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\(\mathrm{e}^{\pi}>\pi^{\mathrm{e}}\;\;\;or\;\;\;\mathrm{e^{\pi } < \pi ^ { \mathrm { e } }}\)


Let’s solve the problem,
\[\begin{aligned} &\frac{\pi^{e}}{e^{\pi}}=\frac{\pi^{e}}{e^{\pi-e} e^{e}}\\\\ &=\frac{1}{e^{\pi-e}}\left(\frac{\pi}{e}\right)^{e}\\\\ &=\frac{1}{e^{\pi-e}}\left(1+\frac{\pi}{e}-1\right)^{e}\\\\ &=\frac{1}{e^{\pi-e}}\left(1+\frac{\pi-e}{e}\right)^{e}\\\\ &=\frac{1}{e^{\pi-e}}\left(1+\frac{1}{\frac{e}{\pi-e}}\right)^{\frac{e}{\pi-e} e \frac{\pi-e}{e}}\\\\ &=\frac{1}{e^{\pi-e}}\left(\left(1+\frac{1}{\frac{e}{\pi-e}}\right)^{\frac{e}{\pi-e}}\right)^{\pi-e} \end{aligned}\]
Let \(f\) the function defined by,
We know that \(f\) is an increasing function for \(x>0\) and \[\lim _{x \rightarrow\infty}\left(1+\frac{1}{x}\right)^{x}=e\]
\[\begin{aligned} &\Rightarrow\left(1+\frac{1}{\frac{e}{\pi-e}}\right)^{\frac{e}{\pi-e}} <e\\\\ &\left.\Rightarrow \frac{1}{e^{\pi-e}}\left(\left(1+\frac{1}{\frac{e}{\pi-e}}\right)^{\frac{e}{\pi-e}}\right)^{\pi-e}<\frac{1}{e^{\pi-e}} e^{\pi-e}=1\right.\\\\ &\Rightarrow \frac{\pi^{e}}{e^{\pi}}<1 \end{aligned}\]
\[\huge \Rightarrow e^{\pi}>\pi^{e}\]
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