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e^x+e^(-x)=8(e^x-e^(-x))

 Oct 16, 2016
 #1
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plz tell me how to calculate the x

 Oct 16, 2016
 #2
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Solve for x over the real numbers:
e^(-x)+e^x = 8 (e^x-e^(-x))

8 (e^x-e^(-x)) = 8 e^x-8 e^(-x):
e^(-x)+e^x = 8 e^x-8 e^(-x)

Multiply both sides by e^x:
1+e^(2 x) = 8 e^(2 x)-8

Subtract 1+8 e^(2 x) from both sides:
-7 e^(2 x) = -9

Divide both sides by -7:
e^(2 x) = 9/7

Take the natural logarithm of both sides:
2 x = log(9/7)

Divide both sides by 2:
Answer: |x = 1/2 log(9/7)

 Oct 16, 2016

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