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Is it correct that if $$3 < a < b$$ then $$a^b > b^a?$$

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  • $\begingroup$ Yes, I think it is correct. $\endgroup$ – user209663 May 16 at 15:29
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    $\begingroup$ Maybe rewriting a^b as e^{b\ln(a) and b^a as e^{a\ln(b) will help $\endgroup$ – user209663 May 16 at 15:31
  • $\begingroup$ yeah it did help $\endgroup$ – MrAnonymous May 16 at 15:53
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You need to check $a^{1/a}>b^{1/b}$, so can you prove that $x^{1/x}$ is decreasing in $(3,\infty)$?.

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  • $\begingroup$ Could you elaborate? Should I differentiate a wrt x? or what? Sorry I'm confused $\endgroup$ – MrAnonymous May 16 at 15:47
  • $\begingroup$ Yes, that's the idea. If you can prove that $\frac{d}{dx}(x^{1/x})<0$ when $x\in(3,\infty)$, then you will prove that $x^{1/x}$ is decreasing in $(3,\infty)$. As someone already suggested, expressing $x^{1/x}=e^{\ln x/x}$ may help to find the derivative. $\endgroup$ – Julian Mejia May 16 at 15:49
  • $\begingroup$ thank you for your help $\endgroup$ – MrAnonymous May 16 at 15:53
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Because $\ln$ is strictly increasing function, if we prove that $\ln a^b > \ln b^a$ we prove that $a^b > b^a$.

$\ln a^b > \ln b^a \rightarrow \frac{b}{\ln b} > \frac{a}{\ln a}$ which is true because $(\frac{x}{\ln x})'=\frac{\ln x -1}{\ln^2 x}>0$ if $x > e$

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