SOLUTION: 2^x-2^(-x)=4

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Question 368115: 2^x-2^(-x)=4
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
2^x-2^(-x)=4
Multiply by 2^x
2^(2x) - 1 = 4*2^x
Sub y for 2^x
y^2 - 1 = 4y
y^2 - 4y - 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 1x%5E2%2B-4x%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=%28-4%29%5E2-4%2A1%2A-1=20.

Discriminant d=20 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+20+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+20+%29%29%2F2%5C1+=+4.23606797749979
x%5B2%5D+=+%28-%28-4%29-sqrt%28+20+%29%29%2F2%5C1+=+-0.23606797749979

Quadratic expression 1x%5E2%2B-4x%2B-1 can be factored:
1x%5E2%2B-4x%2B-1+=+%28x-4.23606797749979%29%2A%28x--0.23606797749979%29
Again, the answer is: 4.23606797749979, -0.23606797749979. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B-1+%29

y = 1 ± sqrt(5)
2^x = 1 + sqrt(5) --- Ignore the negative answer
x*log(2) = log(1 + sqrt(5))
x = log(1 + sqrt(5))/log(2)
x =~ 1.69242