SOLUTION: I am struggling with the following problem: 2e^2x+5e^x-12=0 I know that we replace e^ with u, then use quadratic formula to solve. We get u=6 or u=-16. Then, I guess we sc

Question 468194: I am struggling with the following problem:
2e^2x+5e^x-12=0
I know that we replace e^ with u, then use quadratic formula to solve. We get u=6 or u=-16.
Then, I guess we scratch the -16 because it's negative. From there I'm not sure how we are supposed to solve for x. I know that the answer is ln (3/2), but I'm not sure of the steps that go into solving for that and why the answer isn't ln3/2 or ln3/ln2, etc. Obviously you get here by setting 6=2e^2x, but I'm not sure why. Just need a brush-up on some rules here.
Thanks,
J
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
2e^2x+5e^x-12=0
let e^x=u
2u^2-5u-12=0
2u^2-8u+3u-12=0
2u(u-4)+3(u-4)=0
(u-4)(2u+3)=0
e^x=4
ln(e^x)= ln(4)
x= ln(4)
e^x= -3/2
x=ln(-3/2) not defined
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