This is not an absolute rule as you easily find counter examples. However, in $[\![0;100]\!]$ there are 51 integers which follow this rule. Many times, the number of digits in $2^{n+1}$ is twice the number of digits in $2^{n}$. Examples: $$ \begin{split} 2^{13} & = 8192\\ 2^{13 \times 2} = 2^{26} & = 67108864 \\ 2^{26 \times 2} = 2^{52} & = 4503599627370496\\ 2^{52 \times 2} = 2^{104} & = 20282409603651670423947251286016 \end{split} $$ Again, we don't observe a universal rule. However, a large part of integers behave this way. I was wondering if it is easily explainable, or if it's just a coincidence.
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1A number $n$, written in base $10$ has roughly $\log_{10}(n)$ digits (more precisely, $n$ has $\lfloor \log_{10}(n) \rfloor$ decimal digits). What is $\log_{10}(2^n)$? What does that tell you about the number of digits in the decimal expansion of $2^n$? – Xander Henderson Jan 15 '23 at 21:38
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Doubling the exponent is the same as squaring the number because $2^{2n}=(2^n)^2$. When you square a number of $k$ digits you get a number with $2k-1$ digits if there is no carry in the high position and a number with $2k$ digits if there is. A carry will happen if the leading digits are greater than 3162, which comes from $\sqrt {10} \approx 3.162$.
Ross Millikan
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