SOLUTION: solve the equation. log(x+2)= -log(x-1) +1

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Question 185230: solve the equation. log(x+2)= -log(x-1) +1
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
solve the equation. log(x+2)= -log(x-1) +1
Assuming all are "common logs", base 10
log(x+2) + log(x-1) = 1
log((x+2)*(x-1)) = log(10)
(x+2)*(x-1) = 10
x^2 + x - 2 = 10
x^2 + x - 12 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-12+=+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=%281%29%5E2-4%2A1%2A-12=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+49+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+49+%29%29%2F2%5C1+=+3
x%5B2%5D+=+%28-%281%29-sqrt%28+49+%29%29%2F2%5C1+=+-4

Quadratic expression 1x%5E2%2B1x%2B-12 can be factored:
1x%5E2%2B1x%2B-12+=+%28x-3%29%2A%28x--4%29
Again, the answer is: 3, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-12+%29

Ignore the -4
x = 3