Question

pls answer my question
no.of digits in
${4}^{2018}$

Answer 1

Given 4^ 2018
Let us take log 4^2018
= log[(2)²]^20138   [∵ (a^m)^n = a^mn]
= log 2^ 4036        
= 4036 log2                [∵ logm^n = nlogm] 
= 4036 × 0.301           [∵ log2 = 0.301]
= 1214.836
∴ 4^2018 has 4 digits before decimal