Avoiding matching first digit of $a^n$ with $b^n$ For any given pairs of positive integers $a$ and $b$, is it possible that the first digit of $a^n$ never matches the first digit of $b^n$ for any positive integer $n$?
(If $a=2$ and $b=5$ the only possible matching first digit is "3".) 
Edit: I meant to say $a=2$ in the parenthetical comment above.  I had previously written $a=3$, which does not work as several comments indicate.

I'm not answering the question, but justifying the claim in the last sentence of the question: when $a=2$ and $b=5$ the only shared leading digit of $a^n$ and $b^n$ is 3.
Suppose $2^n$ and $5^n$ share leading digit $d$ and have $r+1$ and $s+1$ digits, resp.  For $n>3$ we have $d10^r < 2^n < (d+1)10^r$ and $d10^s < 5^n < (d+1)10^s$.  Therefore $d^2 10^{r+s} < 10^n < (d+1)^2 10^{r+s}$ so $d^2 < 10^{n-r-s} < (d+1)^2$.  Since $d$ is a digit, $d \geq 1$ and $(d+1) \leq 10$; these restrictions force $n-r-s = 1$, so $d^2 < 10 < (d+1)^2$ meaning only $d=3$ works.  And since $2^5 = 32$ and $5^5 = 3125$ it does in fact occur as a leading digit.
