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Question Number 69412 by ahmadshah last updated on 23/Sep/19

Commented by Rasheed.Sindhi last updated on 23/Sep/19

$$x=4$$

Answered by mr W last updated on 23/Sep/19

$$generallysolve{x}^{\sqrt{x}}=a>0$$$$x=a^{\frac{1}{\sqrt{x}}}$$$${\left(\sqrt{x}\right)}^2=a^{\frac{1}{\sqrt{x}}}$$$$\sqrt{x}=a^{\frac{1}{2\sqrt{x}}}$$$$\sqrt{x}=e^{\frac{\ln a}{2\sqrt{x}}}$$$$\frac{\ln a}{2}=\frac{\ln a}{2\sqrt{x}}{e}^{\frac{\ln a}{2\sqrt{x}}}$$$$\Rightarrow \frac{\ln a}{2\sqrt{x}}=W\left(\frac{\ln a}{2}\right)$$$$\Rightarrow \sqrt{x}=\frac{\ln a}{2W\left(\frac{\ln a}{2}\right)}$$$$\Rightarrow x={\left[\frac{\ln a}{2W\left(\frac{\ln a}{2}\right)}\right]}^2$$$$$$$$witha=16:$$$$\ln a=4\ln 2$$$$x={\left[\frac{2\ln 2}{W\left(2\ln 2\right)}\right]}^2={\left(\frac{2\ln 2}{0.69317418}\right)}^2=4$$$$$$$$witha=5:$$$$x={\left[\frac{\ln 5}{2W\left(\frac{\ln 5}{2}\right)}\right]}^2={\left(\frac{\ln 5}{2\times 0.492004741}\right)}^2\approx 2.675161$$

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