Prove ( using calculus) that $20.17^{20.16}<20.16^{20.17}$.
How do I do that?
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Martin Sleziak
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1You can probably find a few related posts on this site. You can check, for example, [Fastest way to check if $x^y > y^x$?](http://math.stackexchange.com/q/517555) and other posts [linked there](http://math.stackexchange.com/questions/linked/517555). – Martin Sleziak Feb 03 '17 at 17:10
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Take the logarithm. Divide each side by $20.16 \times 20.17$. So the question becomes:
Prove $$ \frac{\ln 20.17}{20.17} < \frac{ \ln 20.16}{20.16}$$
However, the derivative of $ f(x)=\frac{\ln x}{x}$ is $$f'(x)=\frac{1-\ln x}{x^2}$$ From the product rule. So the fucntion $f(x)=\frac{\ln x}{x}$ is decreasing if $x>e$. The result follows as. $$20.17>20.16>e$$As seen here.
