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Question Number 102234 by Harasanemanabrandah last updated on 07/Jul/20

Answered by OlafThorendsen last updated on 07/Jul/20

$$f\left(x\right)=x-e\ln x,x>0$$$$f^{\prime }\left(x\right)=1-\frac{e}{x}$$$$f^{\prime }\left(x\right)=0\iff x=e$$$$f\left(e\right)=e-e\ln e=e-e=0$$$$\mathrm{And}\mathrm{for}x\in ]e;+\mathrm{\infty }[f^{\prime }\left(x\right)>0$$$$\mathrm{Then}f\left(\pi \right)>0$$$$\iff \pi -e\ln \pi >0$$$$\iff \pi >e\ln \pi$$$$\iff \pi \ln e>e\ln \pi$$$$\iff \ln e^{\pi }>\ln {\pi }^e$$$$\iff e^{\pi }>{\pi }^e$$

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