Existence of Rational Powers of Irrational Numbers/Proof 3

Theorem
There exist irrational numbers $a$ and $b$ such that $a^b$ is rational.

Proof
Consider the equation:
 * $\left({\sqrt 2}\right)^{\log_{\sqrt 2} 3} = 3$

We have that $\sqrt 2$ is irrational and $3$ is (trivially) rational.

It remains to be proved $\log_{\sqrt 2} 3$ is irrational.

We have:

Assume, for the sake of establishing a contradiction, that $2 \, \log_2 3 = \dfrac m n$ for some positive integers $m$ and $n$.

Then:

But $3^{2n} = 2^m$ would imply that there exists a number whose prime factorization is not unique.

This contradicts the Fundamental Theorem of Arithmetic.

Hence the assumption that $2 \, \log_2 3 = \dfrac m n$ must be incorrect.

That is, $\log_{\sqrt 2} 3 = 2 \, \log_2 3$ is irrational.

Thus there exist two irrational numbers $a = \sqrt 2$ and $b = \log_{\sqrt 2} 3$ such that $a^b$ is rational.