Proof of logarithm power change I am not too sure how to explain this in words. So the question is proofing that 
$a ^{\log_bc} = c ^{\log_b a}$
So far what I have done was:

I cannot think of anything else, I mean if I do the same for $c ^{\log_b a}$, I would be getting $\log_b a^c$ which I believe is not enough to proof. What steps am I missing here?
 A: Recall how logarithms and exponents are inverse operations:

$$ b^{\log_b x} = x \tag{1}$$

Furthermore, recall how powers interact with logarithms:

$$ \log_bx^k = k\log_b x \tag{2}$$

With that in mind, observe that:
$$ \begin{align*}
a^{\log_b c} &= b^{\log_b (a^{\log_b c})} & \text{by (1), where } x=a^{\log_b c}\\
&= b^{(\log_b c)(\log_b a)} & \text{by (2), where }x=a \text{ and }k=\log_b c\\
&= b^{(\log_b a)(\log_b c)} \\
&= b^{\log_b (c^{\log_b a})} & \text{by (2), where }x=c \text{ and }k=\log_b a\\
&= c^{\log_b a} & \text{by (1), where }x=c^{\log_b a}\\
\end{align*} $$
as desired.
A: You must rely on the logarithm properties:
$$\forall\,a,b,x,y>0\;,\;\;a,b\ne 1:\;\;\color{red}{\log_ax=\frac{\log_bx}{\log_ba}\;,\;\;x^y=a^{y\log_ax}}$$
So
$$\begin{align*}a^{\log_bc}&=\underbrace{b^{\log_bc\log_ba}}\\
c^{\log_ba}&=\overbrace{b^{\log_ba\log_bc}}^{||}\end{align*}$$
and we're done...
A: I would take logarithms to base $b$ straight away
$$log_b{(a ^{\log_bc})}=\log_b{c}\log_b{a}$$ $$\log_b{(c^{\log_b a})}=\log_b a\log_b c$$
