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.
$\endgroup$ 73 Answers
$\begingroup$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 $$d = 1 + \lfloor \log_{10}(n) \rfloor$$
$\endgroup$ $\begingroup$Solve
$$ 10^{d} = 2^{1000}.$$
You should get an answer of approximately $301.03,$ which when rounded up gives $302.$
$\endgroup$ $\begingroup$$2^{10}=4$ digits
$2^{20}=7$ digits
from above we know that
$2^{10} = k$
$2^{20} = k+3$
Since $2^{10n}=\text{previous digits}+3$, following the process then $$
2^{1000} \text{ has }302\text{ digits}
$$
