Question-109336

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}

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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