2015/080) If $log_4 5 =a$ and $log_5 6 =b$, then $log_3 2 =$

From the given condition

$5= 4^a$
$6 = 5^b = (4^a)^b = (2^2)^{ab} = 2^{2ab}$

so $3 *2 = 2^{2ab}$

so $3 = \frac{2^{2ab}}{2} = 2^{2ab-1}$
or $2 = 3^{\frac{1}{2ab-1}}$

hence $log_3 2$=$\dfrac{1}{2ab-1}$