# Comparing $\pi^e$ and $e^\pi$ without calculating them

How can I

compare (without calculator or similar device) the values of $$\pi^e$$ and $$e^\pi$$ ?

Another proof uses the fact that $$\displaystyle \pi \ne e$$ and that $$e^x > 1 + x$$ for $$x \ne 0$$.

We have $$e^{\pi/e -1} > \pi/e,$$

and so

$$e^{\pi/e} > \pi.$$

Thus,

$$e^{\pi} > \pi^e.$$

Note: This proof is not specific to $$\pi$$.

• This is from The Book! Impeccable proof. Commented Sep 14, 2013 at 22:20
• if x is negative than would your first equation which is e^x > 1 + x be true ? Commented Jan 15, 2015 at 17:36
• @MurtuzaVadharia: Yes, it is true. Consider $f(x) = e^x -1 -x$. It's derivative is $e^x -1$ which is $\lt 0$ for $x \lt 0$ and $\gt 0$ for $x \gt 0$, so $f$ decreases from $-\infty$ to $0$, and increases from $0$ to $\infty$. Since $f(0) = 0$... Commented Jan 21, 2015 at 8:11
• I think it's worth mentioning that this works for all positive numbers which aren't $e$. The proof has nothing to do with $\pi$. Commented Feb 7, 2015 at 20:48
• @David: Sorry to be of no help :(. No idea what the thought process was, it has been quite a while. Commented Aug 3, 2020 at 20:06

This is an old chestnut. As a hint, it's easier to consider the more general problem: for which positive $x$ is $e^x>x^e$?

• one more step, why not considering y^x and x^y Commented Nov 24, 2020 at 17:01

Alternatively, we can compare $e^{1/e}$ and $\pi^{1/\pi}$.

Let $f(x) = x^{1/x}$. Then $f'(x) = x^{1/x} (1 - \log(x))/x^2$. Since $\log(x) > 1$ for $x > e$, we see that $f'(x) < 0$ for $e < x < \pi$. We conclude that $\pi^{1/\pi} < e^{1/e}$, and so $\pi^e < e^\pi$.

The same calculation shows that $f(x)$ reaches its maximum at $e^{1/e}$, and so in general $x^e < e^x$.

• Correction: $\text{log}(x)<1$ for $x<e$. Commented Oct 26, 2010 at 15:39
• Your argument appears to need $x>e$, but it is easy to also include $x<e$ as well i the $x^e<e^x$ inequality. Commented May 22, 2013 at 12:43
• It doesn't really need $\log x > 1$. All you need is that the only solution of $\log x = 1$ is $x = e$. Commented May 22, 2013 at 14:29

From Proofs without Words.

• Which software did you use to plot this figure? Commented Aug 5, 2013 at 16:29

Let $f (x) =$ $x^\frac1x$

Find value of $x$ such that function gets maximum value

For this functions for $x=e$ function will get the maximum value

so $e^\frac1e$ is greater than $\pi^\frac1\pi$

so $e^\pi$ is greater than $\pi ^e$.

Elaborating Robin's answer take $f(x) = \log{x} - \frac{x}{e}$. We have $$f'(x)= \frac{e-x}{xe}$$ Thus $f'(x)>0$ for $0 < x < e$ and $f'(x) <0$ if $x > e$. Consequently, we have $f(x) < f(e)$ if $x \neq e$.

Exercise: Try to prove this using the same methods: $2^{\sqrt{2}} < e$.

Hint:

Prove that the function $f(x)=\frac{e^x}{x^e}, x\geq e$ is strictly increasing on the interval $x\in \left [ e,\pi \right ]$. What is $f(e)$ and $f(\pi)$?

Another visual proof. Recently published in arXiv

Yet another line of thought would be this: \begin{align} e^{\pi} > \pi^e &\iff \pi \ln(e)>e\ln(\pi)\\ &\iff \pi >e\ln(\pi) \\ &\iff \ln(\pi)>\ln(e)+\ln (\ln (\pi)) \\ &\iff \ln(\pi)>1+\ln (\ln (\pi)) \end{align} By concavity of $\ln$, $x-1>\ln(x)$ for all $x\neq 1$. With $x=\ln(\pi)$ we get the inequality wanted.

\begin{align} &e^\pi>\pi^e \\[5pt] \iff&\exp(\pi)>\exp(e\log\pi) \\[5pt] \iff&\pi>e\log\pi \\[5pt] \iff&\frac{\pi}{\log\pi}>e \\[5pt] \iff&\frac{\pi}{\log\pi}>\frac{e}{\log e} \end{align} The final line is true because the function $$\dfrac{x}{\log x}$$ is strictly increasing on $$[e,\infty) \, ,$$ and the result follows.

Denote $n = e^\pi$, $m = \pi^e$ and $s = \log \pi$. Then $\log n = \pi = e^s$ and $\log m = e \log \pi = es$. Then $$\log \frac {n} {m} = \log n - \log m = e (e^{s - 1} - s).$$ By Taylor expansion, we have $$e^{s - 1} = 1 + (s - 1) + \cdots > s.$$ Then $$\log \frac {n} {m} = e (e^{s - 1} - s) > 0.$$ Hence, $n > m$.

Just adding a recent one. Another visual proof. I am doubtful if it has open access.

Not that this question needs another answer, but here is a proof of $$e^\pi > \pi^e$$ using the Mean Value Theorem applied to $$\ln x$$ on the interval $$(e,\pi)$$, along with the assumptions that $$e<\pi$$, $$\ln$$ is increasing, and $$\frac{d}{dx} \ln x = \frac{1}{x}$$.

By the MVT, there exists $$c \in (e,\pi)$$ such that $$\frac{\ln \pi - \ln e}{\pi - e} = \frac{1}{c}$$ We can increase the right hand side by replacing $$c$$ with the smaller number $$e$$, and so we have $$\frac{\ln \pi - \ln e}{\pi - e} < \frac{1}{e}$$ and thus $${\ln \pi - \ln e} < \frac{1}{e}(\pi-e)$$ or $${\ln \pi - 1} < \frac{\pi}{e}-1$$ which gives $$e \ln \pi < \pi \ln e$$ and therefore $$e^\pi < \pi^e$$

Let

$$f(x) = e^x$$

$$G(x) = x^e$$

We can simply show that

$$f(e)=G(e)$$

$$f'(e)=G'(e)$$

For $$x > e$$ the $$f(x)$$ will grow faster than $$G(x)$$

Then

$$e^{\pi} > \pi^{e}$$

In fact it is true that $$e^x > x^e$$ for all $$x > e.$$ Here's a proof using the fundamental theorem of calculus.

Note that $$e^{x-2} > x^{e-2}$$ for all $$x > e.$$ This can be easily verified by noting that one of these is convex and the other is concave and they are equal at $$x=e$$ and evaluating the derivatives at this point.

Then, $$e^{x-1} - (e-1) x^{e-2} \geq e(e^{x-2} - x^{e-2}) > 0$$ for $$x > e.$$

But then let $$g(x) = e^{x-1} - x^{e-1}$$ and noting that $$g'(x) = e^{x-1} - (e-1) x^{e-2} > 0$$ we thus have for all $$x > e,$$ $$g(x) - g(e) = g(x) = \int_{e}^{x} [ e^{x-1} - (e-1) x^{e-2} ] dx > 0$$

And then letting $$f(x) = e^x - x^e$$ we have $$f'(x) = e^x - e x^{e-1} = e g(x) > 0$$

And by similar argument conclude that $$f(x) - f(e) = f(x) = \int_{e}^{x} e g(x) dx > 0$$ for all $$x \geq e.$$