Prove \(a^{\log_b{c}} = c^{\log_b{a}}\) where \(b \neq 1\).

To start we have:

\[a^{\log_bc}\]

Using the identity \(\log_b{a} = \frac{\log_c{a}}{\log_c{b}}\) we can obtain:

\[a^{\log_bc} = a^{\frac{\log_a{c}}{\log_a{b}}}\]

Next we can use routine exponential identities to rearrange our terms:

\[a^{\log_bc} = a^{\frac{\log_a{c}}{\log_a{b}}} = (a^{\log_a{c}})^{\frac{1}{\log_a{b}}}\]

Based on the identity \(a = b^{\log_b{a}}\):

\[a^{\log_bc} = a^{\frac{\log_a{c}}{\log_a{b}}} = (a^{\log_a{c}})^{\frac{1}{\log_a{b}}} = c^{\frac{1}{\log_a{b}}}\]

Lastly we use the identity \(\log_b{a} = \frac{1}{log_a{b}}\) to complete the proof:

\[a^{\log_bc} = a^{\frac{\log_a{c}}{\log_a{b}}} = (a^{\log_a{c}})^{\frac{1}{\log_a{b}}} = c^{\frac{1}{\log_a{b}}} = c^{\log_b{a}}\]