How many digits in $2^{738}-1$. I don't exactly know how to approach this problem. Could the expression somehow be rewritten in the form $a*10^{n-1}$, where $n$ would be the number of digits?
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3$\lceil 738\cdot\log_{10}2\rceil=223$ – J. W. Tanner Jan 26 '20 at 04:07
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1Welcome to Mathematics Stack Exchange. Does this help answer your question? – J. W. Tanner Jan 26 '20 at 04:17
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Yes, thank you. – CN05 Jan 26 '20 at 04:18
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$a$ has $n $ digits $\iff 10^{n-1}\le a<10^n\iff n-1\le \log_{10} a\lt n\iff n-1=\lfloor \log_{10}a\rfloor$
$\iff n=\lfloor \log_{10}a\rfloor+1.$
Therefore, the number of digits of $2^{738}$ is $\lfloor 738\log_{10}2\rfloor+1$, which can be calculated to be $223$.
$a$ has the same number of digits as $a-1$, unless $a$ is a power of $10$, which $2^{738}$ clearly is not.

J. W. Tanner
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