# How do we prove that $a^{\log{b}} = b^{\log{a}}$ for $a > 1$ and $b > 1$?

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?

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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*}

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Let $x=a^{\log b}$ and $y=b^{\log a}$. Then $\log x= \log a$. $\log b= \log y$. So, $x=y$.

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Let (capital) $B$ be the base of the logarithms. Then $$a^{\log_B b} = \Big(B^{\log_B a}\Big)^{\log_B b} = B^{(\log_B a)(\log_B b)},$$ and that is clearly symmetric in $a$ and $b$. Or you could just go on from there: $$\cdots= \Big(B^{\log_B b}\Big)^{\log_B a} = b^{\log_B a}.$$

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