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Question Number 164923 by mathlove last updated on 23/Jan/22

Commented by MJS_new last updated on 24/Jan/22

$$\mathrm{obviously}{x}_1=\frac{1}{16}\wedge x_2=\frac{1}{4}$$

Answered by Eulerian last updated on 23/Jan/22

$$\mathbf{Solution}:$$$$\mathrm{let}:\sqrt{\mathrm{x}}=\mathrm{k}\to \mathrm{x}={\mathrm{k}}^2$$$$$$$$\therefore$$$${\mathrm{k}}^{2\mathrm{k}}=\frac{1}{2}$$$${\mathrm{k}}^{\mathrm{k}}=\frac{\sqrt{2}}{2}$$$$\mathrm{k}\cdot \ln \left(\mathrm{k}\right)=-\ln \left(\sqrt{2}\right)$$$$\mathrm{By}\mathrm{applying}\mathrm{the}\mathrm{productlog}\mathrm{on}\mathrm{both}\mathrm{sides}:$$$$\ln \left(\mathrm{k}\right)=\mathrm{W}(-\ln \left(\sqrt{2}\right))$$$$\mathrm{k}={\mathrm{e}}^{\mathrm{W}(-\ln \left(\sqrt{2}\right))}=\frac{1}{2}$$$$\therefore$$$$\mathrm{x}=\frac{1}{4}$$

Answered by mr W last updated on 23/Jan/22

$${\left(\sqrt{x}\right)}^{2\sqrt{x}}=\frac{1}{2}$$$${\left(\sqrt{x}\right)}^{\sqrt{x}}={\left(\frac{1}{2}\right)}^{\frac{1}{2}}$$$$\Rightarrow \sqrt{x}=\frac{1}{2}$$$$\Rightarrow x=\frac{1}{4}$$$$or$$$$e^{\ln \sqrt{x}}\ln \sqrt{x}=-\frac{\ln 2}{2}$$$$\ln \sqrt{x}=W(-\frac{\ln 2}{2})$$$$\sqrt{x}=e^{W(-\frac{\ln 2}{2})}$$$$\Rightarrow x=e^{2W(-\frac{\ln 2}{2})}=\{\begin{array}{l}\approx 0.0625\\ 0.25\end{array}$$

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