proving $ e^{\pi} > \pi ^{e}$

Question

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

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

Srijan · Accepted Answer · 2013-04-20T17:58:19.427

4

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?

edited Apr 20 '13 at 17:58

answered Apr 20 '13 at 17:40

Srijan

score 3 · Answer 2 · answered Apr 20 '13 at 17:31

3

Look at the function $$f(x)=\frac{\log x}x$$

You're looking at $x^y<y^x$, or equivalently $f(x)<f(y)$.

answered Apr 20 '13 at 17:31

Pedro

2 Answers2