How to prove $\log_23$ is irrational?

Question

I think using contradiction is good.

Assume $\log_23$ is rational

Then $\exists p\in \Bbb{Z}, q\in \Bbb{Z}^*: \log_23 = \frac{p}{q}$ ###$p, q$ has no common factors.

Then $3^{q}=2^{p}$

...

Here I failed to continue. Could someone give any advice?

Answer 1

$$ \log_2 3 = \frac p q $$ $$ 2^{p/q} = 3 $$ $$ 2^p = 3^q $$ $$ \text{An even number} = \text{an odd number}. $$

(Here we rely on the fact that this number is positive, so $p$ and $q$ are either both positive or both negative, and in the latter case, we just multiply them both by $-1$ and get positive numbers. Thus $2^p$ and $3^q$ are integers.)