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Question Number 124203 by n0y0n last updated on 01/Dec/20

$$2^{2-2^{2-2^{2-2^{2-.....}}}=?}$$

Commented by n0y0n last updated on 01/Dec/20

$$\mathrm{pls}\mathrm{details}$$

Commented by MJS_new last updated on 01/Dec/20

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{an}\mathrm{answer},\mathrm{this}\mathrm{user}\mathrm{is}\mathrm{not}\mathrm{able}\mathrm{to}$$$$\mathrm{post}\mathrm{his}\mathrm{own}\mathrm{questions}...$$

Commented by Ar Brandon last updated on 01/Dec/20

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Answered by MJS_new last updated on 01/Dec/20

$$x=2^{2-2^{2-2^{...}}}$$$$\ln x=(2-x)\ln 2$$$$\mathrm{let}t=\frac{\ln x}{\ln 2}$$$$2^t+t-2=0$$$$\mathrm{approximating}\mathrm{I}\mathrm{get}t\approx .543000440865$$$$\Rightarrow x\approx 1.45699955913$$

Answered by mr W last updated on 01/Dec/20

$$x=2^{2-x}$$$$x{2}^x=2^2$$$${xe}^{x\ln 2}=4$$$$x\ln 2e^{x\ln 2}=4\ln 2$$$$x\ln 2=W\left(4\ln 2\right)$$$$\Rightarrow x=\frac{W\left(4\ln 2\right)}{\ln 2}\approx 1.457$$

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