Problem: Prove $log$$_2$($3$) is irrational using the Fundamental Theorem of Arithmetic.
What I have so far:
Proof: Suppose $log$$_2$($3$) $\in$ $\mathbb Q$
Then there are $p,q$ $\in$ $\mathbb Z$ , $q$$\neq$$0$ s.t. $log$$_2$($3$)=$\frac pq$
Then $2$$^p$ $=$ $3$$^q$
This is where I get stuck because I'm not sure how to incorporate the Fundamental Theorem of Arithmetic.
The fundamental theorem of arithmetic tells you that the prime factorisation of any positive integer is unique.
Applying this to the positive integer $n = 2^p = 3^q$, you see that $2^p$ and $3^q$ are two candidate prime factorisations for the same integer $n$.
The only way these prime factorisations can actually be the same is if $p = 0$ and $q = 0$. This is absurd, because $q \neq 0$.