Question 1000717
e^(x)+e^(-x)=3
x=0.9624 and -0.9624 but please explain how to get there.
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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.
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e^x = u
x = ln(u)
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As the other tutor mentioned, this is the hyperbolic cosine.
The hyperbolic sine is {{{(e^(x)-e^(-x))/2}}}