SOLUTION: e^(x)+e^(-x)=3 x=0.9624 and -0.9624 but please explain how to get there. Thanks

Question 1000717: e^(x)+e^(-x)=3
x=0.9624 and -0.9624 but please explain how to get there.
Thanks
Found 2 solutions by fractalier, Alan3354:
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
You need the hyperbolic cosine function...
cosh x = (1/2)(e^x + e^(-x))
so here, if you plug in what you are given you get
cosh x = (1/2)(3) = 1.5
Now take the inverse hyperbolic cosine and get
x = .9624 and x = -.9624
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
e^(x)+e^(-x)=3
x=0.9624 and -0.9624 but please explain how to get there.
------------
e^(x)+e^(-x)=3
Multiply thru by e^x
e^(2x)+ 1 = 3e^x
e^2x - 3e^x + 1 = 0
Sub u for e^x
u^2 - 3u + 1 = 0
Solve for u.
--------
e^x = u
x = ln(u)
========================
As the other tutor mentioned, this is the hyperbolic cosine.
The hyperbolic sine is
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