# Is it correct that if (1) then (2)?

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

• Yes, I think it is correct. – user209663 May 16 at 15:29
• Maybe rewriting a^b as e^{b\ln(a) and b^a as e^{a\ln(b) will help – user209663 May 16 at 15:31
• yeah it did help – MrAnonymous May 16 at 15:53

You need to check $$a^{1/a}>b^{1/b}$$, so can you prove that $$x^{1/x}$$ is decreasing in $$(3,\infty)$$?.
• 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. – Julian Mejia May 16 at 15:49
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 $$x > e$$