Solve the equation $e^x=x^e$

Question

Solve the equation: $$e^x=x^e$$ with domain $(0,\infty)$.

Should I find first and second derivative so I can find max limit? I have not done anything yet. Any help is appreciated.

Answer 1

Note that

$$e^x=x^e\iff \log e^x=\log x^e\iff x= e\log x$$

Now consider

$$f(x)=x-e\log x \implies f'(x)=1-\frac{e}{x}=0 \implies x=e$$

and

$$f''(x)=\frac{e}{x^2}>0$$

thus since $f(e)=0$ is a minimum $x=e$ is the unique solution.

Answer 2

Hint: Write $e^x=x^e$ as $f(x)=f(e)$, where $f(x)=x^{1/x}$.