Observe the following equation:
$$\log_2(m)=\log_3(n).$$
It's clear that there are no integer solution other that $(m, n)=(2^k, 3^k)$ for some integer $k\ge 0$, however, it is not clear to me how to prove this.
Other similar diophantine equations such as $\sqrt{m}=\sqrt[3]{n}$ are trivial to solve by algebraic manipulation (raising everything to the sixth power), which makes me think that the first equation should be easy to solve too. Similar ideas applied to the first equation don't seem to lead anywhere.
I'm unable to prove something trivial such as $\log_2(3)\neq\log_5(7)$. Of course, one can estimate the left-hand-side and right-hand-side to see that they must differ by some small amount. What I'm looking for is a general procedure for proving that inequality with any four different positive integers (for simplicity, we can look at only different primes).
Is there an elementary proof of this fact?
