Proof of algebraic equation I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:
$$a^{\log_b c} = c^{\log_b a}$$
I understand that $a^{\log_a b} = b$ (and vice versa), but I must be missing something. 
 A: Two positive numbers are equal if and only if their logarithms (in the same base) are equal.
Compute the logarithm in base $b$ of both:


*

*$\log_b(a^{\log_bc})=\log_b c\log_b a$

*$\log_b(c^{\log_ba})=\dotsb$
Done.
A: First Prove and convince yourself that for $x> 0; y > 0; n > 0; n \ne 1$ then $x = y \iff \log_n x = \log_n y$.
Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $\log_n x = \log_n y$
And if $\log_n x = \log_n y = k$ then $n^k = x$... and $n^k = y$.  So $x = y$.
(For this to be acceptable it is ESSENTIAL that you  accept for all $n > 0$ and $n \ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$.   That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)
.....
Okays so $a^{\log_b c} = c^{\log_b a} \iff \log_b (a^{\log_b c})  = \log_b (c^{\log_b a})$.
And $\log_b  (a^{\log_b c}) = \log_b c\log_b a$ (because $\log_n a^m = m\log_n a$).  
And $\log_b (c^{\log_b a})=\log_b a\log_b c$
And that's that. 
....
Another way of thinking about it is:
$a^{\log_b c} = (b^{\log_b a})^{\log_b c} = b^{\log_b a\cdot \log_b c}=(b^{\log_b c})^{\log_b a} = c^{\log_b a}$.
A: $$a^{\log_bc}=a^{\frac{\log_ac}{\log_ab}}\\ = c^{\frac{1}{\log_ab}}\\=c^{\frac{1}{{\left(\frac{\log b}{\log a}\right)}}}\\=c^{\frac{\log a}{\log b}}\\=c^{\log_ba}$$
The key to the answer is based on the rule that $$\log_xy = \frac{\log_ky}{\log_kx}$$
This is the Change-of-Base Formula.
