2-x-x-2-hence-x-2-prove-

Question Number 25787 by ibraheem160 last updated on 14/Dec/17

2^{x} = x^{2}, h e n c e x = 2 . p r o v e

Commented by mrW1 last updated on 14/Dec/17

{I t}^{'} s n o t t r u e, S i r .

I f 2^{x} = x^{2}, t h e n x = 2 o r x = 4 o r x = - 0.77 .

Commented by mrW1 last updated on 15/Dec/17

s o l u t i o n o f 2^{x} = x^{2} :

i f x > 0 :

2^{\frac{x}{2}} = x

e^{\frac{x \ln 2}{2}} = x

1 = {x e}^{- \frac{x \ln 2}{2}}

- \frac{\ln 2}{2} = (- \frac{x \ln 2}{2}) \times e^{- \frac{x \ln 2}{2}}

\Rightarrow - \frac{x \ln 2}{2} = W (- \frac{\ln 2}{2})

\Rightarrow x = - \frac{W (- \frac{\ln 2}{2})}{\frac{\ln 2}{2}} = - \frac{W (- \ln \sqrt{2})}{\ln \sqrt{2}}

i f x < 0 :

2^{\frac{x}{2}} = - x

e^{\frac{x \ln 2}{2}} = - x

1 = - {x e}^{- \frac{x \ln 2}{2}}

\frac{\ln 2}{2} = (- \frac{x \ln 2}{2}) \times e^{- \frac{x \ln 2}{2}}

\Rightarrow - \frac{x \ln 2}{2} = W (\frac{\ln 2}{2})

\Rightarrow x = - \frac{W (- \frac{\ln 2}{2})}{\frac{\ln 2}{2}} = - \frac{W (\ln \sqrt{2})}{\ln \sqrt{2}}

a l l t o g e t h e r :

\Rightarrow x = - \frac{W (\pm \ln \sqrt{2})}{\ln \sqrt{2}}

= {\begin{cases} - \frac{0.2657}{0.3466} = - 0.7666 \\ - \frac{- 0.6932}{0.3466} = 2 \\ - \frac{1.3864}{0.3466} = 4 \end{cases}

i n g e n e r a l t h e s o l u t i o n o f

a^{x} = x^{n}

i s x = - \frac{W (- \frac{l n a}{n})}{\frac{\ln a}{n}}

o r x = - \frac{W (\pm \frac{l n a}{n})}{\frac{\ln a}{n}} i f n i s e v e n

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