solve for x, e^x+1=0

solve for x, e^x+1=0

Have you studied d'Moivre's theorem?


e^{i \theta} = \cos \theta + i \sin \theta


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

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!

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.