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Question Number 187856 by Michaelfaraday last updated on 23/Feb/23

find x  2^(√x) =8x

$${find}\:{x} \\ $$$$\mathrm{2}^{\sqrt{{x}}} =\mathrm{8}{x} \\ $$

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)) :}

$${t}=\sqrt{{x}}\:>\mathrm{0} \\ $$$$\mathrm{2}^{{t}} =\mathrm{8}{t}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\frac{{t}}{\mathrm{2}}} =\mathrm{2}\sqrt{\mathrm{2}}{t} \\ $$$${e}^{\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =\mathrm{4}\sqrt{\mathrm{2}}×\frac{{t}}{\mathrm{2}} \\ $$$$−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}{e}^{−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}={W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$$${t}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$$${x}={t}^{\mathrm{2}} =\left[−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right)\right]^{\mathrm{2}} \\ $$$$\approx\begin{cases}{\left[−\frac{\mathrm{2}×\left(−\mathrm{3}.\mathrm{29038565}\right)}{\mathrm{ln}\:\mathrm{2}}\right]^{\mathrm{2}} =\mathrm{90}.\mathrm{136912}}\\{\left[−\frac{\mathrm{2}×\left(−\mathrm{0}.\mathrm{141101}\right)}{\mathrm{ln}\:\mathrm{2}}\right]^{\mathrm{2}} =\mathrm{0}.\mathrm{165756}}\end{cases} \\ $$

Commented by Michaelfaraday last updated on 25/Feb/23

thanks sir

$${thanks}\:{sir} \\ $$

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