Is the reasoning/algebra for my proof correct? (musical tuning theory proof) This isn't for a class, I was just wondering if I would be able to work out a proof for something like this myself for fun, and wanted to verify that my methods are correct. Basically, what I'm trying to prove, in terms of music theory is: Prove that it is impossible to stack a number of pure 5ths in just intonation, and to end up with a perfectly tuned octave, or multiple of octaves. Or more formally...
Let $ (\frac{3}{2})^m = (\frac{1}{2})^n $, where n and m are positive integers. Show that there exists no positive integers m and n such that the equation is true.
Assume the equation is true:
$ (\frac{3}{2})^m = (\frac{1}{2})^n $
$ (\frac{3}{2})^m = \frac{1}{2^n} $
$ (\frac{3}{2})^m = 2^{-n} $
$ \log{_2}[(\frac{3}{2})^m] = -n $
$ n = -\log{_2}[(\frac{3}{2})^m] $
$ n = -m\log{_2}[\frac{3}{2}] $
$ n \not\in Z^{+} $
$ \therefore $ by contradiction, there exists no n, m $\in Z^{+}$ such that $ (\frac{3}{2})^m = (\frac{1}{2})^n $.
Just for reference, showing a more elegant proof of the same thing would also be appreciated (maybe from using other proven theorems in mathematics). 

Yes, your proof is correct.
Here's another proof:
$$\left(\dfrac{3}{2}\right)^m=\left(\dfrac{1}{2}\right)^n\\
\implies 3^m=2^{m-n}\\
\implies m-n=m\log_23\not\in\mathbb{Z}^+\text{ as $m$ is an integer.}\\
\implies n=m(1-\log_23)\not\in\mathbb{Z}^+\text{ as $m$ is an integer.}\\
\implies \{n,m\}\not\subset\mathbb{Z}^+$$
Contradiction!
