How many digits does $2^{25964951}-1$ have? Mersenne prime numbers are prime numbers of the form $2^n - 1 \,(\mathbb{N}\ni n>1)$such as
$2^2 -1$ or $2^{25964951}-1$: How many digits does the latter have?
I found this here:
$$\log_{10}(2^{25964951}-1)+1 =  7 816 230$$
Is this the way how to proof this or do I have to do something other?
 A: Taking the 10-logarithm is basically the easiest way to do it. Remember to always round down, no matter what the decimal part is (for instance, $\log_{10}(99)+1\approx2.996$, and we want the answer to be $2$).
However, if all you have is a classic calculator that won't work with such large numbers, or you want to get an approximate answer with just mental calculations, some more manipulations must be done first.
Start with noting that $2^{25\,964\,951}-1$ and $2^{25\,964\,951}$ have the same number of digits. From there, we know that
$$
\log_{10}(2^{25\,964\,951})={25\,964\,951}\cdot\log_{10}(2)
$$
And then either let the calculator do the work, or if you go the mental route, know that $\log_{10}(2)\approx 0.3$ (from $2^{10}\approx 10^3$) and just get
$$
{25\,964\,951}\cdot\log_{10}(2)\approx 7\,800\,000
$$
For the mental calculation, adding 1 here (although it's technically what you're supposed to do) makes no sense in my opinion.
A: Firstly, we have to make sure that there are no positive integer value of $n$ such that:
$$2^{25 964 951} = 10^n$$
Cause if $2^{25 964 951}$ is equal $10...0$ then after minus $1$, number of digits will be different.
And that's true, cause:
$$log_{10}(2^{25 964 951}) = n$$
$$25964951\times log_{10}(2) = n$$
And $n$ will not be an integer.
So, $2^{25 964 951}$ and $2^{25 964 951} - 1$ have the same number of digits.
With the equation above, $n \approx 7816229$.
Then adding $1$, we have the results.
