Originally Posted by

**Soroban** Hello, jimbutler83!

You're on the right track . . .

$\displaystyle \text{Let: }\:X^{\log(Y)} \:=\:P$ .[1]

Take logs: .$\displaystyle \log\left(X^{\log(Y)}\right) \:=\:\log(P) \quad\Rightarrow\quad \log(Y)\!\cdot\!\log(X) \:=\:\log(P)$

We have: .$\displaystyle \log(X)\!\cdot\!\log(Y) \:=\:\log(P) \quad\Rightarrow\quad \log\left(Y^{\log(X)}\right) \:=\:\log(P)$

Hence: .$\displaystyle Y^{\log(X)} \:=\:P$ .[2]

Equate [1] and [2]: . $\displaystyle X^{\log(Y)} \;=\;Y^{\log(X)} $