Question

If ${\log }_{c}2.{\log }_{b}625={\log }_{10}16.{\log }_{c}10$, where $c>0;c\ne 1;b>1;b\ne 1$, then the value of $b$ is

• A. $25$
• B. $5$
• C. $625$
• D. $16$

Answer 1

The correct option is B $5$

$\left({\log }_{c}2\right)\left({\log }_{b}625\right)=\left({\log }_{10}16\right)\left({\log }_{c}10\right)$
$\Rightarrow \frac{{\log }_{c}2}{{\log }_{c}10}\times {\log }_{b}625={\log }_{10}16[\because {\log }_{a}b=\frac{\log b}{\log a}]$
$\Rightarrow {\log }_{10}2\times {\log }_{b}625={\log }_{10}{2}^4$
$\Rightarrow {\log }_{10}2\times {\log }_{b}625=4{\log }_{10}2$
$\Rightarrow {\log }_{b}625=4$
$\Rightarrow b^4=625$

$\Rightarrow b^4=5^4$

$\Rightarrow b=5$
Ans: B