Question

e^2x +4e^x-5=0 solve for x

Answer 1

Well, let's do a substitution and say that e^x = u. Because e^(2x) is the same as (e^x)^2 = u^2, we can rewrite this as
\[u ^{2}+4u - 5 = 0\]

Now factor it as if it were a normal quadratic with x's. Once you finish and solve for u, just replace u again with e^x.

Answer 2

So when you factor it you get (u-1)(u+5)=0 where do you go from that

Answer 3

Well, you could solve for u first or do it later, but remember, we just said that u = e^x. So replace u with e^x.

Answer 4

Ok so (e^x -1)(e^x+5)?

Answer 5

Correct. Now solve for e^x.

Answer 6

e^x=1,-5

Answer 7

Mhm. So now since you actually need x and not just e^x
e^x = 1
e^x = -5
Do you know how to actually solve for x now from here?

Answer 8

Natural log both sides and you get 0 and ln(5)

Answer 9

Well, ln(-5), which is undefined. So your only answer is x = 0.

Answer 10

Thank you:)

Answer 11

yep yep