Question

The number of digits in ${20}^{301}$ is

• A. $602$
• B. $301$
• C. $392$
• D. $391$

Answer 1

The correct option is C $392$

Let $y={20}^{301}$
Number of digits $=$ Integral part of $\left(301{\log }_{10}20\right)+1$
$=$ Integral part of $\left[301({\log }_{10}10+{\log }_{10}2)\right]+1$
$=$ Integral part of $\left[301(1+0.3010)\right]+1$
$=$ Integral part of $[301\times 1.3010]+1$
$=$ Integral part of $\left[391.601\right]+1$
$=391+1=392$