I got a question to solve. Find the number of digits in $\displaystyle 6^{100} $

Value of $\displaystyle \log_{10} 2 = 0.30102 \and \log_{10} 3 = 0.47712 \was \given$

I used logarithm.

$\displaystyle \log y = 100log(2*3) $

$\displaystyle log y = 100 (log2 + log3)$

$\displaystyle log y = 100( 0.30102 + 0.47712)$

$\displaystyle logy = 100 * 0.77814$

$\displaystyle logy = 77.814$

So will the number of digits will be 78 ? because whatever power 10 is raised to. the number of digits is (power + 1). Am i correct. If not then what should be the answer.