solve for x e^(2x) + 2e^x + 1 = 0


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Question Number 38957 by Rio Mike last updated on 01/Jul/18

$s o l v e f o r x$ $e^{2 x} + 2 e^{x} + 1 = 0$
Commented by math khazana by abdo last updated on 02/Jul/18

$i f x f r o m R (e) \Leftrightarrow {(e^{x} + 1)}^{2} = 0 \Leftrightarrow e^{x} + 1 = 0$ $\Leftrightarrow e^{x} = - 1 ? \Rightarrow n o ? s o l u t i o n .$ $i g x f r o m C (e) \Leftrightarrow e^{x} + 1 = 0 \Leftrightarrow e^{x} = - 1 \Leftrightarrow$ $e^{x} = e^{i (2 k + 1) π} \Leftrightarrow x_{k} = i (2 k + 1) π k \in Z$ $t h e c o m p l e x x_{k} a r e s o l u t i o n s .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18
	$(e^x +1)^2 =0 when e^(x ) =−1 e^x =e^(iΠ) x=iΠ {e^(iΠ) =cosΠ+isinΠ=cosΠ+isinΠ=−1}$
	${(e^{x} + 1)}^{2} = 0 w h e n e^{x} = - 1$ $e^{x} = e^{i Π}$ $x = i Π$ ${e^{i Π} = c o s Π + i s i n Π = c o s Π + i s i n Π = - 1}$
	Commented by MJS last updated on 01/Jul/18

	$true but not complete$ $x = (2 z + 1) π i; z \in Z$
	Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jul/18

	$y e s s i r i p u t t h e p r i n i c i p a l v a l u e o n l y$
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