TITLE: percentage of numbers starting with $2$ in $\{2^n\}$
QUESTION [7 upvotes]: I have once heard a professor telling (during a course on Fourier theory) that there is a way to determine the numbers starting with a $2$ in the sequence $\{2^n\colon n\in\mathbb{N}\}$. I asked him about it, but he could not remember how. The proof involved the theory of equidistributed sequences the Weyl criterion.
Is there someone who knows how to prove this? Or is there a nice reference (preferably a book)?

REPLY [4 votes]: $2^n$ begins with a $2$
$\iff 2^n /10^k \in [2;3)$ with $k \in \Bbb N$
$\iff n\log 2- k \log 10 \in [log 2; \log 3)$ with $k \in \Bbb N$
$\iff \left\{n\frac{\log 2}{\log 10}\right\} \in \left[\frac{\log 2}{\log 10}; \frac{\log 3}{\log 10}\right)$, where $\{x\} \in [0;1)$ is the fractional part of $x$.
Since $\frac{\log 2}{\log 10}$ is irrational, the sequence $\left\{n\frac{\log 2}{\log 10}\right\}$ is equidistributed in $[0;1)$, and so it lands in $\left[\frac{\log 2}{\log 10}; \frac{\log 3}{\log 10}\right)$ with "probability" $\frac{\log 3 - \log 2}{\log 10}$.