Prove that $\log_2 3$ is irrational!

Question

So, I have been told that for every $x\in\mathbb{N}$, $\ln{x}$ is irrational by using the fact that $e$ is transcendental number.

But, how to prove that $\log_{2}{3}$ is irrational number? My idea is to rewrite the form

$$\log_2 3 = \frac{\ln 3}{\ln 2}$$

But, if both $x, y$ irrational, it isn't necessary that $\frac x y$ irrational. Please, help me!

Answer 1

Suppose $\log_2 3 = \dfrac m n$ for some $m,n\in\{1,2,3,\ldots\}.$

Then $2^{m/n} = 3.$

Therefore $2^m = 3^n.$

Therefore an even number equals an odd number.

Answer 2

More generally the logs of primes $\rm\,p_i\,$ are linearly independent over $\,\Bbb Q\,$ since if

$$\rm\ \: c_1 \log\,2 \, +\, c_2 \log\, 3\, + \cdots +\, c_n\log\,p_n =\ 0,\ \: c_i\in\mathbb Q\:,\: $$

then multiplying by a common denominator we can assume all $\rm\ c_i \in \mathbb Z,\:$ so, exponentiating $$\rm\: 2^{\,\large c_1} 3^{\,\large c_2}\cdots p_n^{\,\large c_n} = 1\ \Rightarrow\ c_i = 0\ \text{ for all }\rm\:i$$

OP is the special case $\rm\, n = 2\ $ since $\rm\,c = \log_2 3 = \log 3/\log 2\,\Rightarrow\, c\log 2 - \log 3 = 0$