solve for x by using lambert function x^2 =16^x


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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

	$t h a n k s s i r$
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