find-x-if-x-2-16-x-

Question Number 58832 by otchereabdullai@gmail.com last updated on 30/Apr/19

find x if x^{2} = 16^{x}

Commented by tanmay last updated on 01/May/19

f (x) = x^{2} - 16^{x}

f (0) < 0

f (1) < 0

s o n o r o o t b e t w e e n (0, 1)

f (- 1) = 1 - \frac{1}{16} > 0

s o r o o t l i e s b e t w e e n (- 1, 0) \leftarrow l o c a t i o n o f r o o t

f (x) = x^{2} - {(4^{x})}^{2} = (x + 4^{x}) (x - 4^{x})

o n t r i a l i f w e p u t x = - 0.5

f (- 0.5) = (- 0.5 + 4^{- 0.5}) (- 0.5 - 4^{- 0.5})

= {{(- 0.5)}^{2} - {(4^{- 0.5})}^{2}}

= {0.25 - (4^{- 1})}

= 0.25 - 0.25 = 0

s o x = - 0.5 i s t h e s o l u t i o n

i h a v e l o c a t e d t h e l o c a t i o n o f r o o t a n d f o u n d

b y t r i a l m e t h o d \dots T h e r e m u s t b e s o m e d i r e c t

m e t h o d t o f i n d t h e r o o t .

Commented by otchereabdullai@gmail.com last updated on 30/Apr/19

thanks alot prof tanmay

Commented by George Mark Samuel last updated on 30/Apr/19

x^{2} = 16^{x}

x^{2} = 4^{(2) x}

x^{2} = 2^{2 (2 x)}

x^{2} = 2^{4 x}

x = 2 . ∣ 2 = 4 x

∣ divide both sides by 4

∣ \frac{2}{4} = \frac{4 x}{4}

∣ x = \frac{1}{2}

∴ x = 2 or x = \frac{1}{2 .}

Commented by MJS last updated on 01/May/19

{it}^{'} s wrong

x = 2 \Rightarrow x^{2} = 4 \land 16^{x} = 256 but 4 \neq 256

x = \frac{1}{2} \Rightarrow x^{2} = \frac{1}{4} \land 16^{x} = 4 but \frac{1}{4} \neq 4

Answered by MJS last updated on 30/Apr/19

obviously

x ⩾ 0 \Rightarrow 16^{x} > x^{2}

x ⩽ - 1 \Rightarrow x^{2} > 16^{x}

\Rightarrow - 1 < x < 0

trying we find

x = - \frac{1}{2} \Rightarrow {(- \frac{1}{2})}^{2} = 16^{- \frac{1}{2}} = \frac{1}{4}

Commented by otchereabdullai@gmail.com last updated on 30/Apr/19

thanks prof

Answered by mr W last updated on 01/May/19

d i r e c t m e t h o d u s i n g L a m b e r t f u n c t i o n :

x^{2} = 16^{x}

\Rightarrow x = \pm 16^{\frac{x}{2}}

\Rightarrow x = \pm e^{\frac{x \ln 16}{2}}

\Rightarrow {x e}^{- \frac{x \ln 16}{2}} = \pm 1

\Rightarrow (- \frac{x \ln 16}{2}) e^{- \frac{x \ln 16}{2}} = \mp \frac{\ln 16}{2}

\Rightarrow - \frac{x \ln 16}{2} = W (\mp \frac{\ln 16}{2})

\Rightarrow x = - \frac{2}{\ln 16} \times W (\mp \frac{\ln 16}{2}) \leftarrow n o r e a l s o l u t i o n f o r W (- \frac{\ln 16}{2})!

\Rightarrow x = - \frac{2 \times 0.69314718}{\ln 16} = - 0.5

B o n u s :

i n g e n e r a l, i f x^{2} = a^{x}

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

e . g . a = 2, i . e . x^{2} = 2^{x}

\Rightarrow x = - \frac{2}{\ln 2} \times W (\pm \frac{\ln 2}{2}) = {\begin{cases} - \frac{2}{\ln 2} \times W (\frac{\ln 2}{2}) = - \frac{2}{\ln 2} \times 0.26570574 = - 0.766665 \\ - \frac{2}{\ln 2} \times W (- \frac{\ln 2}{2}) = {\begin{cases} - \frac{2}{\ln 2} \times (- 0.69314718) = 2 \\ - \frac{2}{\ln 2} \times (- 1.38629436) = 4 \end{cases} \end{cases}

Commented by otchereabdullai@gmail.com last updated on 01/May/19

thank you soo much prof W

Commented by malwaan last updated on 02/May/19

H o w c a n I c a l c . W f u n c t i o n ?

Commented by mr W last updated on 02/May/19

i f y o u {c a n}^{'} t f i n d a p r o p e r c a l c u l a t o r,

y o u c a n g e t t h e f u n c t i o n v a l u e i n

t h i s w a y :

t h e r o o t (s) o f e q n .

x e^{x} = a

i s (a r e) t h e f u n c t i o n v a l u e (s) o f W (a) .

Commented by malwaan last updated on 03/May/19

I w i l l t r y

t h a n k y o u s i r

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