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Question Number 211675 by MrGaster last updated on 16/Sep/24

$$$$$$$$$${\mathit{x}}^2=2^{\mathit{x}}$$$$\mathrm{Find}\mathrm{more}\mathrm{than}\mathrm{three}\mathrm{solutions}$$$$$$

Commented by mr W last updated on 16/Sep/24

$$butthereareonlythreesolutions!$$

Answered by mr W last updated on 16/Sep/24

$$x^2=2^x$$$$\Rightarrow x=\pm 2^{\frac{x}{2}}$$$$\Rightarrow (-\frac{x}{2})\times 2^{-\frac{x}{2}}=\pm \frac{1}{2}$$$$\Rightarrow (-\frac{x}{2})\left(\ln 2\right)\times e^{(-\frac{x}{2})\left(\ln 2\right)}=\pm \frac{\ln 2}{2}$$$$\Rightarrow (-\frac{x}{2})\left(\ln 2\right)=\mathbb{W}(\pm \frac{\ln 2}{2})$$$$\Rightarrow x=-\frac{2}{\ln 2}\times \mathbb{W}(\pm \frac{\ln 2}{2})=\{\begin{array}{l}2\\ \{\begin{array}{l}4\\ -0.766664696\end{array}\end{array}$$

Answered by Frix last updated on 16/Sep/24

$$x^2=2^x$$$$2\ln x=x\ln 2$$$$-\frac{\ln x}{x}=-\frac{\ln 2}{2}$$$$x={\mathrm{e}}^{-t}$$$${\mathrm{e}}^{t}t+\frac{\ln 2}{2}=0$$$$t=a+b\mathrm{i}$$$$({\mathrm{e}}^a(a\cos b-b\sin b)+\frac{\ln 2}{2})+{\mathrm{e}}^a(a\sin b+b\cos b)\mathrm{i}=0$$$$\{\begin{array}{l}{\mathrm{e}}^a(a\cos b-b\sin b)+\frac{\ln 2}{2}=0\\ {\mathrm{e}}^a(a\sin b+b\cos b)=0\Rightarrow a=-b\cot b\end{array}$$$$\Rightarrow$$$$\frac{2b}{\sin b}-{\mathrm{e}}^{b\cot b}\ln 2=0\left[\mathrm{approximate}\right]$$$$t=-b(\cot b-\mathrm{i})$$$$x={\mathrm{e}}^{b\cot b}(\cos b-\mathrm{i}\sin b)$$$$$$$$2\mathrm{solutions}\notin \mathbb{R}:$$$$b\approx \pm 7.45408757255$$$$x\approx 9.09072986075\pm 21.5079503505\mathrm{i}$$

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