find x^2 =2^× findx?


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Question Number 96241 by joki last updated on 30/May/20

$find x^{2} = 2^{\times} findx ?$
	Answered by 1549442205 last updated on 31/May/20
	$It is easy to see that x∈{0,1} don′t satisfy and x∈{2,4} satisfying the given equation We prove that 2^x >x^2 for any x ≥5 Indeed,consider the function f(x)= 2^x −x^2 .We have f ′ (x)=2^x ln2−2x, f ′′(x)=2^x ln^2 2−2>32ln^2 2−2>0 Hence,f ′(x) is increasing function on ]5;+∞)⇒f ′(x)>f ′(5)=32ln2−10>0 This show that the function f(x)=2^x −x^2 is increasing on [5;+∞)⇒f(x)>f(5)=2^5 −25>0 ⇒2^x −x^2 >0⇒2^x >x^2 for every x≥5. Thus,the numbers x satisfy the given equation are x∈{2;4}$
	$It is easy to see that x \in {0, 1} {don}^{'} t satisfy$ $and x \in {2, 4} satisfying the given equation$ $We prove that 2^{x} > x^{2} for any x ⩾ 5$ $Indeed, consider the function f (x) =$ $2^{x} - x^{2} . We have f^{'} (x) = 2^{x} \ln 2 - 2 x,$ $f^{″} (x) = 2^{x} \ln^{2} 2 - 2 > {32 \ln}^{2} 2 - 2 > 0$ $Hence, f^{'} (x) is increasing function on$ $] 5; + \infty) \Rightarrow f^{'} (x) > f^{'} (5) = 32 \ln 2 - 10 > 0$ $This show that$ $the function f (x) = 2^{x} - x^{2} is increasing$ $on [5; + \infty) \Rightarrow f (x) > f (5) = 2^{5} - 25 > 0$ $\Rightarrow 2^{x} - x^{2} > 0 \Rightarrow 2^{x} > x^{2} for every x ⩾ 5 .$ $Thus, the numbers x satisfy the given$ $equation are x \in {2; 4}$
	Answered by mr W last updated on 31/May/20
	$x^2 =2^x ⇒x=±2^(x/2) =±e^((x/2)ln 2) ⇒xe^(−(x/2)ln 2) =±1 ⇒(−(x/2)ln 2)e^(−(x/2)ln 2) =±((ln 2)/2) ⇒−(x/2)ln 2=W(±((ln 2)/2)) with W()=Lambert W function ⇒x=−((2/(ln 2)))W(±((ln 2)/2))= { ((−(2/(ln 2))W(((ln 2)/2))≈−(2/(ln 2))×0.26570574=−0.766665)),((−(2/(ln 2))W(−((ln 2)/2))= { ((−(2/(ln 2))×(−0.693147)=2)),((−(2/(ln 2))×(−1.386294)=4)) :})) :} that means there are three roots.$
	$x^{2} = 2^{x}$ $\Rightarrow x = \pm 2^{\frac{x}{2}} = \pm e^{\frac{x}{2} \ln 2}$ $\Rightarrow {x e}^{- \frac{x}{2} \ln 2} = \pm 1$ $\Rightarrow (- \frac{x}{2} \ln 2) e^{- \frac{x}{2} \ln 2} = \pm \frac{\ln 2}{2}$ $\Rightarrow - \frac{x}{2} \ln 2 = W (\pm \frac{\ln 2}{2}) w i t h W () = L a m b e r t W f u n c t i o n$ $\Rightarrow x = - (\frac{2}{\ln 2}) W (\pm \frac{\ln 2}{2}) = {\begin{cases} - \frac{2}{\ln 2} W (\frac{\ln 2}{2}) \approx - \frac{2}{\ln 2} \times 0.26570574 = - 0.766665 \\ - \frac{2}{\ln 2} W (- \frac{\ln 2}{2}) = {\begin{cases} - \frac{2}{\ln 2} \times (- 0.693147) = 2 \\ - \frac{2}{\ln 2} \times (- 1.386294) = 4 \end{cases} \end{cases}$ $t h a t m e a n s t h e r e a r e t h r e e r o o t s .$
	Commented by mr W last updated on 31/May/20

	$t h e c u r v e o f x^{2} - 2^{x} s h o w s t h a t i t h a s$ $i n d e e d t h r e e r o o t s :$
	Commented by mr W last updated on 31/May/20

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