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Question Number 96321 by Mathudent last updated on 31/May/20

$$Itisgiventhat{x}^2=2^x.Findx.$$

Commented by bemath last updated on 31/May/20

$$\mathrm{look}\mathrm{qn}96241$$

Answered by selea last updated on 31/May/20

$$2^{\mathrm{x}}={\mathrm{x}}^2$$$$\ln \left(2^{\mathrm{x}}\right)=\ln \left({\mathrm{x}}^2\right)$$$$\mathrm{xln}\left(2\right)=2\ln \mid \mathrm{x}\mid$$$$\frac{\mathrm{lnx}}{\mathrm{x}}=\frac{\mathrm{l}}{2}\ln \left(2\right)$$$$\ln \left(\mathrm{x}\right){\mathrm{e}}^{\ln \left(\frac{1}{\mathrm{x}}\right)}=\ln \sqrt{2}$$$$\ln \left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{lnx}}=\ln \sqrt{2}$$$$-\ln \left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{lnx}}=-\ln \sqrt{2}$$$$\mathrm{since}$$$$\mathrm{w}\left({\mathrm{xe}}^{\mathrm{x}}\right)=\mathrm{x}\mathrm{where}\mathrm{w}\left(\right)\mathrm{is}\mathrm{lambert}\mathrm{function}$$$$\mathrm{w}(-{\mathrm{lnxe}}^{-\mathrm{lnx}})=\mathrm{w}(-\ln \sqrt{2})$$$$\mathrm{lnx}=-\mathrm{w}(-\ln \sqrt{2})$$$$\mathrm{x}={\mathrm{e}}^{-\mathrm{w}(-\ln \sqrt{2})}\mathrm{for}\mathrm{x}>0$$$$$$$$$$$$$$$$$$$$$$

Commented by Mathudent last updated on 31/May/20

$$CanyousolvewithoutusingDe{l}^{\prime }Ambertfunction?$$

Commented by selea last updated on 31/May/20

$$\mathrm{U}\mathrm{can}\mathrm{use}\mathrm{Newton}\mathrm{Raphson}\mathrm{Method}$$

Commented by mr W last updated on 31/May/20

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