Proof of why the solution is $e$

Question

$$x^e = e^x$$

Hello, I had this question in my calc 2 exam, I needed to write the proof of why the only real solution to the equation was $e$.

Answer 1

First note that the left side is only defined for $x\geq 0$ and the equation is not satisfied at $x=0$. So, consider only $x>0$:

• Write $x^e = e^{e\ln x}$
• $e^{e\ln x} = e^x \Leftrightarrow e\ln x = x$
• Now, consider $f(x) = x- e \ln x$
• $\Rightarrow f'(x) = 1- \frac{e}{x}$
• $\Rightarrow f$ has the global minimum $f(e) =0$ since $f$ is strictly decreasing for $0<x<e$ and strictly increasing for $x>e$
• $\Rightarrow x=e$ is the only solution to the equation.

Answer 2

Hint: Write your equation in the form $$e=\frac{x}{\ln(x)}$$ so we get with $$f(x)=\frac{x}{\ln(x)}$$ $$f'(x)=\frac{\ln(x)-1}{(\ln(x))^2}$$