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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.

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  • $\begingroup$ Typically problems like this may be tackled by taking logarithms. Have you tried that? Did you try something else that would help your Readers understand where you are stuck on this? $\endgroup$ – hardmath Apr 27 '15 at 0:43
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    $\begingroup$ I wanted to use the formal definition of a^b, but I'm not sure I can do that with b=e or a=e. It's for calculus, so probably I do have to use logarithms. $\endgroup$ – Edgar Apr 27 '15 at 0:46
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$$\sqrt 5^e >< e^{\sqrt5} \text{ iff } e\ln\sqrt 5 ><\sqrt 5 \ln e \text{ iff } \frac{\ln\sqrt 5}{\sqrt 5} >< \frac{\ln e} e$$

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 $$ \frac{\ln\sqrt 5}{\sqrt 5} < \frac{\ln e} e \implies\sqrt 5 ^e < e^\sqrt 5. $$

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

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  • $\begingroup$ $x^{1/x}$ is increasing for $1 < x < e$ and decreasing for $x > e$. So it is not monotone. Note that $\sqrt{5} < e$. $\endgroup$ – marty cohen Apr 27 '15 at 1:18
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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 $x=e$.

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