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Question Number 50972 by Necxx last updated on 22/Dec/18

Answered by mr W last updated on 22/Dec/18

2^(2x) =8x e^(2xln 2) =8x ((ln 2)/4)e^(2xln 2) =2xln 2 ((ln 2)/4)=(2xln 2)e^(−2xln 2) −((ln 2)/4)=(−2xln 2)e^(−2xln 2) ⇒−2xln 2=W(−((ln 2)/4)) ⇒Lambert function ⇒x=−(1/(2ln 2))W(−((ln 2)/4))= { ((−((−0.2148)/(2 ln 2))=0.1549)),((−((−2.7726)/(2 ln 2))=2)) :}

$$\mathrm{2}^{\mathrm{2}{x}} =\mathrm{8}{x} \\ $$$${e}^{\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{8}{x} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}{e}^{\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{2}{x}\mathrm{ln}\:\mathrm{2} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}=\left(\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}\right){e}^{−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} \\ $$$$−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}=\left(−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}\right){e}^{−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}={W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}\right)\:\Rightarrow{Lambert}\:{function} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{1}}{\mathrm{2ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}\right)=\begin{cases}{−\frac{−\mathrm{0}.\mathrm{2148}}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}}=\mathrm{0}.\mathrm{1549}}\\{−\frac{−\mathrm{2}.\mathrm{7726}}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}}=\mathrm{2}}\end{cases} \\ $$

Commented by Necxx last updated on 23/Dec/18

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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