Write the logarithmic form of 92^1/5 = 2.47.

$a^b=x\iff {\log }_{a}x=b$, where a $\mathrm{a} \ne 1, \mathrm{and} \mathrm{a} > 0 \mathrm{and} \mathrm{x} >0.$

Given: 92^1/5 = 2.47
As we know that, $a^b=x\iff {\log }_{a}x=b$
Comparing ${92}^{1/5}=2.47$  with $a^b=x$  we have,
Here,  $a=92, b=1/5\text{ and }x=2.47$
So, the logarithmic form of  ${92}^{1/5}=2.47\text{ is }{\log }_{92}2.47=\frac{1}{5}$.