Which is bigger, $e^{\pi}$ or $\pi^{e}$?

How to work it out

1. 1

   Take the natural log of both numbers, since ln is increasing: compare $\pi \ln e = \pi$ with $e \ln \pi$.

   $$e^{\pi} \text{ vs } \pi^{e} \iff \pi \text{ vs } e\ln\pi \iff \frac{\ln e}{e} \text{ vs } \frac{\ln \pi}{\pi}$$
2. 2

   The function $f(x) = \frac{\ln x}{x}$ has its maximum at $x = e$, because its derivative $\frac{1 - \ln x}{x^{2}}$ is zero there and changes from positive to negative.

3. 3

   So $f(e) > f(\pi)$, which means $\frac{\ln e}{e} > \frac{\ln \pi}{\pi}$, and therefore $e^{\pi}$ > $\pi^{e}$. The same argument shows $a^{b} > b^{a}$ whenever $e \le a < b$.

The two numbers (computed)

Questions

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