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

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 $$\log(10)/\log(4)$$?

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.