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Show that $e^x \geq x^e$ for $ 0 \lt x \lt \infty $.

I tried to apply the normal logarithm here, which yields $x \geq e\times \ln(x)$ Still, I am kind of stuck here, anyone mind giving me a hand?

Martin Sleziak
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MikhaelM
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3 Answers3

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Hint

Consider the function $f(x) = e^x - x^e$ and find its minimum over positive $x$. If you do it correctly (e.g. solve $f'(x) = 0$, etc) you will find that it is non-negative. Hence, the claim will follow.

UPDATE

Easier to take logs: $e^x \ge x^e$ iff $x \ge e \ln x$, and now look at $f(x) = x - e \ln x$, which should be elementary

gt6989b
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  • f′(x) = $e^x$ - e$x^{e-1}$ which would give $e^x$ - e$x^{e-1}$ = 0. The equation becomes e($e^{x-1}$ - $x^{e-1}$) = 0

    Which would mean that $e^{x-1}$ - $x^{e-1}$ = 0

    How do I find the root here? I mean, I can see the roots are x = 1 and x = e, but I can't seem to solve it.

    I understand that once I solve it and prove that the local minimum point for the function is positive, then the original claim is true.

     – MikhaelM Jun 11 '15 at 19:25
  • @MikhaelM please see update, taking logs making things very basic – gt6989b Jun 11 '15 at 19:46
  • Yup, thank you, I missed that somehow. I understand now! – MikhaelM Jun 11 '15 at 20:22
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Hint: $e^{x/e}$ is convex and $y=x$ is its tangent at $x=e$.

Macavity
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Consider the function $f(x)=x-eln(x)$. $f'$ exists, $$ f'(x) = 1-\frac{e}{x}. $$ That function has a root, Which are easily found: $x=e$. Now, $f'$ is negative in $e-h$ and positive in $e+h$ with appropriate $h$, so that critical point is a minimum. When plugged into the original function, we have $0$, so the argument is indeed true.

Eemil Wallin
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