Which Is Greater, a^b or b^a?

This post was translated from Korean into English by AI.

One day, out of the blue, I found myself wondering whether $e^\pi$ or $\pi^e$ was greater. Of course, it is easy to find out with a calculator, but I wanted to prove which one was greater without using one.

So, on a whim, I proved it.

Here is the proof.

Let us generalize the question and compare $a^b$ and $b^a$ for positive numbers $a$ and $b$.

$$a^b>b^a\\ \leftrightarrow b\log a > a \log b (\because x>y\leftrightarrow\log x> \log y)\\ \leftrightarrow \frac{b\log a}{ab}>\frac{a \log b}{ab} (\because ab>0)\\ \leftrightarrow \frac{\log a}{a}>\frac{\log b}{b}$$

Now, let

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

Then

$$f'(x)=\frac{1-\log x}{x^2}$$

• Therefore, if $x\geq e$, we can see that $f(x)$ is monotonically decreasing.
• Thus, at least when $a,b\geq e$, the ordering of $a^b$ and $b^a$ is the reverse of the ordering of $a$ and $b$.
• Since $e\geq e$, $\pi \geq e$, and $\pi>e$, it follows that $\pi^e<e^\pi$.

Calculating the values confirms this: $e^\pi\sim23.14$ and $\pi^e\sim22.45$, so $e^\pi$ is slightly greater.