find x 2^(√x) =8x


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

$f i n d x$ $2^{\sqrt{x}} = 8 x$
	Answered by mr W last updated on 23/Feb/23
	$t=(√x) >0 2^t =8t^2 2^(t/2) =2(√2)t e^((tln 2)/2) =4(√2)×(t/2) −((tln 2)/2)e^(−((tln 2)/2)) =−((ln 2)/(4(√2))) −((tln 2)/2)=W(−((ln 2)/(4(√2)))) t=−(2/(ln 2))W(−((ln 2)/(4(√2)))) x=t^2 =[−(2/(ln 2))W(−((ln 2)/(4(√2))))]^2 ≈ { (([−((2×(−3.29038565))/(ln 2))]^2 =90.136912)),(([−((2×(−0.141101))/(ln 2))]^2 =0.165756)) :}$
	$t = \sqrt{x} > 0$ $2^{t} = 8 t^{2}$ $2^{\frac{t}{2}} = 2 \sqrt{2} t$ $e^{\frac{t \ln 2}{2}} = 4 \sqrt{2} \times \frac{t}{2}$ $- \frac{t \ln 2}{2} e^{- \frac{t \ln 2}{2}} = - \frac{\ln 2}{4 \sqrt{2}}$ $- \frac{t \ln 2}{2} = W (- \frac{\ln 2}{4 \sqrt{2}})$ $t = - \frac{2}{\ln 2} W (- \frac{\ln 2}{4 \sqrt{2}})$ $x = t^{2} = {[- \frac{2}{\ln 2} W (- \frac{\ln 2}{4 \sqrt{2}})]}^{2}$ $\approx {\begin{cases} {[- \frac{2 \times (- 3.29038565)}{\ln 2}]}^{2} = 90.136912 \\ {[- \frac{2 \times (- 0.141101)}{\ln 2}]}^{2} = 0.165756 \end{cases}$
	Commented by Michaelfaraday last updated on 25/Feb/23

	$t h a n k s s i r$
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