Find the total number of digits in the number $6^{100}$ (Given ${log}_{10} 2 = 0.010; {log}_{10} 3 = 0.47771$ )

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Solution

Let $x = 6^{100}$
$\Rightarrow {log}_{10} x = 100 {log}_{10} 6$
$\Rightarrow {log}_{10} x = 100 [{log}_{10} 2 + {log}_{10} 3]$
$= 100 [0.3010 + 0.47771]$
$= 100 (0.7781)$
$= 77.81$
$∴$ The number of digit $= 78$ .

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