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?

Consider the case $\log_27=p/q$ where $p, q<0$. You'll still have
$$
7^q=2^p
$$
and thus you'll have
$$
\frac{1}{7^{-q}}=\frac{1}{2^{-p}}
$$
which is the same as saying
$$
7^{-q}=2^{-p}
$$
and since $-q$ and $-p$ are both positive, you're back in the case where both exponents are positive.
