I dont get this problem ..theres gonna be one like this on final?

Question

e^2x + 6e^x +8=0

Answer 1

No solutions my ascii, haven't these people ever heard of complex numbers?

As has been previously mentioned, for the first part of the equation you can set U=e^x and solve for U. Thus:
U^2+6U+8=0
(U+2)(U+4)=0
U=-2 or -4
e^x=-2 or -4

Now, here's where we demonstrate that this equation has solutions. Remember Euler's identity: e^(iπ)+1=0. Thus, e^(iπ)=-1. Thus:
e^x=-2
(-1)e^x=2
e^(iπ)e^x=2
e^(x+iπ)=2
x+iπ=ln 2
x=ln 2+πi

And from e^x=-4, we get x=ln 4+πi. So x=ln 2+πi or x=ln 4+πi. And that's your solution.

Actually, for the sake of completeness, since e^(2πi)=1, x could also be any of these solutions plus an integral multiple of 2πi. Thus, x=ln 2 + (2K+1)πi or x=ln 4 + (2K+1)πi, where K is an arbitrary integer.

Answer 2

Let Y = e^x.

Then the eq'n becomes: Y^2 +6Y +8 = 0

=> (Y+4)(Y+2) = 0
=> Y = -4, -2

=> e^x = -4, -2

which has no real solution. You cannot take the ln of a negative number. Look at the graph of e^x. It never has a negative value.

Answer 3

e^(2x) + 6e^x + 8 = 0
think of this like this
x^2 + 6x + 8 = 0
Factor this and you get
(x + 4)(x + 2) = 0
x = -4 or -2

this becomes

e^x = -4 or e^x = -2
ln both sides
x = ln(-4) or ln(-2)

Which has no solution

Answer 4

replace e^x with v. Now you have a quadratic equation in v. Solve it using the quadratic formula (you don't need me to tell you how) and the solutions are whichever value(s) of v are positive (as e^x is always positive). Take the log of these solution(s) and you've got the answer for solution(s) in x.

edit: in this case all your solutions are negative in v (-2 and -4) so their won't be any real solutions for x. You can find out the complex solutions but their are an infinite amount of them. (i.e. (2k-1).i.pi+ln2 and (2k-1).i.pi+ln4 where k is an integer)

Answer 5

try (e^x)^2+6(e^x)+8=0

or y^2+6y+8=0 (y=-2 and y=-4)

then y=e^x .... (no solutions in this case, e^x>0, -2<0, -4<0)


and ... YES

Answer 6

Simple factoring: (e^x + 4)(e^x + 2) = 0

Answer 7

does e^2x equal e^(2x) or (e^2)x? I'm pretty sure that it means e^(2x).

replace e^x with "a".

a²+6a+8 = 0
(a+2)(a+4) = 0
e^x could equal -2 or -4, so x cannot be a real number.

Answer 8

take e^x= y
then you can write this as:

y^2+ 6y+8=0
solve this quadratic equation
you get y= -4 and -2
so x= ln(-4) and ln(-2)
ohh..this is not possible...
are you sure this is the question..you may have something like this but this is the approach..

Answer 9

tough luck, math is hard the best thing to do is keep asking and look in your math book for selected answers or where it teaches you how to do the problem.