SOLUTION: Solve for x algebraically: ((e^x)-(e^-x))/2=5. Consider exponential equations of quadratic type. I think the answer is supposed to be x=ln(5+sqrt(26)) but I can't figure out the

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve for x algebraically: ((e^x)-(e^-x))/2=5. Consider exponential equations of quadratic type. I think the answer is supposed to be x=ln(5+sqrt(26)) but I can't figure out the      Log On


   

Question 974322: Solve for x algebraically: ((e^x)-(e^-x))/2=5. Consider exponential equations of quadratic type.
I think the answer is supposed to be x=ln(5+sqrt(26)) but I can't figure out the steps to get it. So far I have (1/2)((e^x)-(e^-x))-5=0. From here I'm lost.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x algebraically: ((e^x)-(e^-x))/2=5
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e%5Ex+-+e%5E%28-x%29+=+10
Multiply by e^x
e%5E%282x%29+-+1+=+10e%5Ex
e%5E%282x%29+-+10e%5Ex+-+1+=+0
Sub u for e^x
u%5E2+-+10u+-+1+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-10x%2B-1+=+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%2A1%2A-1=104.

Discriminant d=104 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--10%2B-sqrt%28+104+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-10%29%2Bsqrt%28+104+%29%29%2F2%5C1+=+10.0990195135928
x%5B2%5D+=+%28-%28-10%29-sqrt%28+104+%29%29%2F2%5C1+=+-0.0990195135927845

Quadratic expression 1x%5E2%2B-10x%2B-1 can be factored:
1x%5E2%2B-10x%2B-1+=+%28x-10.0990195135928%29%2A%28x--0.0990195135927845%29
Again, the answer is: 10.0990195135928, -0.0990195135927845. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-10%2Ax%2B-1+%29

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The answer matches.