Question Number 102234 by Harasanemanabrandah last updated on 07/Jul/20
Answered by OlafThorendsen last updated on 07/Jul/20
![f(x) = x−elnx, x>0 f′(x) = 1−(e/x) f′(x) = 0 ⇔ x = e f(e) = e−elne = e−e = 0 And for x∈]e;+∞[ f′(x)>0 Then f(π) > 0 ⇔ π−elnπ>0 ⇔ π>elnπ ⇔ πlne>elnπ ⇔ lne^π >lnπ^e ⇔ e^π >π^e](https://www.tinkutara.com/question/Q102236.png)
$${f}\left({x}\right)\:=\:{x}−{e}\mathrm{ln}{x},\:{x}>\mathrm{0} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{1}−\frac{{e}}{{x}} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{0}\:\Leftrightarrow\:{x}\:=\:{e} \\ $$$${f}\left({e}\right)\:=\:{e}−{e}\mathrm{ln}{e}\:=\:{e}−{e}\:=\:\mathrm{0} \\ $$$$\left.\mathrm{And}\:\mathrm{for}\:{x}\in\right]{e};+\infty\left[\:{f}'\left({x}\right)>\mathrm{0}\right. \\ $$$$\mathrm{Then}\:{f}\left(\pi\right)\:>\:\mathrm{0} \\ $$$$\Leftrightarrow\:\pi−{e}\mathrm{ln}\pi>\mathrm{0} \\ $$$$\Leftrightarrow\:\pi>{e}\mathrm{ln}\pi \\ $$$$\Leftrightarrow\:\pi\mathrm{ln}{e}>{e}\mathrm{ln}\pi \\ $$$$\Leftrightarrow\:\mathrm{ln}{e}^{\pi} >\mathrm{ln}\pi^{{e}} \\ $$$$\Leftrightarrow\:{e}^{\pi} >\pi^{{e}} \\ $$