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Question Number 116779 by mathdave last updated on 06/Oct/20

solve  x^x =2

$${solve} \\ $$$${x}^{{x}} =\mathrm{2} \\ $$

Answered by Dwaipayan Shikari last updated on 06/Oct/20

xlogx=log(2)  logxe^(logx) =log(2)  logx=W_0 (log(2))  x=e^(W_0 (log(2))) =1.559610

$${xlogx}={log}\left(\mathrm{2}\right) \\ $$$${logxe}^{{logx}} ={log}\left(\mathrm{2}\right) \\ $$$${logx}={W}_{\mathrm{0}} \left({log}\left(\mathrm{2}\right)\right) \\ $$$${x}={e}^{{W}_{\mathrm{0}} \left({log}\left(\mathrm{2}\right)\right)} =\mathrm{1}.\mathrm{559610} \\ $$

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