SOLUTION: Solve for x. {{{ e^(2x) - 3e^x - 40 = 0 }}}

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Question 476214: Solve for x.
+e%5E%282x%29+-+3e%5Ex+-+40+=+0+
Found 2 solutions by Tatiana_Stebko, lwsshak3:
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
+e%5E%282x%29+-+3e%5Ex+-+40+=+0+
Let t=e%5Ex t%3E0
t%5E2-3t-40=0
%28t-8%29%28t%2B5%29=0
t-8=0 or t%2B5=0
t=8 or t=-5%3C0 exteraneous root
e%5Ex=8
x=ln8

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x. e^(2x) - 3e^x - 40 = 0
**
e^(2x) - 3e^x - 40 = 0
let u=e^x
u^2=e^(2x)
..
u^2-3u-40=0
(u-8)(u+5)=0
u=8
or
u=-5
..
e^x=8
e^x=-5(reject,e^x>0)
xlne=ln(8)
x=2.0794
..
Check:
e^(2x) - 3e^x - 40
e^4.1588-3*e^2.0794-40=64-24-40=0