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If the function $f(x)=\frac{ln(x)}{x}$ where $x>0$, has a maximum at $(e,\frac{1}{e})$ compare the numbers $e^{\pi}$ and ${\pi}^e$.

***$ln(x)$ is the natural logarithmic function i.e the logarithmic function with base $e$.

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marked as duplicate by Dietrich Burde, Andrew D. Hwang, Claude Leibovici calculus Nov 2 '16 at 10:42

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Well, since $f$ has a max at $x = e$, then

$$ f( e) \geq f(x) $$

for all $x$ in the domain of $f$. In particular, if $x = \pi $, then

$$ f(e) > f( \pi ) \implies \frac{ \ln e }{e} > \frac{\ln \pi }{\pi} \implies \pi \ln e > e \ln \pi \implies \boxed{ e^{\pi} > \pi^e} $$

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