In a perfect number $2^{p−1} \times (2^p − 1)$, the ratio of $p$ to the digits in its perfect number approaches $\log(10) / \log(4)$?

Question

I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} \times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $\log(10) / \log(4) = 1.6609640474\ldots$" but I cannot find the proof that shows this to be true. Why is it $\log(10)/\log(4)$?

Answer 1

Because the number of digits in an integer $N$ is approximately $\log_{10}N$. The precise formula is $\lfloor\log_{10}N\rfloor+1$, but when $N$ is large this rounding is insignificant.

For $N=2^{p−1}\times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives: $$\begin{aligned} \log_{10}N&=\log_{10}\left(2^{p−1}\times(2^p−1)\right)=\\ &=\log_{10}\left(2^{p−1}\right)+\log_{10}\left(2^p−1\right)\approx\\ &\approx\log_{10}\left(2^{p−1}\right)+\log_{10}\left(2^p\right)=\\ &=(p-1)\log_{10}2+p\log_{10}2=\\ &=(2p-1)\log_{10}2\approx\\ &\approx2p\log_{10}2=\\ &=p\log_{10}4, \end{aligned}$$ and then the change of base formula finishes the proof.