Doubt in this question involving logarithm If $\log2= 0.301$, then how many number of digits are in $2^{64}$?
What I did:
$$\log(2)^{64}=\log2^{64}=64\log2=64\log2=19.264$$
Number of digits comes out to be $5$. But answer is $20$? I have written $\log 2$ raise to the power $64$

Because number of digits it's$$[64\log2]+1=20,$$ which gives the answer.
In the general case if $n$ is a number of digits of natural $N$ then $n=[\log{N}]+1$
Indeed, let $N=a_0\cdot10^{n-1}+a_1\cdot10^{n-2}+...+a_{n-1}$, where $a_i\in\{0,1,...,9\}$ and $a_0\neq0$. 
Hence, $n$ is a number of digits. 
We see that $10^{n-1}\leq N<10^n$ or $n-1\leq\log{N}<n$, 
which says $[\log{N}]=n-1$ and we are done!
