Question

If ${\log }_{4}5=a$ and ${\log }_{5}6=b$ then ${\log }_{3}2$ is equal to

• A. $\frac{1}{2a+1}$
• B. $\frac{1}{2b+1}$
• C. $2ab+1$
• D. $\frac{1}{2ab-1}$

Answer 1

The correct option is D $\frac{1}{2ab-1}$

Since, ${\log }_{4}5=a$ ...(1)
and ${\log }_{5}6=b$ ...(2)
Multiplying equations (1) & (2), we get

$ab=\left(\frac{\log 5}{\log 4}\right)\left(\frac{\log 6}{\log 5}\right)$
$\Rightarrow ab=\frac{\log 6}{\log 4}$

$\Rightarrow ab=\frac{\log 3\times 2}{\log 2^2}$

$\Rightarrow ab=\frac{\log 3+\log 2}{2\log 2}$
$\Rightarrow 2ab=1+{\log }_{2}3$
$\Rightarrow {\log }_{3}2=\frac{1}{2ab-1}$.