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Having this function, $f:(0,oo)->RR;f(x)=xlogp-plogx;p inRR,p>0$ how to demonstrate that $e^x>=x^e forallx in(0,oo)$ ?

1 Answer

Apr 29, 2017

From

https://socratic.org/questions/having-function-f-0-oo-rr-f-x-xlnp-plnx-pinrr-p-0-how-to-find-p-0-sush-that-f-x-?answerEditSuccess=1

we know that

$xlogp-plogx ge 0-> x^(1/x) le e^(1/e)$ and also

$x^(1/x) le e^(1/e)->x^e le e^x$

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