powers of 2 in base 3 What is the asymptotic limit of the ratio of $1s$ to $2s$ in the first digits of $2^n$ in base $3$? 
If the $2^{nd},3^{rd}$ digits etc. were random but equally likely to be $0,1$ or $2$ then the ratio would be $2:1$ since if $2^n$ begins with $2$ then 2^{n+1} begins with $1$, and if $2^n$ begins with $11..10$ then $2^n+1$ begins with $2$, and if $2^n$ begins with $11...12$ then $2^{n+1}$ begins with 1. 
 A: Here is an elementary way to look at it: There are exactly $\lfloor n\log 3 / \log 2\rfloor = \lfloor n\log_2 3\rfloor$ powers of $2$ less than $3^n$. Of these, exactly $n$ base-$3$ representations start with a $1$, because for each $1 \le k < n$, (i) the first power of $2$ that is $k$ digits long must start with a $1$; and (ii) there can't be two successive $k$-digit powers of $2$ that start with a $1$.  
So the proportion of powers of $2$ that begin with $1$ is exactly $n/\lfloor n\log_2 3\rfloor$, which tends to $1/\log_2 3 = \log_3 2$.
Updated to add: This method doesn't generalise very well, except to show the following:  
Given positive integers $a,b$ with $a < b$, then the proportion of powers of $a$ whose first base-$b$ digit is less than $a$ is $\log_b a$.  
A: $\log2$ is irrational, so $n\log2$ is uniformly distributed modulo 1, but $n\log2=\log2^n$. This means you get a Benford's Law distribution (which see - surely there is a lot about this on the web). In particular, the proportion of times the first digit is a 1 is $\log_32$, but you can work out the proportion of times any given finite string appears as the first (base 3) digits. 
