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Question Number 216266 by mathlove last updated on 02/Feb/25

$$\sqrt{2}{}^{\sqrt{2}{}^{\sqrt{2}{}^{\text{iddots}}}}=?$$

Answered by BaliramKumar last updated on 02/Feb/25

$${\left(\sqrt{2}\right)}^x=x$$$$2^{\frac{x}{2}}=x$$$$2^x=x^2$$$$x=2or4$$

Commented by Frix last updated on 02/Feb/25

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{the}\mathrm{value}\mathrm{can}\mathrm{be}4$$$$y=x^{x^{x...}}$$$$y=x^y$$$$\frac{\ln y}{y}=\ln x$$$$\mathrm{This}\mathrm{has}2\mathrm{solutions}\mathrm{for}0<x<{\mathrm{e}}^{\frac{1}{\mathrm{e}}}\approx 1.44467$$$$\mathrm{exactly}1\mathrm{solution}\mathrm{for}x={\mathrm{e}}^{\frac{1}{\mathrm{e}}}\mathrm{and}\mathrm{no}\mathrm{solution}$$$$\mathrm{for}x>{\mathrm{e}}^{\frac{1}{\mathrm{e}}}:x\to 0^{+}\Rightarrow y_1\to 1\wedge y_2\to \mathrm{\infty }\mathrm{which}$$$$\mathrm{means}{y}_2\mathrm{is}\mathrm{obviously}\mathrm{wrong}.\mathrm{Using}\mathrm{a}\mathrm{good}$$$$\mathrm{calculator}\mathrm{we}\mathrm{see}\mathrm{that}$$$$1.1^{1.1^{...}}\approx 1.11178\mathrm{how}\mathrm{could}\mathrm{it}\mathrm{be}38.2287?$$$$1.{01}^{1.{01}^{...}}\approx 1.01010\mathrm{how}\mathrm{could}\mathrm{it}\mathrm{be}651.100?$$$$...$$$$\Rightarrow$$$$y_2\mathrm{is}\mathrm{always}\mathrm{wrong}$$

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