# How is $e$ the only number $n$ for which $n^x > x^n$ is satisfied for all values of $x$?

So I was watching this video and at 1:35, I found out that:

$$e^x > x^e,\quad{}x > 0$$

is a unique property of $e$. No other number does that.

It seems legit, and probably is, anyway. But I find it a bit weird because $\pi$ seems to fit in place of $e$ just fine. In fact, many numbers do. For example, $3^4 > 4^3$. But at some point, there is an exception like if the inequality is flipped, $4^3 < 3^4$ . Another one would be $2^5 < 5^2$. But I can't seem to be able to find out why ${\pi}$ seems to fit in too.

In fact, my understanding is that it indeed should. What makes $e$ special? It's pretty similar to $\pi$. Then why is $e$ supposed to be the only number with that property?

Also, I stumbled upon this answer according to which $e$ has another unique property:

$$e^x\ge x+1,\quad\text{for all }x$$

which again $\pi$ seems to fulfill too. So what exactly am I missing here?

• what is "all" $x$? Do you notice that $x^e$ is not necessarily real for $x<0$? – user251257 Dec 30 '16 at 19:06
• @user251257 I really don't know much about all this... Yeah the video says any positive number... My bad I missed it... hold on, I'll fix it. – Farhan Anam Dec 30 '16 at 19:08
• As a counterexample to your claim that this works for $\pi$, $\pi^3 \ngtr 3^{\pi}$. – DooplissForce Dec 30 '16 at 19:09
• @DooplissForce I see... exactly what I was looking for. Thanks. But what exactly makes $e$ special? Just lying between $2$ and $3$ shouldn't make it special, should it? And for that matter, $\pi$ is irrational too... just like $e$. – Farhan Anam Dec 30 '16 at 19:12

[Corrected]

Hint: can you show that the maximal value of $f(x) = \frac{\ln{x}}{x}$ on $(0, \infty)$ is attained for $x=e$?

Because $a^x \geqslant x^a$ for all $x>0$ $\Longleftrightarrow$ $x\ln{a} \geqslant a\ln{x}$ for all $x > 0$ $\Longleftrightarrow$ $\frac{\ln{a}}{a} \geqslant \frac{\ln{x}}{x}$ for all $x > 0$.

[There's a typo in the title and text: it should be $n^x \geqslant x^n$ not $n^x > x^n$.]

• I think you mean maximal instead of minimal. – JimmyK4542 Dec 31 '16 at 0:17
• @JimmyK4542 Of course! Thanks for the comment. – Catalin Zara Dec 31 '16 at 0:21

I stumbled upon this answer according to which $e$ has another unique property:

$e^x \ge x+1$, for all $x$

which again $\pi$ seems to fulfill too.

If you plot $\pi^x$ and $x+1$ together and look closely, you'll see that the two curves have two intersections. All exponentials $a^x$ go through point $(0,1)$, just like $x+1$. If you know a little calculus, you may then observe that the derivative of $a^x$ is $a^x \ln(a)$, which means that for $x=0$, if the exponential has to be tangent to $x+1$, $a$ must equal $e$.

If $a > e$, as in the case of $a=\pi$, there is going to be an intersection at $x=0$ and one for some negative value of $x$; if $a < e$, the second intersection will be for some positive value of $x$.

I don't intend this to be an outright answer, but simply a graphic to aid intuition. This is a 3D plot of the surface $z = x^y - y^x$. As noted we are only in the first quadrant since negative inputs yield not necessarily real $z$ values. We want to look at cross sections of the surface for fixed $x$ values and see which ones are always greater than 0. In this graphic the red curve is a parameterization of the cross section for $x = e$ and the blue plane is $z=0$. We see that the parameterized line at $x=e$ is the only one that will satisfy as moving left or right dips below the $z = 0$ plane.

Hope this visualization helps aid your intuition with more formal justifications! If $a$ is a number such that $$a^x \geq x^a \forall x >0$$ then setting $x=e$ we get $$a^e \geq e^a$$

But we know that $$e^a \geq a^e$$ with equality if and only if $a=e$.

This proves that $a=e$.

As for why does this happen: In many of these proofs we use at some point or another either that $(e^x)'=e^x$ or that $(\frac{d}{dx}e^x)_{x=0}=1$. These two properties are equivalent to the definition of $e$.