# Logarithm formula proof Prove: x^(log(y))=y^(log(x))

Logarithm formula proof
Prove:
${x}^{\mathrm{log}\left(y\right)}={y}^{\mathrm{log}\left(x\right)}$
I have been trying this for the past 1 hour, still cant prove it.
I started with
${\mathrm{log}}_{b}\left(y\right)=m$
${\mathrm{log}}_{b}\left(x\right)=n$
To show:
${x}^{m}={y}^{n}$
How do i proceed?
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$\mathrm{log}\left(x\right)\mathrm{log}\left(y\right)=\mathrm{log}\left(x\right)\mathrm{log}\left(y\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{e}^{\mathrm{log}\left(x\right)\mathrm{log}\left(y\right)}={e}^{\mathrm{log}\left(x\right)\mathrm{log}\left(y\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\left({e}^{\mathrm{log}x}\right)}^{\mathrm{log}y}={\left({e}^{\mathrm{log}y}\right)}^{\mathrm{log}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}^{\mathrm{log}y}={y}^{\mathrm{log}x}$
Identities used:
${e}^{\mathrm{log}x}=x$
$\left({a}^{m}{\right)}^{n}=\left({a}^{n}{\right)}^{m}={a}^{mn}$
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benatudq
Since we can take logs on both sides because taking the log preserves the equality i.e.
$\mathrm{log}a=\mathrm{log}b\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}a=b$
in short, we can still prove the statement if are working the logarithm of the equality.
Anyway,
$\mathrm{log}\left({x}^{\mathrm{log}y}\right)=\mathrm{log}\left(y\right)\mathrm{log}\left(x\right)$
similarly
$\mathrm{log}\left({y}^{\mathrm{log}x}\right)=\mathrm{log}\left(x\right)\mathrm{log}\left(y\right)$
here I used $\mathrm{log}{a}^{b}=b\mathrm{log}a$.