# e^pi > pi^e

Let’s prove that $e^{\pi} > \pi^e$ without a calculator. If you haven’t seen this before, give it a try before you read any further.

Consider the function $f(x) = x/\ln(x)$ on the interval $(1,\infty)$. This function shoots off to positive infinity as $x$ tends towards either endpoint of the domain. Let’s find its minimum value via an application of good clean differential calculus.

Differentiating (remember the quotient rule?) with respect to $x$ gives us that

$\displaystyle f'(x) = \frac{\ln x - 1}{(\ln x)^2}.$

Then, we have that $f'(x) = 0$ when $\ln x = 1$, so clearly the stationary point of this function occurs at $x = e$. It is also easy to check that this is a minimum using your favourite high-school test.

Therefore, as the minimum value of $f(x) = x/ \ln(x)$ occurs when $x=e$, then if we choose any number $a>1$ which is not equal to $e$ we have

$\displaystyle \frac{a}{\ln a} > \frac{e}{\ln e}.$

For instance, we can choose $a =\pi$. Using the fact that $\ln(e)=1$, this gives us that

$\displaystyle \frac{\pi}{\ln \pi} > e.$

This inequality can then be manipulated to get that $e^{\pi} > \pi^e$. The sharp reader will notice that any real number greater than 1 and not equal to $e$ can replace $\pi$ in the inequality and it shall still be true!

1. Mom. says:

I disagree.