solve-2-x-2-40x-0-Q-172086-reposted-

Question Number 172244 by mr W last updated on 24/Jun/22

$${solve}\: \\ $$$$\mathrm{2}^{{x}^{\mathrm{2}} } −\mathrm{40}{x}=\mathrm{0} \\ $$$$ \\ $$$$\left[{Q}#\mathrm{172086}\:{reposted}\right] \\ $$

Commented by mr W last updated on 24/Jun/22

$$\mathrm{2}^{{x}^{\mathrm{2}} } −\mathrm{40}{x}=\mathrm{0} \\ $$$$\mathrm{2}^{{x}^{\mathrm{2}} } =\mathrm{40}{x}\:>\mathrm{0} \\ $$$$\left(\mathrm{2}^{{x}^{\mathrm{2}} } \right)^{\mathrm{2}} =\left(\mathrm{40}{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{2}{x}^{\mathrm{2}} } =\mathrm{40}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$−\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{40}^{\mathrm{2}} }=\left(−\mathrm{2}{x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{2}\right){e}^{−\mathrm{2}{x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{2}} \\ $$$$−\mathrm{2}{x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{2}={W}\left(−\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{40}^{\mathrm{2}} }\right) \\ $$$${x}^{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{40}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{x}=\sqrt{−\frac{\mathrm{1}}{\mathrm{2ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{40}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\:=\begin{cases}{\sqrt{−\frac{−\mathrm{9}.\mathrm{27886375}}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}}}=\mathrm{2}.\mathrm{587138}}\\{\sqrt{−\frac{−\mathrm{0}.\mathrm{00086718566}}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}}}=\mathrm{0}.\mathrm{025011}}\end{cases} \\ $$

Commented by Tawa11 last updated on 24/Jun/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by peter frank last updated on 27/Jun/22

$$\mathrm{thank}\:\mathrm{you} \\ $$

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