Number of digits in the decimal representation of the powers of 2. I am trying to prove that as $n$ runs through the positive integers, the number of digits in the decimal representation of the sequence of powers of $2$, the set $2, 2^2, 2^3, \cdots$ can assume every possibile positive integral value. It seems that the number of digits increases by $1$ at every fourth power. This seems believable because ${{\log_{10}}}2$ is about $0.3$. But I am trying to find a rigorous proof of the above statement.
Why is $2$ special? Why is this not the case for other positive integers like $13$ for instance.
 A: Let $d$ the number of digits of $2^n$. That means that $10^{d-1} \le 2^n < 10^d$.
Taking $\log_{10}$ we get $d-1 \le n \log_{10}(2)< d$ which leads to $d = \lfloor n \log_{10}(2)\rfloor + 1$ .
That explains the behavior you're observing. A similar formula can be derived for others numbers, there's nothing special about $2$ (or the base $10$).
A: Assume some fixed length $d$ representing the number of digits, and let's see if there exists some $k$ such that $2^k$ that has length $d$ when expanded.
The number of digits of $2^k$ is equal to $\lfloor\log_{10}(2^k)\rfloor + 1$.
$\lfloor\log_{10}(2^k)\rfloor + 1 = d$
$\lfloor\log_{10}(2^k)\rfloor = d - 1$
$d - 1 \leq \log_{10}(2^k) \lt d$
$d - 1 \leq k \log_{10}(2) \lt d$
$\frac{d - 1}{\log_{10}(2)} \leq k \lt \frac{d}{\log_{10}(2)}$
So $k$ can take on an integral value for sure if the size of that range is at least of size $1$.
And since $\frac{d}{\log_{10}(2)} - \frac{d - 1}{\log_{10}(2)} = \log_2(10) = 3.32...$, the threshold will always be greater than $1$, which means there will exist an integral value for $k$ regardless of what $d$ is, assuming we're dealing with base $2$ numbers.
It would stop working for base $b$ once we have $\log_{b}(10) < 1$, which begins once $b > 10$.
