Proving $x^e = e^x$ has only one positive solution

Question

Prove that $x^e = e^x$ has only one positive solution.

Let us consider the general equation $x^b=b^x$, where $b\in \mathbb{R}$. The obvious solution is $x=b$. When $b=2$, $x^2=2^x$ has at least two solutions: $x=2$ and $x=4$. In this case, the other solution is greater than $b$. When $b=4$, $x^4=4^x$ also has at least two solutions: $x=4$ and $x=2$. In this case, the other solution is less than $b$. We assume that there must be a value $b$ where these two solutions merge into a single solution, and that $b$ must be between $2$ and $4$. How do we prove that $b$ is $e$?

Answer 1

The equation is equivalent to

$$e^{x/e}-x=0.$$

Then the function $f(x)=e^{x/e}-x$ has a single minimum when $e^{x/e}/e=1$, i.e. $x=e$.

Then we precisely have $f(e)=0$, and the function has a double root.

Answer 2

You equation can be written as

$f(x)=x-e\ln(x)=0$

$f$ is differentiable at $(0,+\infty)$ and

$f'(x)=\frac{x-e}{x}$ $\implies f$ strictly decreasing at $(0,e)$ and strictly increasing at $(e,+\infty)$

and since $f(e)=0$, $e$ is the unique root.