Daily Archives: September 2, 2020

“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)


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}


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.