frac{\log _{14}7\cdot 4}{\log _{14}7\cdot 5}=</tex> <tex>\frac{\log _{14}7+\log _{14}4}{\log _{14}7+\log _{14}5}=</tex> <tex>\frac{a+2\log _{14}2}{a+b}</tex> <hr/> <tex>\frac{\log _{14}28}{\log _{14}35}=</tex> <tex>\frac{\log _{14}14\cdot 2}{\log _{14}7\cdot 5}=</tex> <tex>\frac{\log _{14}14+\log _{14}2}{\log _{14}7+\log _{14}5}=</tex> <tex>\frac{1+\log _{14}2}{a+b}</tex> <hr/> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \\ \log _{35}28=\dfrac{1+\log _{14}2}{a+b} \end{cases} </tex> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \\ 2\log _{35}28=\dfrac{2\left(1+\log _{14}2\right)}{a+b} \end{cases} </tex> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \\ 2\log _{35}28-\log _{35}28=\dfrac{2\left(1+\log _{14}2\right)}{a+b}-\dfrac{a+2\log _{14}2}{a+b} \end{cases} </tex> <tex> \begin{cases} \log _{35}28=\dfrac{a+2\log _{14}2}{a+b} \\ \\ \log _{35}28=\dfrac{2-a}{a+b} \end{cases} </tex> Finally, <tex>\log _{35}28=\frac{2-a}{a+b}</tex> </answer> </carryout> <lookback> <question> Can we derive the result differently? <hint> How can we express <tx>\log _{14}2</tx> in terms of <tx>a</tx>? </hint> </question> <answer> We can write <tex>\log _{35}28=</tex> <tex>\frac{\log _{14}28}{\log _{14}35}=</tex> <tex>\frac{\log _{14}7\cdot 4}{\log _{14}7\cdot 5}=</tex> <tex>\frac{\log _{14}7+\log _{14}4}{\log _{14}7+\log _{14}5}=</tex> <tex>\frac{a+2\log _{14}2}{a+b}</tex> <hr/> <tex>a=\log _{14}7=\log _{14}\frac{14}{2}=</tex> <note> We use the logarithmic identity <fl>\log _{C}\frac{A}{B}=\log _{C}A-\log _{C}B</fl> when <tx>A>0, B>0</tx> where <tx>A=14, B=2, C=14</tx> </note> <tex>\log _{14}14-\log _{14}2=</tex> <note> We note that <tx>\log _{14}14=1</tx> </note> <tex>1-\log _{14}2</tex> So, <tex>\log _{14}2=1-a</tex> <hr/> <tex>\log _{35}28=</tex> <tex>\frac{a+2\log _{14}2}{a+b}=</tex> <tex>\frac{a+2\left( 1-a \right)}{a+b}=</tex> <tex>\frac{a+2-2a}{a+b}=</tex> <tex>\frac{2-a}{a+b}</tex> </answer> <question> How can we check the result? </question> <answer> We can use a calculator or we can estimate the values of <tx>\log _{35}28</tx> and <tx>\frac{2-a}{a+b}</tx> and show that they are close. Let's use Thales's theorem to estimate these values. <center> <img style="height:320px" height="320" alt="approximation using thales theorem" src="https://cdn.dzdx.xyz/thales.png"> </center> Let's estimate <tx>\log _{35}28</tx>. We can note that <tx>35^0=1</tx> and <tx>35^1=35</tx>, so <tex>0 < \log _{35}28 < 1</tex> Let <tx>x'</tx> be an approximate value of <tx>\log _{35}28</tx> <tex>\frac{x'-0}{1-0}=\frac{28-1}{35-1}</tex> <tex>x'=\frac{27}{34}\approx 0.8</tex> Therefore, <tex>\log _{35}28\approx 0.8</tex> Let's estimate the value of <tx>a</tx>. Let's express the logarithm <tx>\log _{14}7</tx> in terms of a logarithm with a lower base to get a more accurate estimate of its value <tex>\log _{14}7=</tex> <note> We use a special case of the change-of-base formula <fl>\log _{A}B=\frac{1}{\log _{B}{A}}</fl> </note> <tex>\frac{1}{\log _{7}14}</tex> We can note that <tx>7^1=7</tx> and <tx>7^2=49</tx>, so <tex>1 < \log _{7}14 < 2</tex> <note> Let <tx>a'</tx> be an approximate value of <tx>\log _{7}14</tx>. <tex>\frac{a'-1}{2-1}=\frac{14-7}{49-7}</tex> <tex>a'=1+\frac{7}{42}=\frac{49}{42}</tex> </note> <tex>\log _{7}14\approx \frac{49}{42}</tex> So, <tex>a=\log _{14}7\approx \frac{42}{49} \approx 0.9</tex> Let's estimate the value of <tx>b</tx>. Let's express the logarithm <tx>\log _{14}5</tx> in terms of a logarithm with a lower base to get a more accurate estimate of its value <tex>\log _{14}5=</tex> <note> We use a special case of the change-of-base formula <fl>\log _{A}B=\frac{1}{\log _{B}{A}}</fl> </note> <tex>\frac{1}{\log _{5}14}</tex> We can note that <tx>5^1=5</tx> and <tx>5^2=25</tx>, so <tex>1 < \log _{5}14 < 2</tex> <note> Let <tx>b'</tx> be an approximate value of <tx>\log _{5}14</tx>. <tex>\frac{b'-1}{2-1}=\frac{14-5}{25-5}</tex> <tex>b'=1+\frac{9}{20}=\frac{29}{20}</tex> </note> <tex>\log _{5}14\approx \frac{29}{20}</tex> So, <tex>b=\log _{14}5\approx \frac{20}{29} \approx 0.7</tex> Now, we can evaluate <tx>\frac{2-a}{a+b}</tx> where <tx>a=0.9</tx> and <tx>b=0.7</tx> <tex>\frac{2-a}{a+b}=\frac{2-0.9}{0.9+0.7}=\frac{1.1}{1.6}\approx 0.7</tex> We see that the estimated value of <tx>\frac{2-a}{a+b}\approx 0.7</tx> is close to that of the <tx>\log _{35}28\approx 0.8</tx> You can read more about Thales's theorem <a target="_blank" href="https://en.wikipedia.org/wiki/Intercept_theorem">here</a> </answer> </lookback></howto>