How many digits does $2^{1000}$ contain?

Question

I tried this way, I only need to know if this is correct or if there are better ways to solve this:

$2^{1000}$ does not have a factor of $5$ obviously therefore we can assume

$$ 10^{m} < 2^{1000} < 10^{m+1}$$ for some $m$

Assume $ k = 2^{1000}$, then take log on both sides $\log k = 1000 \log 2 \approx 301.02999 > 301$

Therefore $2^{1000}$ has $302$ digits.

Answer 1

Recall that $10^{d-1}$ has $d$ digits. So for any number $n,$ the number of digits of $n$ is given by solving $ 10^{d-1} = n,$ or $$d = 1 + \lfloor \log_{10}(n) \rfloor$$