Question

The number of digits in ${20}^{301}$ (given, ${\log }_{10}\left(2\right)=0.3010$) is

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

Answer 1

The correct option is C

$392$

Explanation for the correct option:

Find the number of digits in the given expression.

An expression ${20}^{301}$ is given.

We know that the number of digits in the expression like $a^b$ can be given by $b·{\log }_{10}\left(a\right)+1$.

Therefore, the number of digits in the given expression is given by:

$$\begin{gathered} 301·{\log }_{10}\left(20\right)+1=301·\left\{{\log }_{10}(2·10)\right\}+1 \\ \Rightarrow 301·{\log }_{10}\left(20\right)+1=301·\left\{{\log }_{10}\left(2\right)+{\log }_{10}\left(10\right)\right\}+1 \\ \Rightarrow 301·{\log }_{10}\left(20\right)+1=301·\left\{0.3010+1\right\}+1 \\ \Rightarrow 301·{\log }_{10}\left(20\right)+1=301·\left\{1.3010\right\}+1 \\ \Rightarrow 301·{\log }_{10}\left(20\right)+1=391.601+1 \\ \Rightarrow 301·{\log }_{10}\left(20\right)+1=392.601 \end{gathered}$$

Therefore, the number of digits in ${20}^{301}$ is $392$.

Hence, option $C$ is the correct answer.