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 $64$
 A: Hint
Let's call $n$ the number of digits of $2^{64}$ then
$$10^{n-1}\le2^{64}<10^n$$
so,
$$n-1\le\log 2^{64}<n$$ $$n-1\le 64\log 2<n$$
Can you finish?
A: 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!
A: See that $100 = 10^2$ and $1000 = 10^3$ and so on.  The number of digits in both those numbers is one more than the exponent on $10$.  If we write any 3-digit number as a power of 10, then the exponent will be between $2$ and $3$.  For instance,  $423 = 10^{2.6263\ldots},$  where, of course $2.6263\ldots$ is the base-10 log of $423.$  
Using this principle we can do this:
$$2^{64}= \left(10^{\log 2}\right)^{64} = 10^{64\log } = 10^{19.26\ldots}. $$
So we know the number has 20 digits.
