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

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.

• Mar 22 '13 at 11:20
• Ah, I see this has been asked before (I didn't find it when I searched initially).
– ivan
Mar 22 '13 at 12:31
• It is a bit hard to search for, I agree, but it is kind of a classic. I would have been surprised if it had never been asked before on this website ^^ Mar 22 '13 at 12:38
• @ivan: have you had calculus?
– robjohn
Mar 22 '13 at 18:33
• @robjohn: yeah, but it was quite a while ago. I started getting into statistics, and found I needed a calc review. I started reviewing calc and found I could really use some precalc review. Etc... :)
– ivan
Mar 22 '13 at 19:07

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

• Wow! Very nice. +1
– A.S
Mar 22 '13 at 4:45
• He's working from a college algebra textbook, there is no reason to suspect that he even knows what differentiation is. Mar 22 '13 at 4:58
• @deftfyodor Why not? We cover differentiation in grade 10 here. At college with those types of bonus questions, you would assume that differentiation would have been taught already. Mar 22 '13 at 5:36
• I have taken calculus, but I'm reviewing algebra in order to shore up the fundamentals before reviewing calculus. The author keeps stopping short of using calc, so I'm pretty sure there's another way to solve this. I like your solution though.
– ivan
Mar 22 '13 at 12:36
• @user1887919: the derivative of $x^\frac{1}{x}$ is $> 0$ for $x < e$ and $< 0$ for $x > e$. So the global maximum at $x=e$ is unique.
– yori
Mar 8 '20 at 14:53

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 $$e^{\frac{\pi}{e}-1}>1+\frac{\pi}{e}-1\Leftrightarrow \frac{e^{\frac{\pi}{e}}}{e}>\frac{\pi}{e}\Leftrightarrow e^{\frac{\pi}{e}}>\pi\Leftrightarrow e^{\pi}>\pi^e$$

• I would have never thought $e^x>1+x$ was enough, +1. Mar 22 '13 at 4:44
• And why the downvote? Mar 22 '13 at 4:54
• Wow, I like. I'm not familiar with Taylor expansion, but that's awesome!
– ivan
Mar 22 '13 at 12:39
• @ivan If you really want to, you can prove that $e^x>1+x \, \forall x>0$ without using differentiation or Taylor series. Mar 22 '13 at 13:29
• Indeed I found this proof even more interesting than the others.
– Tapu
Mar 23 '13 at 6:17

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 $$f(\pi) < f(e) \implies \dfrac{\log(\pi)}{\pi} < \dfrac{\log(e)}{e} \implies \pi^e < e^{\pi} \implies \pi^{1/\pi} < e^{1/e}$$

• In fact, $e^x\gt x^e$ for all $x\ge0$ and $x\ne e$ (+1)
– robjohn
Mar 22 '13 at 17:19

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

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

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 $e^\pi > \pi^e$.

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!)

• You can get that $\pi>e\ln(\pi)$ without using a calculator by showing that $x/e\geq\ln(x)\forall x$, which is straightforward when looking at the derivatives.
– Jens
Mar 22 '13 at 9:46
• Yes, this is kind of a non-answer at the moment. You still need to find a method to show that $\pi > e \ln{(\pi)}$, as Jen proposes above, simply saying it is so is equivalent to saying that $e^{\pi} > \pi^e$ is so. Unless this is somehow an obvious fact that everyone just knows, but I'm not seeing it. Mar 22 '13 at 12:43
• @Thomas - Yes, I know there are ways suggested that work, but the OP indicated that this problem was an exercise in a College Algebra textbook, and was perplexed as to how this can be shown using material one would have encountered at that point, which certainly precedes calculus and derivatives. Mar 22 '13 at 13:03