If $a$ is a positive integer and $p$ is prime, then $\log_{p}(a)$ is rational if and only if $a$ is a power of $p$. 
If $a$ is a positive integer and $p$ is prime, then $\log_{p}(a)$ is rational if and only if $a$ is a power of $p$.

We are asked to used the fundamental theorem of arithmetic to prove this statement. I began my proof as follows:

"Let's assume that $\log_pa$ is rational, and that $\log_pa=x$, for a positive integer $x$. By the definition of a logarithm, we know that $p^x=a$, meaning that $a$ is a product of $p$ times itself $x$ times. This suggests that the only prime factor of $a$ is $p$. By the fundamental theorem of arithmetic, we can conclude that $a$ is an integer because it has a unique prime factorization."

I'm not sure if this makes sense, or how to proceed. Thank you for your assistance.
 A: This is a reasonable outline to start with.
The major problem with what you've written is that you assume that $\log_pa$ is rational and then assign an integer value to it.  You should actually say that there are relatively prime integers $x,y\in\mathbb Z$ such that $\log_pa=\frac xy$ and then show that it must be that $y=1$.  Your strategy of thinking about the prime factorization of $a$ is a good idea here that will help you here.
You also need to note that you are asked to prove the equivalence of the statements, so you also need to prove the converse of the statement you proved.  This is not difficult, but it is essential in "if and only if" proofs.
A: You're on the right track but you write something incorrect in the very beginning of your proof.  You begin with the hypothesis that $\log_pa$ is rational, and then you say that therefore we can write $\log_p a = x$ for a positive integer $x$.  You do not yet know that this is the case.  All you know is that $x$ is a rational number, not an integer.
So instead, you should say that since $\log_p a$ is rational, you can write $\log_p a = \frac{m}{n}$ for some integers $m$ and $n$.  This means that $p^{m/n} = a$.  See if you can proceed from here.
Also, for some reason in your proof you make the conclusion that $a$ must be an integer.  This does not add anything to the proof.  You were already told that $a$ was an integer!
