# Is there a proof that $n^xm^x = (n^x)^{(\log(mn)/\log(n))}$?

This isn't a homework question, just something I'm curious about, but you can treat it that way if you like.

So the other day I was playing with my calculator and I noticed that

$$2^x10^x = (2^x)^{(\log(20)/\log(2))}$$

I tried it out with some other numbers and came to the conclusion that

$$n^xm^x = (n^x)^{(\log(nm)/\log(n))}$$

So I wanted to see if there is a way to prove that.

I already know that $m = n^{(\log(m)/\log(n))}$ and I figured that there must be a relation. So from that I can see that $mn = n^{(\log(nm)/\log(n))}$. However I don't understand why that would mean that $(nm)^x = (n^x)^{\log(nm)/\log(n)}$.

Is what I say actually true? How do the powers fit into the proof?

It's true, and you're almost done. Probably recalling $$a^{x\cdot y} = (a^x)^y = (a^y)^x$$ is all you need.
$$n^x m^x=(n^x)^{\log mn/\log n}$$ will be true
$$\iff x\log n+x\log m=\frac{x(\log m+\log n)}{\log n}\log n$$ which is true
using $\displaystyle \log (a^x)=x\log a$ and $\displaystyle \log ab=\log a+\log b$