A question comparing $\pi^e$ to $e^\pi$ [duplicate]

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I was doing an algebra problem set following a chapter on logarithms and exponentiation, and it presented this "bonus question":

Without using your calculator, determine which is larger: $e^\pi$ or $\pi^e$.

I wasn't able to come up with anything, and I'm just curious how you might tackle this, keeping in mind it came out of a college algebra textbook, less than halfway through, so I don't imagine the author had any super-advanced tactics in mind.

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6 Answers

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Consider the function $x^{\frac{1}{x}}$. Differentiating gives $x^{\frac{1}{x}}(\frac{1}{x^2})(1-\ln x)$, so the function attains its global maximum at $x=e$.

Thus $e^{\frac{1}{e}} \geq \pi^{\frac{1}{\pi}}$, and it is clear that the inequality is strict, so $e^{\pi}>\pi^{e}$.

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If you know Taylor expansion: then $$e^x=1+x+\frac{x^2}{2!}+...$$ We can get (Or you may take derivative to prove it) $$e^{x}>1+x, \forall x>0$$ Then set $$x=\frac{\pi}{e}-1>0$$ We get $$e^{\frac{\pi}{e}-1}>1+\frac{\pi}{e}-1\Leftrightarrow \frac{e^{\frac{\pi}{e}}}{e}>\frac{\pi}{e}\Leftrightarrow e^{\frac{\pi}{e}}>\pi\Leftrightarrow e^{\pi}>\pi^e$$

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Note that if $f(x) = \dfrac{\log(x)}x$, we have $$f'(x) = \dfrac{x \times \dfrac1x - \log(x)}{x^2} \begin{cases} < 0 & \forall x > e\\ > 0& \forall x < e\end{cases} $$ Hence, $f(x)$ is a strictly montone decreasing function for $x \geq e$ and strictly montone increasing function for $x \leq e$. Since $\pi > 3 > e$, we have $$f(\pi) < f(e) \implies \dfrac{\log(\pi)}{\pi} < \dfrac{\log(e)}{e} \implies \pi^e < e^{\pi} \implies \pi^{1/\pi} < e^{1/e}$$

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So, we want to prove $e^\pi>\pi^e$. Taking $\log$ (where $\log$ means $\ln$) of both sides tells us this is equivalent to showing that

$$\pi\log(e)>e\log(\pi)$$

or

$$\frac{\log(e)}{e}-\frac{\log(\pi)}{\pi}>0$$

But,

$$\frac{\log(e)}{e}-\frac{\log(\pi)}{\pi}=-\int_{e}^{\pi}\frac{d}{dx}\left(\frac{\log(x)}{x}\right) dx$$

but,

$$-\frac{d}{dx}\left(\frac{\log(x)}{x}\right)=\frac{\log(x)-1}{x^2}$$

Thus, we see that

$$\frac{\log(e)}{e}-\frac{\log(\pi)}{\pi}=\int_e^\pi \frac{\log(x)-1}{x^2} dx\,$$

Since, for $x\in(e,\pi)$ we have that

$$\frac{\log(x)-1}{x^2}>0$$

it follows that

$$0<\int_e^\pi \frac{\log(x)-1}{x^2}dx=\frac{\log(e)}{e}-\frac{\log(\pi)}{\pi}$$

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Another way of doing it:

Since $\ln x$ is monotonically increasing for $x>0$, by taking the log we see that $e^\pi > \pi^e$ iff $\pi > e\ln\pi$. Taking the natural log again we see this is true iff $\ln\pi > 1 + \ln\ln\pi$.

To show this, consider the function $f(x) = \ln x - 1 - \ln\ln x$. Then $f'(x) = \frac1x - \frac{1}{x\ln x}$. This derivative is continuous for $x>0$ and has one zero at $x = e$. It is easy to see that $f' < 0 $ for $x < e$ and $f' > 0$ for $x > e$, so $x = e$ is a minimum.

Therefore $f(\pi)$ > $f(e) = 0$, which means $\ln\pi > 1 + \ln\ln\pi$, which by the above iffs means $e^\pi > \pi^e$.

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This may not help much, but if you know your $\ln$-values well, and/or have encountered $\ln(\pi)$:

Note that $\ln(e^\pi) = \pi$ and $\ln(\pi^e) = e\ln(\pi)$, and we have that $\pi > e\ln(\pi)$.

Hence $\;e^\pi > \pi^e$.

But admittedly, the inequality isn't immediately obvious! (which can be seen if you do approximate with a calculator!)

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