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Question Number 116347 by Khalmohmmad last updated on 03/Oct/20

Answered by MJS_new last updated on 03/Oct/20

$$x^{-x^x}=x^{(-x^x)}=\frac{1}{x^{\left(x^x\right)}}$$$$\Rightarrow x^{\left(x^x\right)}=2^{-\sqrt{2}}$$$$x^x\ln x=-\sqrt{2}\ln 2$$$$x^x\mathrm{defined}\mathrm{for}x>0\Rightarrow \ln x<0\Rightarrow 0<x<1$$$$\mathrm{I}\mathrm{think}\ln x=a\ln 2\Rightarrow x=2^a\wedge a<0$$$$\mathrm{trying}\mathrm{leads}\mathrm{to}a=-2\Rightarrow x=\frac{1}{4}$$

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