# “Mine’s Bigger!”

Question: Which one is bigger, $e^\pi$ or $\pi^e?$

Consider function

$f(x) = \frac{\log(x)}{x}$.

We have

$f'(x) = \frac{1-\log(x)}{x^2} \implies f'(e)=\frac{1-\log(e)}{e^2} \overset{\log(e)=1}{=} 0\quad\quad\quad(1)$

and

$f''(x) = \frac{2\log(x)}{x^3}-\frac{3}{x^3}\implies f''(e) = \frac{2\log(e)}{e^3}-\frac{3}{e^3}= -\frac{1}{e^3}<0.\quad\quad\quad(2)$

From (1) and (2), we see that

$\frac{\log(x)}{x}$ attains its global maximum at $x=e$

which means

$\forall x >0, x \neq e \implies \frac{\log(x)}{x} < \frac{\log(e)}{e}$.

When $x=\pi$,

$\frac{\log(\pi)}{\pi} < \frac{\log(e)}{e}$

or,

$e\log(\pi) < \pi\log(e)$.

It follows that

$\log(\pi^e) < \log(e^\pi).\quad\quad\quad(3)$

Since $\log(x)$ is a monotonic increasing function (see Exercise-1), we deduce from (3) that

$e^\pi > \pi^e$

Exercise-1 Show that $\log(x)$ is a monotonic increasing function.