Is it correct? n^(log x)=x^(log n) Can you proof and describe that, for any base? Please explain completely. Thank you.

Is it correct? ${n}^{\left(\mathrm{log}\phantom{\rule{thinmathspace}{0ex}}x\right)}={x}^{\left(\mathrm{log}\phantom{\rule{thinmathspace}{0ex}}n\right)}$?
Can you proof and describe that, for any base? Please explain completely. Thank you.
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Danielle Gilbert
${n}^{\left(\mathrm{log}x\right)}=\left({e}^{\mathrm{log}n}{\right)}^{\left(\mathrm{log}x\right)}={e}^{\left(\mathrm{log}n\right)\left(\mathrm{log}x\right)}={e}^{\left(\mathrm{log}x\right)\left(\mathrm{log}n\right)}=\left({e}^{\mathrm{log}x}{\right)}^{\left(\mathrm{log}n\right)}={x}^{\left(\mathrm{log}n\right)}$
If you use $\mathrm{log}$ to a different base $b$, then use $b$ instead of $e$
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rialsv
Hint: compute the logarithm of both sides. Since the logarithm function is one-to-one this will tell you if they are equal.