Logarithm proof problem: $a^{\log_b c} = c^{\log_b a}$ I have been hit with a homework problem that I just have no idea how to approach. Any help from you all is very much appreciated. Here is the problem
Prove the equation: $a^{\log_b c} = c^{\log_b a}$
Any ideas?
 A: If you apply the logarithm with base $a$ to both sides you obtain,
$$\log_a a^{\log_b c} = \log_a c^{\log_b a}$$
$$\log_b c = \log_b a \log_a c$$
$$\frac{\log_b c}{\log_b a} = \log_a c$$
however this last equality is the change of base formula and hence is true.  Reversing the steps leads to the desired equality.
A: $a^{\log_b c} =c^{\log_b a}$
let $k= a^{log_b c}$
$\log k=\log( a^{\log_b c})$
$\log k=\log_b c \log a$
$\log k=\frac{\log c}{\log b} \log a$
$\log k=\frac{\log a}{\log b} \log c$
$\log k=\log_b a \log c$
$\log k=\log c^{\log_b a}$
substituting value of k
$\log a^{\log_b c}=\log c^{\log_b a} $
therefore
$a^{\log_b c} =c^{\log_b a}$
A: $\large{a^{\log_b c}=e^{\ln a \cdot\log_b c}= e^{\ln a\cdot\ln c/\ln b}}=c^{\ln a/\ln b}=c^{\log_b a}$.
A: In log there is a property that :$x^{\large {\log _x  y}}=y$,$\log_x y=\log_w y\times\log_x w$
where $w$ can be any hold any possible value that is valid for a log base.
so $$a^{\large{\log_b c}}\implies a^{\large{\log_a c\times \log_b a}}$$
since $a^{\large{\log_a c}}=c$
so $$a^{\large{\log_a c\times \log_b a}}\implies c^{\large{\log_b a}}$$
