Is my method correct for to prove $a^{\log_b c} = c^{\log_b a}$? My problem is to prove this equality:
$$a^{\log_b c} = c^{\log_b a}$$
My method:

$$\begin{cases}
\log_b a=m\\
\log_b c=n\\
\end{cases} \Rightarrow \begin{cases}
a=b^m\\
c=b^n\\
\end{cases} 
 \Rightarrow 
\begin{cases}
a^n=(b^m)^n\\
c^m=(b^n)^m\\
\end{cases} \Rightarrow a^n=c^m \Rightarrow a^{\log_b c} = c^{\log_b a} $$

Is this method correct or is there a more elegant solution?
Thank you!
 A: Yes, it is correct. You can also consider
$$
\log_b(a^{\log_bc})=\log_bc\cdot\log_ba
=
\log_ba\cdot\log_bc=\log_b(c^{\log_ba})
$$
Since the function $\log_b$ is invertible, you conclude
$$
a^{\log_bc}=c^{\log_ba}
$$
A: I think your way is better than any "elegant" ways because elegant ways look slick but they rely are rote recitation early reliance upon which can hinder comprehension.
Slick:
$a^{\log_b c} = (b^{\log_b a})^{\log_b c} = (b^{\log_b c})^{\log_b a} = c^{\log_b a}$ 
but that... well, it looks like slick sliding symbols around.  It's great and elegant but...
Also if we take for granted that $b^x = b^y \iff x =y$ and that therefore that $x=y \iff \log_b x = \log_b y$  (because $x = b^{\log_b x}$ and $y = b^{\log_b y})$ then
$a^{\log_b c} = c^{\log_b a} \iff$
$\log_b a^{\log_b c} = \log_b c^{\log_b a} \iff$
$\log_b c*\log_b a = \log_b a*\log_b c$
... but that really feels like sliding symbols around.
(In fact, in typing it, I made a typo and got the obviously incorrect result $\log_b c*\log_a b = \log_b a* \log_b c$ and it took me seven minutes of proof-reading to find my error as I was lost in symbols rather than meaning.)
....
Third way.
$a^{\log_b c} = c^{\log_b a} \iff$
$\log_a a^{\log_b c} = \log_a c^{\log_b a} \iff$
$\log_b c = \log_b a *\log_a c \iff$
$b^{\log_b c} = b^{ \log_b a *\log_a c} \iff$
$c = (b^{ \log_b a})^{\log_a c} \iff $
$c = a^{\log_a c} \iff$
$c =c $.
[Note: Everything assume $b,a \ne 1$ and $a,b,c > 0$.]
