Is my proof that $\log_23$ is transcendental correct?

Could anyone please check my following proof that $\log_23$ is transcendental? If it is not correct, could you note how to prove that $\log_23$ is transcendental in another way?

Theorem: $\log_23$ is transcendental.

Lemma $1$: $\log_23$ is irrational.

Prove the lemma by contradiction:

Suppose $\log_23$ is rational. Then it can be written as a ratio of two positive integers $p, q$: $\log_23 = \frac{p}{q}$ (clearly $\log_23$ is positive). Then, from the definition of logarithms, $2^{(p/q)}=3$. Now, if we raise both sides of the equation to the power of $q$, we get $2^p = 3^q$. But $2$ to the power of any positive integer is an even number and $3$ to the power of any positive integer is an odd number, so $2^p = 3^q$ has no solution for positive integers $p, q$ (contradiction). So $\log_23$ cannot be rational, it must be irrational. This proves the lemma.

We will now prove the theorem that $\log_23$ is transcendental by contradiction. Suppose $\log_23$ is algebraic. We already proved it is irrational, so we suppose it is algebraic irrational. Then, via Gelfond-Schneider we know that if $a$ is a positive integer larger than $1$ and $b$ is algebraic irrational, then $a^b$ is transcendental. So if $\log_23$ is algebraic irrational, then any positive integer larger than $1$ raised to the power of $\log_23$ is transcendental. But from the definition of logarithms, $2^{\log_23} = 3$ which obviously isn't transcendental (contradiction). So $\log_23$ cannot be algebraic, it must be transcendental.

Solution 1:

You've successfully proved the the Lemma, and then the Theorem. And should feel proud that you succeeded.

I post this "community post" in part, so that an answer exists, and is posted objectively, (no one has anything to gain), hence removing one potential "unanswered" questions.