Question

The value of $\frac{\left({\log }_3\left(5\right)\times {\log }_{25}\left(27\right)\times {\log }_{49}\left(7\right)\right)}{{\log }_{81}\left(3\right)}$ is

• A. $1$
• B. $6$
• C. $\frac{2}{3}$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct answer:

Compute the required value:

$S=\frac{\left({\log }_3\left(5\right)\times {\log }_{25}\left(27\right)\times {\log }_{49}\left(7\right)\right)}{{\log }_{81}\left(3\right)}$

We know that, ${\log }_b\left(a\right)=\frac{\log \left(a\right)}{\log \left(b\right)}$

$\Rightarrow$ $S=\frac{{\displaystyle \frac{\log \left(5\right)}{\log \left(3\right)}}\times {\displaystyle \frac{\log \left(27\right)}{\log \left(25\right)}}\times {\displaystyle \frac{\log \left(7\right)}{\log \left(49\right)}}}{{\displaystyle \frac{\log \left(3\right)}{\log \left(81\right)}}}$

$\Rightarrow$ $S=\frac{{\displaystyle \frac{\log \left(5\right)}{\log \left(3\right)}}\times {\displaystyle \frac{\log \left(3^3\right)}{\log \left(5^2\right)}}\times {\displaystyle \frac{\log \left(7\right)}{\log \left(7^2\right)}}}{{\displaystyle \frac{\log \left(3\right)}{\log \left(3^4\right)}}}$ $\left[\because \log \left(a^m\right)=m\log \left(a\right)\right]$

$\Rightarrow$ $S=\frac{{\displaystyle \frac{\log \left(5\right)}{\log \left(3\right)}}\times {\displaystyle \frac{3\log \left(3\right)}{2\log \left(5\right)}}\times {\displaystyle \frac{\log \left(7\right)}{2\log \left(7\right)}}}{{\displaystyle \frac{\log \left(3\right)}{4\log \left(3\right)}}}$

$\Rightarrow$ $S=\frac{{\displaystyle \frac{1}{1}}\times {\displaystyle \frac{3}{2}}\times {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{4}}}$

$\Rightarrow$ $S=\frac{{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{1}{4}}}$

$\Rightarrow$ $S=3$

Hence, the value of $\frac{\left({\log }_3\left(5\right)\times {\log }_{25}\left(27\right)\times {\log }_{49}\left(7\right)\right)}{{\log }_{81}\left(3\right)}$ is $3$.

Hence, option $\left(D\right)$ is the correct answer.