# compare the numbers $e^\pi$ and ${\pi}^e$ [duplicate]

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 $e$.

## marked as duplicate by Dietrich Burde, Andrew D. Hwang, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 2 '16 at 10:42

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