Question

Find the value of $\dfrac{{{\log }_{8}}17}{{{\log }_{9}}23}-\dfrac{{{\log }_{2\sqrt{2}}}17}{{{\log }_{3}}23}$

Answer 1

Hint: In this problem we need to calculate the value of the given expression which contains the logarithmic functions with different base values. So, we will consider each term individually in the first given fraction. Now we will simplify the value by using the logarithmic formula ${{\log }_{{{a}^{n}}}}b=\dfrac{1}{n}{{\log }_{a}}b$. By using this formula, we will simplify terms in the first fraction and apply them in the given expression and apply some basic mathematical operations to get the required result.

Complete step by step solution:
Given expression is $\dfrac{{{\log }_{8}}17}{{{\log }_{9}}23}-\dfrac{{{\log }_{2\sqrt{2}}}17}{{{\log }_{3}}23}$.
Considering the first fraction in the given expression which is $\dfrac{{{\log }_{8}}17}{{{\log }_{9}}23}$.
In the above fraction we have ${{\log }_{8}}17$ as numerator and ${{\log }_{9}}23$ as denominator.
Considering the numerator ${{\log }_{8}}17$ and writing the number $8$ as ${{2}^{3}}$, then we will get
${{\log }_{8}}17={{\log }_{{{2}^{3}}}}17$
Applying the formula ${{\log }_{{{a}^{n}}}}b=\dfrac{1}{n}{{\log }_{a}}b$ in the above equation, then we will have
${{\log }_{8}}17=\dfrac{1}{3}{{\log }_{2}}17$

Considering the denominator ${{\log }_{9}}23$ and writing the number $9$ as ${{3}^{2}}$, then we will get
${{\log }_{9}}23={{\log }_{{{3}^{2}}}}23$
Applying the formula ${{\log }_{{{a}^{n}}}}b=\dfrac{1}{n}{{\log }_{a}}b$ in the above equation, then we will have
${{\log }_{9}}23=\dfrac{1}{2}{{\log }_{3}}23$

In the second given fraction $\dfrac{{{\log }_{2\sqrt{2}}}17}{{{\log }_{3}}23}$, considering the numerator ${{\log }_{2\sqrt{2}}}17$ and writing the number $2\sqrt{2}$ as ${{2}^{1+\dfrac{1}{2}}}={{2}^{\dfrac{3}{2}}}$, then we will get
${{\log }_{2\sqrt{2}}}17={{\log }_{{{2}^{\dfrac{3}{2}}}}}17$
Applying the formula ${{\log }_{{{a}^{n}}}}b=\dfrac{1}{n}{{\log }_{a}}b$ in the above equation, then we will have
${{\log }_{2\sqrt{2}}}17=\dfrac{2}{3}{{\log }_{2}}17$

Substituting all the values we have simplified in the given expression, then we will get
$\dfrac{{{\log }_{8}}17}{{{\log }_{9}}23}-\dfrac{{{\log }_{2\sqrt{2}}}17}{{{\log }_{3}}23}=\dfrac{\dfrac{1}{3}{{\log }_{2}}17}{\dfrac{1}{2}{{\log }_{3}}23}-\dfrac{\dfrac{2}{3}{{\log }_{2}}17}{{{\log }_{3}}23}$
Simplifying the above equation by using some basic mathematical operation, then we will have
$\begin{align}
  & \dfrac{{{\log }_{8}}17}{{{\log }_{9}}23}-\dfrac{{{\log }_{2\sqrt{2}}}17}{{{\log }_{3}}23}=\dfrac{2{{\log }_{2}}17}{3{{\log }_{3}}23}-\dfrac{2{{\log }_{2}}17}{3{{\log }_{3}}23} \\
 & \Rightarrow \dfrac{{{\log }_{8}}17}{{{\log }_{9}}23}-\dfrac{{{\log }_{2\sqrt{2}}}17}{{{\log }_{3}}23}=0 \\
\end{align}$

Hence the value of $\dfrac{{{\log }_{8}}17}{{{\log }_{9}}23}-\dfrac{{{\log }_{2\sqrt{2}}}17}{{{\log }_{3}}23}$ is $0$.

Note: In this problem we have the bases of the given logarithmic function as different, so we have simplified the each and every value by using the logarithmic formulas. If we have the base of the given logarithmic function as $10$, then we need to use the logarithmic table and find the values of the terms and simplify the expression by using basic mathematical operations.