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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.

  • 0
    5 years too late but to make this *perfectly* clear: $2^m = (10^{\log_{10}2})^m=10^{m\log_{10} 2}$. $10^k$ will have $k + 1$ digits (1 and k zeros) so $10^x$ with k < x < k+1 will have $x$ rounded up to the next higher integer ($k+1$). So $2^{m}$ will have $\lceil m\log_{10} 2 \rceil= \lceil m*0.30102999566398119521373889472449.... \rceil$ .2017-01-07

4 Answers 4

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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 $d = 1 + \lfloor \log_{10}(n) \rfloor$

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you can simply right it in the form of log as
Log2x=1000
Log x/log2=1000
Taking value of log 2 which is 0.301
Log x=301
Converting in exponents form
10^301=x
It means x has 301 digits

  • 1
    There are most definitely 302 digits. Count them: $107150860718626732094842504906000181056140481170553360744375038837035105112\\ 4936122493198378815695858127594672917553146825187145285692314043598457757469\\ 85748039345677748242309854210746050623711418779541821530464749835819412673\\ 98767559165543946077062914571196477686542167660429831652624386837205668069376.$2015-06-03