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Question Number 109336 by Rasikh last updated on 22/Aug/20

Commented by Ari last updated on 22/Aug/20
x=2
Commented by Ari last updated on 22/Aug/20
x=4
Commented by mr W last updated on 22/Aug/20

$a n d x \approx - 0.7667$
	Answered by Dwaipayan Shikari last updated on 22/Aug/20
	$2^x =x^2 xlog2=2logx e^(logx) =((2logx)/(log2)) e^(−logx) =((log2)/(2logx)) −logxe^(−logx) =−((log2)/2) −logx=W_0 (−((log2)/2)) x=e^(−W_0 (((−log2)/2))) { ((x=2,4)),((x=e^(−W_0 (−((log2)/2))) )) :}$
	$2^{x} = x^{2}$ $x l o g 2 = 2 l o g x$ $e^{l o g x} = \frac{2 l o g x}{l o g 2}$ $e^{- l o g x} = \frac{l o g 2}{2 l o g x}$ $- {l o g x e}^{- l o g x} = - \frac{l o g 2}{2}$ $- l o g x = W_{0} (- \frac{l o g 2}{2})$ $x = e^{- W_{0} (\frac{- l o g 2}{2})}$ ${\begin{cases} x = 2, 4 \\ x = e^{- W_{0} (- \frac{l o g 2}{2})} \end{cases}$
	Commented by mr W last updated on 22/Aug/20

	$i n 2^{x} = x^{2} t h e x c a n b e n e g a t i v e, b u t i n$ $x \log 2 = 2 \log x t h e x m u s t b e p o s i t i v e .$ $i t m e a n s s o m e r o o t s m a y g e t l o s t i n$ $y o u r s o l u t i o n x = e^{- W_{0} (\frac{- \log 2}{2})} . i n f a c t$ $y o u g e t o n l y t w o r o o t s : x = 2, 4 . t h e$ $‘ ‘ {o t h e r s}^{″} y o u m e a n a r e n o t i n c l u d e d i n$ $y o u r s o l u t i o n x = e^{- W_{0} (\frac{- \log 2}{2})} . i n t h i n k$ $t h e c o m p l e t e s o l u t i o n s h o u l d b e$ $x = \mp e^{- W_{0} (\pm \frac{\log 2}{2})} . p l e a s e c h e c k!$
	Commented by Dwaipayan Shikari last updated on 22/Aug/20

	$2^{x} = x^{2}$ $2^{\frac{x}{2}} = \pm x$
	Commented by mr W last updated on 22/Aug/20

	$y e s!$
	Commented by Rasikh last updated on 22/Aug/20

	$thanks$
	Commented by Rasikh last updated on 22/Aug/20

	$thank you sir$
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