Question

The number of digits in $20^{301}$ (given $\log_{10}2 = 0.3010$) is

• A. 602
• B. 301
• C. 392
• D. 391

Answer 1

Answer: (C)

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