SOLUTION: ((e^x)+(4e^-x))/2 = 2
solution in terms of natural log and answer in decimal approx.?????
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Question 694560: ((e^x)+(4e^-x))/2 = 2
solution in terms of natural log and answer in decimal approx.?????
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
((e^x)+(4e^-x))/2 = 2
solution in terms of natural log and answer in decimal approx.?????
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e^x + [4/e^x] = 4
Multiply thru by e^x to get:
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e^(2x) + 4 = 4e^x
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Rearrange to quadratic form:
(e^x)^2 - 4e^x +4 = 0
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Solve using quadratic formula.
e^x = [4 +- sqrt(16-4)]/2
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e^x = [4+- 2sqrt(3)]/2
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e^x = [2 +- sqrt(3)[
---
Separate:
e^x = 2+sqrt(3) or e^x = 2-sqrt(3
-----
x = ln(2+sqrt(3)) or x = ln(2-sqrt(3))
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Approximation answers:
x = 1.3170 or x = -1.3170
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Cheers,
Stan H.
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