Question

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

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

Answer 1

Answer: (D)

$\frac{1}{2ab-1}$

${\log }_{4}5=a$ ...(1)
${\log }_{5}6=b$ ...(2)
Multiplying eq (1) & (2) we get,
$ab=\left(\frac{\log 5}{\log 4}\right)\left(\frac{\log 6}{\log 5}\right)$
$\Rightarrow ab=\frac{\log 6}{\log 4}=\frac{\log 3+\log 2}{2\log 2}$
$\Rightarrow 2ab=1+{\log }_{2}3$
$\Rightarrow {\log }_{3}2=\frac{1}{2ab-1}$
Ans: D