Question

The number of digits in ${20^{301}}$ is
$A.602 \\\ B.301 \\\ C.392 \\\ D.391 \\\$

Answer 1

Hint:In this question apply the base rule of logarithm i.e. $\log {x^2} = 2\log x$ and $\log(ab)$=$\log a$+$\log b$.The value of ${\log _{10}}2 = 3.010$. Use this to find the number of digits in ${20^{301}}$.

Complete step-by-step answer:
According to the question we have to find the number of digits in ${20^{301}}$.
Let $x = {20^{301}}$
Taking $\log$ both sides , we get
$\Rightarrow \log x = 301\log 20$
We know that $\log(ab)$=$\log a$+$\log b$ and $\log10=1$ So we can write,
$\Rightarrow \log x = 301\left( {1og2 + \log 10} \right) \\\ \Rightarrow \log x = 301\left( {.3010 + 1} \right) \\\ \Rightarrow \log x = 391.601 \\\ \\\$
The value of $\log x$ gives the number of digits in ${20^{301}} = 391 + 1 = 392$

Note:Students should remember the important logarithmic properties and formula
i.e $\log(ab)$=$\log a$+$\log b$ ,$\log {x^2} = 2\log x$ for solving these types of problems.In above problem $\log(20)$ we written as $\log(2.10)$=$\log2+\log10$ by using logarithmic property as they given value of $\log2$ and we know value of $\log10=1$ .Finally the value of $\log x$ gives the number of digits.