Suppose $a\_n \to L$ in $\mathbb{R}$. Prove $2^{a\_n} \to 2^L$.

I tried to start from $| 2^{a\_n} - 2^L | < \epsilon$ and work my way to $|a\_n - L| < \epsilon$, but logarithms are showing up and I think the problem is too simply to need to prove an inequality with a logarithm or exponential. 

I cannot use the continuity of $2^n$ in this question.

Any help is appreciated.