Prove that $\log_{2}(7)$ is irrational 
Prove that $\log_{2}(7)$ is an irrational number.

My Attempt:
Assume that $\log_{2}(7)$ is a rational number. Then we can express $\log_{2}(7)$ as $\frac{p}{q}$, where $p,q\in \mathbb{Z}$ and $q\neq 0$. This implies that $7^q = 2^p$, where either $p,q>0$ or $p,q<0$.
My question is this: why can't we count for $p,q<0$? My textbook's author counts only for $p,q>0$. Could someone explain the reasoning being used by the author?

Saying what Neal has already said a bit different: First you realize that $\log_2(7) > 0$. Then if $\log_2(7)$ is rational you can find positive integers $p$ and $q$ such that $$\log_2(7) = \frac{p}{q}.$$
You could, of course, also have chosen $p$ and $q$ both negative, but you can definitely pick then both positive. So then, as you note, $7^q = 2^p$. Now since both $7$ and $2$ are prime, you know that this will never happen when $p$ and $q$ are positive integers.
