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Question Number 180604 by mathlove last updated on 14/Nov/22

Answered by Frix last updated on 14/Nov/22

$$x=\sqrt{2}{}^{\sqrt{2}{}^{\sqrt{2}{}^{...}}}=\sqrt{2}{}^x$$$$\ln x=x\ln \sqrt{2}$$$$\frac{\ln x}{x}=\frac{\ln 2}{2}$$$$x=2$$

Commented by mr W last updated on 14/Nov/22

$$\frac{\ln x}{x}=\frac{\ln 2}{2}=\frac{\ln 4}{4}$$$$x=2or4?$$

Commented by Frix last updated on 14/Nov/22

$$2$$$$1⩽x⩽{\mathrm{e}}^{{\mathrm{e}}^{-1}}\Rightarrow 1⩽x^{x^{x^{...}}}⩽\mathrm{e}$$$$x>{\mathrm{e}}^{{\mathrm{e}}^{-1}}\Rightarrow x^{x^{x^{...}}}=+\mathrm{\infty }$$$$y=x^{x^{x^{...}}}\iff y=x^y\iff x=y^{\frac{1}{y}}$$$$\Rightarrow x^{\prime }=y^{\frac{1}{y}-2}(1-\ln y)>0;0<y<\mathrm{e}$$$$\Rightarrow x=y^{\frac{1}{y}}\mathrm{is}\mathrm{increasing}\mathrm{in}(0;\mathrm{e})\mathrm{and}$$$$\mathrm{decreasing}\mathrm{for}y>\mathrm{e}$$$$\Rightarrow \mathrm{the}\mathrm{maximum}\mathrm{is}\mathrm{at}x={\mathrm{e}}^{\frac{1}{\mathrm{e}}}$$$$\Rightarrow \nexists x\mathrm{with}{x}^{x^{x^{...}}}=y\in \mathbb{R}\wedge y>\mathrm{e}$$

Commented by mathlove last updated on 15/Nov/22

$$thanks$$

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