Show that $\frac{\log 2}{\log 3}$ is irrational. I found this exercise in Beachy and Blair:Abstract algebra. 
Show that
$$
\frac{\log 2}{\log 3}
$$
is not a rational number.
Here is what I did:
Assume that it is rational so
$$
\frac{\log 2}{\log 3}=\frac{a}{b}, \qquad a,b\in\mathbb{Z}, b\neq 0
$$
Then we have the following 
$$
\frac{\log 2}{\log 3}=\frac{a}{b} \iff
$$
$$
b\cdot \log 2=a\cdot \log 3 \iff
$$
$$
\log 2^b=\log 3^a \iff
$$
$$
2^b=3^a
$$
which is trivially false and hence the result follows.
Now my problem is that all the previous exercises are a lot "harder" and this seemed really easy. Am I correct here or missing something? I can't see a problem in my solution. 
 A: Of course the last equality has a solution when $a = b = 0$, but that would have no significance in the first step, hence we don't care about that particular solution.
Other than that, I see no fault in your reasoning. But that is just me, there is a chance that I am making a mistake as well :P
A: By the base change rule (applied in reverse),
$$\frac{\log 2}{\log 3} = \log_3 2$$
Setting that to $\displaystyle \frac ab$, you can get where you ended up a little quicker. But there is nothing wrong with your approach.
(To add on, you might want to add a brief explanation why the last step is "trivially false" - use uniqueness of prime factorisation).
A: It is not the point of the question, but it is not difficult to prove that $\alpha=\frac{\log 2}{\log 3}$ is trascendental, once $\alpha\not\in\mathbb{Q}$ has been proved. If we assume that $\alpha$ is algebraic over $\mathbb{Q}$ from the Gelfond-Schneider theorem we get that $3^\alpha$ is trascendental. But $3^\alpha=2\in\mathbb{Q}$, hence $\alpha$ is trascendental.
