# $e$ is the only number in the universe having this property?

I read that $e$ is the only number in the universe which, if used as a base raised to any power has the highest value compared to when the base and power are exchanged i.e., $$e^p > p^e,\forall p\neq e$$ Is it a true statement? Is there any proof available for this statement or has someone already proved it in this forum? Please provide the link .

Where can one find all the wonderful and unique properties of $e$

• Note that there could not be two distinct numbers with this property (consider $a^b$ and $b^a$ for the two numbers concerned - which would be larger?) Apr 8 '17 at 13:26

Claim : Whenever , we have real numbers $a,b$ with $e\le a<b$, we have $b^a<a^b$

Proof : for real numbers $a,b> 1$ , the condition $$b^a<a^b$$ is equivalent to $$a\ln(b)<b\ln(a)$$ and therefore equivalent to $$\frac{a}{\ln(a)}<\frac{b}{\ln(b)}$$ But $f(x)=\frac{x}{\ln(x)}$ has derivate $f'(x)=\frac{\ln(x)-1}{(\ln(x))^2}$, which is positive for $x>e$ and $0$ for $x=e$. Hence, $f(x)$ is stricly increasing for $x\ge e$. This implies the claim.

This useful generalization can , in particular be used to decide whether $e^{\pi}$ or $\pi^e$ is larger.

As already pointed out in the answer above, the function $f(x)$ has even a global minimum for $b=e$, so if we set $a=e$, we have $b^e<e^b$ for $b\ne e$ and $b>0$. A number $a\ne e$ cannot have the desired property because if we choose $b=e$, we get $a^b<b^a$ contradicting the required inequality $a^b>b^a$

• If we restrict $p$ to be a positive integer, we can replace $e$ by $2.5$ or $2.9$. So, for example $2.5^p>p^{2.5}$ and $2.9^p>p^{2.9}$ is true for every positive integer. Apr 8 '17 at 11:12

yes, the function $f(x)=\frac{\log(x)}{x}$, $x>0$ has a unique maximum (just look at the derivative) for $x=e$ which by taking exponential (strictly monotone) is equivalent to your statement.

$$e\log x \leq x \log e \Leftrightarrow x^e \leq e^x$$

with strict inequality iff $x\neq e$.

• @ResearchEngineer, You question has been answered. it is proven that the inequality above is an equality iff x = e. Hence for all $p \ne e, e^p \gt p^e$ Apr 8 '17 at 10:21
• Let $p:=\pi$ . Apr 8 '17 at 10:29
• Perhaps you are worried about the strict inequality? For this, just note that the maximum is unique (again from looking at the sign of the derivative). Apr 8 '17 at 10:31