Question Number 187856 by Michaelfaraday last updated on 23/Feb/23
$${find}\:{x} \\ $$$$\mathrm{2}^{\sqrt{{x}}} =\mathrm{8}{x} \\ $$
Answered by mr W last updated on 23/Feb/23
![t=(√x) >0 2^t =8t^2 2^(t/2) =2(√2)t e^((tln 2)/2) =4(√2)×(t/2) −((tln 2)/2)e^(−((tln 2)/2)) =−((ln 2)/(4(√2))) −((tln 2)/2)=W(−((ln 2)/(4(√2)))) t=−(2/(ln 2))W(−((ln 2)/(4(√2)))) x=t^2 =[−(2/(ln 2))W(−((ln 2)/(4(√2))))]^2 ≈ { (([−((2×(−3.29038565))/(ln 2))]^2 =90.136912)),(([−((2×(−0.141101))/(ln 2))]^2 =0.165756)) :}](Q187867.png)
$${t}=\sqrt{{x}}\:>\mathrm{0} \\ $$$$\mathrm{2}^{{t}} =\mathrm{8}{t}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\frac{{t}}{\mathrm{2}}} =\mathrm{2}\sqrt{\mathrm{2}}{t} \\ $$$${e}^{\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =\mathrm{4}\sqrt{\mathrm{2}}×\frac{{t}}{\mathrm{2}} \\ $$$$−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}{e}^{−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}={W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$$${t}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$$${x}={t}^{\mathrm{2}} =\left[−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right)\right]^{\mathrm{2}} \\ $$$$\approx\begin{cases}{\left[−\frac{\mathrm{2}×\left(−\mathrm{3}.\mathrm{29038565}\right)}{\mathrm{ln}\:\mathrm{2}}\right]^{\mathrm{2}} =\mathrm{90}.\mathrm{136912}}\\{\left[−\frac{\mathrm{2}×\left(−\mathrm{0}.\mathrm{141101}\right)}{\mathrm{ln}\:\mathrm{2}}\right]^{\mathrm{2}} =\mathrm{0}.\mathrm{165756}}\end{cases} \\ $$
Commented by Michaelfaraday last updated on 25/Feb/23
$${thanks}\:{sir} \\ $$