Logarithms – How to Prove $a^{\log{b}} = b^{\log{a}}$ for $a > 1$ and $b > 1$

Question

I have tried using the change of base formula, but can't quite complete the equality:

$$
a^{\log{b}} \\
a^{\frac{\log_a{b}}{\log_a{a}}}
$$

How do I get the base of the exponent to be b?

Answer 1

Since $x = e^{\log x}$

\begin{align*} a^{\log b} = \left(e^{\log a}\right) ^{\log b} = e^{\log a \log b} = e^{\log b \log a} = \left(e^{\log b}\right)^{\log a} = b^{\log a} \end{align*}