SOLUTION: Solve equation algebraically. 5e^x=10-e^-x

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Question 644596: Solve equation algebraically. 5e^x=10-e^-x
Answer by Alan3354(69443) About Me  (Show Source):
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Solve equation algebraically. 5e^x=10-e^-x
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5e^x=10-e^-x
5e%5Ex+=+10+-+e%5E%28-x%29
Multiply by e^x
5e%5E2x+=+10e%5Ex+-+1
5e%5E2x+-+10e%5Ex+%2B+1+=+0
Sub y for e^x
5y%5E2+-+10y+%2B+1+=+0
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 5x%5E2%2B-10x%2B1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-10%29%5E2-4%2A5%2A1=80.

Discriminant d=80 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--10%2B-sqrt%28+80+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-10%29%2Bsqrt%28+80+%29%29%2F2%5C5+=+1.89442719099992
x%5B2%5D+=+%28-%28-10%29-sqrt%28+80+%29%29%2F2%5C5+=+0.105572809000084

Quadratic expression 5x%5E2%2B-10x%2B1 can be factored:
5x%5E2%2B-10x%2B1+=+%28x-1.89442719099992%29%2A%28x-0.105572809000084%29
Again, the answer is: 1.89442719099992, 0.105572809000084. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5%2Ax%5E2%2B-10%2Ax%2B1+%29

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e^x = 1 + sqrt(0.8)
x = ln(1+sqrt(0.8))
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e^x = 1 - sqrt(0.8)
x = ln(1-sqrt(0.8))
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Just calculator work left.