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Question Number 32897 by artibunja last updated on 06/Apr/18

$$$$ $$e^{\pi }>{\pi }^{e}or{e}^{\pi }<{\pi }^e?$$ $$$$

Answered by MJS last updated on 06/Apr/18

$$e^x=x^e;x>1\Rightarrow x=e\Rightarrow \mathrm{\forall }x>e:e^x>x^e$$ $$\pi >e\Rightarrow e^{\pi }>{\pi }^e$$

Commented byMJS last updated on 06/Apr/18

$$\mathrm{interestingly}\mathrm{for}{a}^x=x^a\mathrm{with}a>1$$ $$\mathrm{and}x>1\mathrm{two}\mathrm{solutions}\mathrm{seem}\mathrm{to}$$ $$\mathrm{exist}(\mathrm{i}.\mathrm{e}.2^x=x^2\Rightarrow x=2\vee x=4;$$ $$3^x=x^3\Rightarrow x=3\vee x\approx 2.47805)\mathrm{but}$$ $$\mathrm{for}a=e{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{only}\mathrm{one}:x=e$$ $$$$ $$\mathrm{can}\mathrm{anybody}\mathrm{further}\mathrm{explain}\mathrm{this}?$$

Commented bymrW2 last updated on 06/Apr/18

$$solutionfor{a}^x=x^a(a>0):$$ $$a^{\frac{x}{a}}=x$$ $$e^{\frac{x}{a}\ln a}=x$$ $$(-\frac{x}{a}\ln a)e^{-\frac{x}{a}\ln a}=-\frac{\ln a}{a}$$ $$\Rightarrow -\frac{x}{a}\ln a=W(-\frac{\ln a}{a})$$ $$\Rightarrow x=-\frac{a}{\ln a}W(-\frac{\ln a}{a})=-\frac{W(-\ln a^{\frac{1}{a}})}{\ln a^{\frac{1}{a}}}$$ $$ifa=e:onesolution$$ $$ifa\ne e:twosolutions$$

Answered by Tinkutara last updated on 06/Apr/18

$$Considerf\left(x\right)=y=x^{\frac{1}{x}}$$ $$\frac{dy}{dx}=\frac{y}{x^2}(1-\ln x)$$ $$Ifx<e,f\left(x\right)isincreasing.$$ $$Ifx>e,f\left(x\right)isdecreasing.$$ $$\therefore e^{\frac{1}{e}}>{\pi }^{\frac{1}{\pi }}$$ $$e^{\pi }>{\pi }^e$$

Answered by artibunja last updated on 06/Apr/18

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