solve-for-x-by-using-lambert-function-x-2-16-x-

Question Number 183756 by Michaelfaraday last updated on 29/Dec/22

s o l v e f o r x b y u s i n g l a m b e r t f u n c t i o n

x^{2} = 16^{x}

Answered by aleks041103 last updated on 30/Dec/22

2 l n x = x l n 16

\Rightarrow l n x = x l n 4

\frac{1}{x} l n (\frac{1}{x}) = - l n (4)

\Rightarrow {y e}^{y} = e^{- l n (4)} = \frac{1}{4} \Rightarrow \frac{1}{x} = W (\frac{1}{4})

\Rightarrow x = \frac{1}{W (1 / 4)}

Commented by Michaelfaraday last updated on 30/Dec/22

s i r, a r e y o u s u r e a b o u t y o u r s o l u t i o n

b e c a u s e w h e n i s o l v e i t, i a r r i v e a t

x = \frac{- W (\frac{- 1}{2} I n 16)}{\frac{1}{2} I n 16}

Commented by mr W last updated on 30/Dec/22

y o u a r e w r o n g, t o o .

x = - \frac{W (\frac{\ln 16}{2})}{\frac{\ln 16}{2}} = - \frac{W (\ln 4)}{\ln 4} = - \frac{1}{2}

Commented by Michaelfaraday last updated on 30/Dec/22

b u t i g o t i t s i r b u t i j u s t d o n t s i m p l i f y i t

t o l o w e s t t e r m .

Commented by mr W last updated on 30/Dec/22

i {d i d n}^{'} t m e a n t h a t! c h e c k a g a i n!

y o u g o t :

x = \frac{- W (\frac{- 1}{2} I n 16)}{\frac{1}{2} I n 16}

t h i s i s w r o n g .

Commented by Michaelfaraday last updated on 30/Dec/22

o h h a v e s e e n i t b u t i m a d e a t y p o i n a n s w e r

t h s t i s w h y, t h a n k s

Answered by MJS_new last updated on 30/Dec/22

x^{2} = 16^{x}

let x = - e^{- t}

e^{- 2 t} = 16^{- e^{- t}}

- 2 t = - e^{- t} 4 \ln 2

e^{t} t = 2 \ln 2

obviously t = \ln 2 is the solution

\Rightarrow x = - \frac{1}{2}

Commented by Michaelfaraday last updated on 30/Dec/22

t h a n k s s i r