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Im trying to study for an exam in my Real Analysis class and we got these problems that my teacher made up for us but I cant find anything similar enough in the book to teach myself with. I was hoping someone could explain how to do prove this and what the thought process is, like how do you know where to start from and that sort of thing.

Show that $e^x \ge x^a$ if $a>0$ for all $x > 0$ iff $a < e$

user326159
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Mr.Rasputin
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  • Related: http://math.stackexchange.com/questions/2077667/how-is-e-the-only-number-n-for-which-nx-xn-is-satisfied-for-all-values/2077678#2077678 – Catalin Zara Feb 27 '17 at 22:58
  • If your teacher phrased the question like that, then he or she needs to learn a lot about clear mathematical writing. "blah if foo for all bah" is rife with ambiguity. Tagging "iff whotsit" onto that just compounds the crime. – Rob Arthan Feb 27 '17 at 23:21
  • I would start by taking the natural log of both sides, and see if you can make an equation like $a < f(x)$ or $1/a > g(x)$. Since it has to hold for all $x$, the min or max of the function you find could put a bound on $a$. – Jonathan Hahn Feb 27 '17 at 23:21

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Too long for a comment: In my opinion the correct statement is $e^x > x^a , x>0 \Longleftrightarrow e > a$

Define $f(x) = (ln(x))/x , x > 0$. Note that $f^{'}(x) = (1 - ln(x))/x^2$ and $$f^{'}(x) = 0 \Longleftrightarrow x = e$$ $$f^{'}(x) > 0 \Longleftrightarrow x < e$$ $$f^{'}(x) < 0 \Longleftrightarrow x > e$$

Therefore $f(x) \leq f(e) = 1 /e$ for all $x>0$ (the global minima of $f$ occurs in $x=e$) Thus $$ \frac{ln(x)}{x} < \frac{1}{e}, x>0. $$

Let $0<a<e$ thus

$$ \frac{ln(x)}{x} < \frac{1}{e} < \frac{1}{a}, x>0. $$

Therefore $$\frac{ln(x)}{x} < \frac{1}{a} $$ which implies $e^x > x^a , x>0$.

math student
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