# How to arrange $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$ in the increasing order?

For these six numbers, $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$, how to arrange them in the increasing order?

This problem is taken from the today test: National Higher Education Entrance Examination.

I think we can consider $$f(x)=\dfrac{\ln{x}}{x}$$

or perhaps other methods, so I am looking forward to seeing other methods to solve this problem. Thank you.

• I don't understand what you are asking. What do you mean by size of the order? Do you mean you want to order the numbers by size?
– 5xum
Jun 7, 2014 at 10:34
• Jun 7, 2014 at 10:40
• @math110 Please check my editing. Is that what you mean? Jun 7, 2014 at 10:52
• @Turk-Fy,Thank you,That's your mean. Jun 7, 2014 at 10:55

We start by assuming that somehow we know that $$\color{green}{e < 3 < \pi}$$. This immediately gives the relations \begin{align*} \color{blue}{e^3 }&\color{blue}{< e^\pi < 3^\pi}\\ \color{blue}{3^e }&\color{blue}{< \pi ^ e < \pi^3} \end{align*} Next, we can consider $$f(x) = \frac {\ln x}x$$ as you stated.

Differentiating with respect to $$x$$ we have $$f'(x) = -\frac{\ln x}{x^2} +\frac 1 {x^2} = \frac{1-\ln x}{x^2} < 0\quad \forall x > e.$$

Therefore, for values of $$x$$ larger than $$e$$, this function is decreasing, which gives

\begin{align*} \frac{\ln 3}{3} &> \frac{\ln \pi}{\pi} &\implies\\ \pi \ln 3 &> 3 \ln \pi &\implies\\ \color{blue}{3^\pi} &\color{blue}{> \pi^3}, \end{align*} and similarly $$\color{blue}{e^3 > 3^e}$$ and $$\color{blue}{e^\pi > \pi^e}$$

The two inequalities we have left are $$e^\pi$$ vs. $$\pi^3$$ and $$\pi^e$$ vs. $$e^3$$. They're taking longer than I thought...

Edit, some years later. (I was browsing through MSE and came across this question again.)

Suppose we can prove that $$\color{green}{6 > e + \pi}$$. Then we can show the following:

\begin{align*} 6 &> e + \pi & \Leftrightarrow \\ 3 - e &> \pi - 3 & \Leftrightarrow \\ 3\cdot\left[1 - \frac{e}{3}\right] &> \pi - 3 & \Leftrightarrow \end{align*}

For all $$x > 0$$, we have $$\log x > 1 - \frac{1}{x}$$. From the preceding, we therefore see that $$3 \log(3/e) > 3\cdot\left[1 - \frac{e}{3}\right] > \pi - 3$$. We thus have

\begin{align*} 3 \log(3/e) &> \pi - 3 &\Leftrightarrow\\ 3 \log(3) &> \pi &\Leftrightarrow\\ 3 ^{3} &> e^{\pi}. \end{align*}

Since $$\pi^3 > 3^3$$, we conclude $$\pi^3 > e^{\pi}$$. To verify the initial inequality, we can pick our favourite upper bounds for $$e$$ and $$\pi$$. Archimedes' $$\pi < {22}/{7}$$ will do for $$\pi$$, while for $$e$$ we can use the fact that

$$e = 3 - \sum_{k=2}^{\infty} \frac{1}{k!(k-1)k} < 3 - \frac{1}{4}.$$

For the last inequality, we start by showing that $$\color{green}{\pi > \frac{e^2}{2e - 3}} \implies \color{blue}{\pi^e > e^3}$$ as long as $$\color{green}{2e > 3}$$:

\begin{align*} \pi &> \frac{e^2}{2e - 3} &\Leftrightarrow \\ \left(2e - 3\right) \pi &> e^2 &\Leftrightarrow \\ -\frac{e^2}{\pi} + e &> 3 - e &\Leftrightarrow\\ e\left(1 - \frac{e}{\pi}\right) &> 3 - e. \end{align*} We now use the same trick as before, that for all $$x$$, $$\log x > 1 - \frac{1}{x}$$ \begin{align*} e\left(1 - \frac{e}{\pi}\right) &> 3 - e &\Rightarrow \\ e\log {\pi}/{e} &> 3-e &\Leftrightarrow\\ e\log \pi &> 3 &\Leftrightarrow\\ \pi^e &> e^3. \end{align*}

To finish, we just need to show the initial inequality. The function $$f : x \mapsto \frac{x^2}{2x -3}$$ is decreasing for $$\frac{3}{2}, so we'll need a lower bound for $$e$$. From the series $$e = \sum_{k=0}^{\infty} \frac{1}{k!}$$ we see $$e > {8}/{3}$$. For $$\pi$$, we need a lower bound, and we can again channel Archimedes who proved $$\pi > \frac{223}{71}$$. We then have

\begin{align*} \pi > \frac{223}{71} &\operatorname{?} \frac{\frac{64}{9}}{\frac{16}{3} - 3} > \frac{e^2}{2e-3}\\ \pi > \frac{223}{71} &\operatorname{?} \frac{64}{21} > \frac{e^2}{2e-3}, \end{align*} and the conclusion follows. The final ordering of the six numbers is therefore $$\color{blue}{3^e< e^3 < \pi^e < e^\pi < \pi^3 < 3^\pi}$$.

As an aside, we've used a number of the properties of $$e$$ and $$\log$$ during the derivations, so an incorrect estimate for $$e$$ could easily throw the order off. However, we haven't used any of the specific properties of $$3$$ or $$\pi$$. Therefore, for any numbers $$x$$, $$y$$, such that

\begin{align*} e &< x < y,\\ \frac{e^2}{2e - y} &< x < 2 y - e,\\ y &< 2e, \end{align*}
the ordering $$x^e< e^x < y^e < e^y < y^x < x^y$$ will hold.

$\bf{My\; Solution\; for \; e^{\pi}\; and \; \pi^e}$

Using $\displaystyle e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+...............+\infty > 1+x\; \forall x>0$

So $e^x > 1+x$ for $x>0$

Now Put $\displaystyle x = \left(\frac{\pi}{e}-1\right)>0$

So $\displaystyle e^{\frac{\pi}{e}-1}>1+\frac{\pi}{e}-1\Rightarrow \frac{e^{\frac{\pi}{e}}}{e}>\frac{\pi}{e}\Rightarrow e^{\pi}>\pi^e$

• This can also be used $e^3$ and $3^e$, right? But, the others are more involved, looks like. Jun 9, 2014 at 4:13

Here's a fairly low-tech argument.

A quick summary is that three inequalities follow from $$u^v > v^u$$ for $$v > u \geqslant e,$$ and the two trickier ones that remain follow from $$e > 2\frac7{10}$$ and: $$\log3 > 1\frac1{12} > 4(2 - \sqrt3) > \frac\pi3 > \sqrt6 - \sqrt2 > 3^{\frac1{39}}.$$

If $$C$$ is a circle of unit radius, then a regular dodecagon inscribed in $$C$$ has perimeter $$6(\sqrt6 - \sqrt2),$$ and a regular dodecagon circumscribed about $$C$$ has area $$12(2 - \sqrt3),$$ therefore: $$$$\label{823770:eq:1}\tag{1} 3(\sqrt6 - \sqrt2) < \pi < 12(2 - \sqrt3).$$$$ Note the implication $$\pi > 3.$$ Of course, this could have been proved more simply by inscribing a regular hexagon, but we will need the better lower bound.

Because $$3\cdot48^2 = 6912 > 6889 = 83^2,$$ we have: $$\sqrt3 > \frac{83}{48} = 2 - \frac{13}{48},$$ and therefore: $$\pi < 3\frac14.$$

If $$|x| < 1,$$ we have: $$$$\label{823770:eq:2}\tag{2} \log\frac{1 + x}{1 - x} = 2\left(x + \frac{x^3}3 + \frac{x^5}5 + \frac{x^7}7 + \frac{x^9}9 + \cdots\right).$$$$ Taking $$x = \frac12,$$ and using only the first two terms of \eqref{823770:eq:2}, we get: $$\log3 > 1\frac1{12}.$$ (Note the implication $$e < 3.$$) Combining this with the upper bound for $$\pi$$ gives: $$3\log3 > 3\frac14 > \pi,$$ whence: $$\boxed{e^\pi < 27 < \pi^3.}$$

The lower bound for $$\log3$$ gives us an upper bound for $$e$$: $$e < 3^{\frac{12}{13}}.$$ In addition to that, we will need this lower bound: $$e > 1 + 1 + \frac12 + \frac16 + \frac1{24} = 2\frac{17}{24} > 2\frac7{10}.$$

By the lower bound in \eqref{823770:eq:1}: $$\pi^3 > 27(12\sqrt6 - 20\sqrt2).$$ Because $$6\cdot24^2 = 3456 < 3481 = 59^2,$$ and $$2\cdot40^2 = 3200 < 3247 = 57^2,$$ we have: $$12\sqrt6 < \frac{59}2 \text{ and } 20\sqrt2 < \frac{57}2,$$ and therefore: $$12\sqrt6 - 20\sqrt2 = \frac{64}{12\sqrt6 + 20\sqrt2} > \frac{64}{58} = 1\frac3{29} > 1.1.$$ But: $$\begin{gather*} 1.1^3 = 1.331 > 1.33, \\ 1.33^2 = 1.7689 > \frac74, \\ \left(\frac74\right)^2 = \frac{49}{16} > 3, \end{gather*}$$ therefore $$1.1^{12} > 3,$$ which gives us the estimate we need: $$\left(\frac\pi3\right)^3 > 3^{\frac1{12}} > 3^{\frac1{13}}.$$ We now have: $$e^{\frac{10}3} < 3^{\frac{40}{13}} = 27\cdot3^{\frac1{13}} < \pi^3,$$ whence: $$\boxed{e^3 < \pi^{\frac{27}{10}} < \pi^e.}$$

Finally, if $$1 \leqslant x < y,$$ then: $$\begin{gather*} e^{y - x} > 1 + (y - x) \geqslant 1 + \frac{y - x}x = \frac{y}x, \\ \therefore\ \frac{x}{e^x} > \frac{y}{e^y} \quad (1 \leqslant x < y). \end{gather*}$$ Writing $$x = \log u,$$ $$y = \log v$$ ($$e \leqslant u < v$$), we get: $$\frac{\log u}u > \frac{\log v}v \quad (e \leqslant u < v).$$ Equivalently: $$u^v > v^u \quad (e \leqslant u < v).$$ Taking in turn: $$(u, v) = (e, 3), \, (e, \pi), \, (3, \pi),$$ we get: $$e^3 > 3^e, \ e^\pi > \pi^e, \ 3^\pi > \pi^3.$$ In conjunction with the earlier inequalities $$e^\pi < \pi^3$$ and $$e^3 < \pi^e,$$ this gives: $$\boxed{3^e < e^3 < \pi^e < e^\pi < \pi^3 < 3^\pi.}$$

• In fact, without any more work: $$e^e < 3^e < e^3 < \pi^e < e^\pi < 3^3 < \pi^3 < 3^\pi < \pi^\pi.$$ Mar 7, 2020 at 15:00
• With hindsight, the slightly irksome calculation of $1.33^2$ can be replaced by: $$\left(\frac43 - \frac1{300}\right)^2 > \frac{16}9 - \frac2{225} > \frac{16}9 - \frac1{36} = \frac74.$$ Mar 7, 2020 at 15:49