Question

The number of digits in ${20}^{301}$ (given ${\log }_{10}2=3.010$) is

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

Answer 1

The correct option is C $392$

Let $x={20}^{301}$

Taking log both sides, we get

$\Rightarrow \log x=301\log 20$

$\Rightarrow \log x=301(\log 2+\log 10)$

$\Rightarrow \log x=301(.3010+1)$

$\Rightarrow \log x=391.601$

Number of digits in ${20}^{301}=391+1=392$