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Question Number 69082 by Henri Boucatchou last updated on 19/Sep/19

$$\mathit{S}\mathit{o}\mathit{l}\mathit{v}\mathit{e}{\mathit{x}}^2={16}^{\mathit{x}}$$

Commented by MJS last updated on 19/Sep/19

$$f\left(x\right)={16}^x-x^2$$$$$$$$f(-1)=-\frac{15}{16}$$$$f\left(0\right)=1$$$$$$$$f(-\frac{1}{2})=0$$$$$$$$f^{\prime }\left(x\right)={16}^{x}4\ln 2-2x>0\mathrm{\forall }x\in \mathbb{R}\Rightarrow$$$$\Rightarrow \mathrm{no}\mathrm{other}\mathrm{real}\mathrm{solution}$$

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

$$x^2={16}^x=2^{4x}$$$$x=\pm 2^{2x}$$$$x=\pm e^{2x\ln 2}$$$${xe}^{-2x\ln 2}=\pm 1$$$$(-2x\ln 2)e^{-2x\ln 2}=\pm 2\ln 2$$$$\Rightarrow -2x\ln 2=W(\pm 2\ln 2)$$$$\Rightarrow x=-\frac{W(\pm 2\ln 2)}{2\ln 2}$$$$realsolution:$$$$\Rightarrow x=-\frac{W\left(2\ln 2\right)}{2\ln 2}=-\frac{0.693147}{2\ln 2}=-\frac{1}{2}$$

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