TITLE: Show that there are infinitely many powers of two starting with the digit 7
QUESTION [38 upvotes]: This is a contest math problem which I was not able to solve. A hint toward the solution would be helpful as well. 
Problem: Show that there are infinitely many powers of 2 starting with the digit 7.
Thanks in advance.

REPLY [4 votes]: Let us demonstrate that we can get arbitrarily close to some $10^n$ factor if we just let $k$ get large enough in $2^k$.
$$\log_{10}({2^k}) = k\log_{10}(2)$$ If we now accept $\log_{10}(2)$ is irrational (easy to show), then this means in practice that doubling does not create any periodicity in it's ten representation.
By selecting $k_1,k_2$ we can then get $\epsilon$ in $(10+\epsilon)^{k_1} = 2^{k_2}$ to be arbitrarily small, so no matter how close a $7a_1a_2\cdots a_n$ is to $80\cdots 0$ we can always find a power of 2 that if multiplied with the previous number would squeeze in between 7 and 8, (although no guarantee of how many new digits on the end).