Is it true that if $\frac{\log m}{\log n}=\frac{\log 3}{\log 2},m,n\in \mathbb N$ then $m=3^k,n=2^k,k\in \mathbb N$?

Question

Is it true that if $\dfrac{\log m}{\log n}=\dfrac{\log 3}{\log 2},m,n\in \mathbb N$ then $m=3^k,n=2^k,k\in \mathbb N$?

Answer 1

Your equation can be rewritten to read $$ \log_nm=x=\log_23 $$ for some real number $x$. So $3=2^x$ and $m=n^x$. If we write $n=2^y$, it follows easily that $m=3^y$. This brings us to the real question of whether both $2^y$ and $3^y$ can be integers for the same choice of $y\notin\Bbb{N}$.

But according to the material covered in this thread that is an open question.