# proving $e^{\pi} > \pi ^{e}$ [duplicate]

I want to show that $e^{\pi} > \pi ^{e}$?

I was trying to make some functional relations to verify this but I am not able to do so . Any help or hints will be helpful for me.

Thanks

## marked as duplicate by Micah, Namaste, Belgi, Git Gud, Andrés E. CaicedoApr 20 '13 at 17:42

• I think this was asked here before. – Pedro Tamaroff Apr 20 '13 at 17:31
• Take logarithms on both the side and see what you got? – srijan Apr 20 '13 at 17:31

Let $a = e^{\pi}$ and $b = \pi ^{e}$.
Taking logarithms we obtain $a > b$ iff $\frac {\log e}{e} > \frac {\log \pi}{\pi}$
Now consider the function $f(x) = \frac{\ln x}{x}$ and check when $f$ is decreasing?
Look at the function $$f(x)=\frac{\log x}x$$
You're looking at $x^y<y^x$, or equivalently $f(x)<f(y)$.