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VICTOR101010101 · Jun 18, 2012, 08:17 AM

solve for x, e^x+1=0

ebaines · Jun 18, 2012, 08:28 AM

Originally Posted by VICTOR101010101

solve for x, e^x+1=0

Have you studied d'Moivre's theorem?

$<br /> e^{i \theta} = \cos \theta + i \sin \theta<br />$

where $\theta$ is a real number. Here you need to find a value for $\theta$ such that $\cos \theta = -1$ and $\sin \theta = 0$ .

cyber_i9 · Jun 19, 2012, 07:59 AM

well, I guess we can Manipulate the equation,
given: e^x+1=o
:- e^x= -1
:- taking log on both sides
:- x = -log 1 ( log and e cancel out)
therefore,
:- x = 0
hope it works, bcoz its just a Hunch!

ebaines · Jun 19, 2012, 09:09 AM

Originally Posted by cyber_i9

well, i guess we can Manipulate the equation,
given: e^x+1=o
:- e^x= -1

You made a mistake there. From e^x+1=0 you get e^x = -1, which leads to x = ln(-1). Hence the need to apply complex numbers.