Comparing the size of $(\sqrt{5})^e$ and $e^{\sqrt{5}}$

Question

So I have to figure out which one is bigger between $(\sqrt{5})^e$ and $e^{\sqrt{5}}$. After some trial and error I've come to the conclusion that $(\sqrt{5})^e > e^{\sqrt{5}}$. But of course I have to supply a formal proof and I'm not sure how to do this.

Thanks for any help.

Answer 1

$$\sqrt 5^e >< e^{\sqrt5} \text{ iff } e\ln\sqrt 5 ><\sqrt 5 \ln e \text{ iff } \frac{\ln\sqrt 5}{\sqrt 5} >< \frac{\ln e} e$$

now we will look at the function $$y = \frac{\ln x}{x}, y' = \frac{1- \ln x}{x^2} $$ has a local max at $x = e, y = 1/e$ now we know that $$ \frac{\ln\sqrt 5}{\sqrt 5} < \frac{\ln e} e \implies\sqrt 5 ^e < e^\sqrt 5. $$

Answer 2

Consider the function $x^{\frac{1}{x}}$ and its monotonicity. ($x^{\frac{1}{x}}>y^{\frac{1}{y}} \implies x^y>y^x$)

Answer 3

The problem is more famous about $e^\pi$ and $\pi^e$.
Rewrite $e^x\lessgtr x^e$ as $\left(e^x\right)^\frac{1}{ex}\lessgtr \left(x^e\right)^\frac{1}{ex}$
$e^\frac{1}{e} \lessgtr x^\frac{1}{x}$
Consider $f(x)=x^\frac{1}{x}=e^{\frac{1}{x}\log x}$ and prove it to have the only maximum at $x=e$.