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Question Number 18430 by aplus last updated on 21/Jul/17

x^2 =3^x find x

$$\mathrm{x}^{\mathrm{2}} =\mathrm{3}^{\mathrm{x}} \\ $$$$\mathrm{find}\:\mathrm{x} \\ $$

Answered by mrW1 last updated on 21/Jul/17

x=±3^(x/2) x=±e^((x/2)ln 3) xe^(−(x/2)ln 3) =±1 (−(x/2)ln 3)e^(−(x/2)ln 3) =∓((ln 3)/2) ⇒−(x/2)ln 3=W(∓((ln 3)/2)) ⇒x=−(2/(ln 3))W(∓((ln 3)/2)) since −((ln 3)/2)<−(1/e) there is only one real root: x=−(2/(ln 3))W(((ln 3)/2))≈−0.69

$$\mathrm{x}=\pm\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$\mathrm{x}=\pm\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}} \\ $$$$\mathrm{xe}^{−\frac{\mathrm{x}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}} =\pm\mathrm{1} \\ $$$$\left(−\frac{\mathrm{x}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}\right)\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}} =\mp\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow−\frac{\mathrm{x}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}=\mathrm{W}\left(\mp\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{x}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}\mathrm{W}\left(\mp\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\mathrm{since}\:−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{2}}<−\frac{\mathrm{1}}{\mathrm{e}} \\ $$$$\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}: \\ $$$$\mathrm{x}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}\mathrm{W}\left(\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{2}}\right)\approx−\mathrm{0}.\mathrm{69} \\ $$

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