SOLUTION: log[3](1-x)+log[3](2-x)=log[3](7) the numbers in brackets are the bases. Please help

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Question 246678: log[3](1-x)+log[3](2-x)=log[3](7) the numbers in brackets are the bases. Please help
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
log[3](1-x)+log[3](2-x)=log[3](7)
log[3](1-x)(2-x)=log[3](7)
(1-x)(2-x)=7
2-x-2x+x^2=7
2-3x+x^2=7
x^2-3x+2=7
x^2-3x-5=0
Applying the quadratic equation will yield:
x = {4.193, -1.193}
Checking for extraneous solutions:
We can throw out the 4.193 solution because it will give us a negative log. So:
x = {-1.193}
.
Details of quadratic to follow:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-3x%2B-5+=+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=%28-3%29%5E2-4%2A1%2A-5=29.

Discriminant d=29 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+29+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+29+%29%29%2F2%5C1+=+4.19258240356725
x%5B2%5D+=+%28-%28-3%29-sqrt%28+29+%29%29%2F2%5C1+=+-1.19258240356725

Quadratic expression 1x%5E2%2B-3x%2B-5 can be factored:
1x%5E2%2B-3x%2B-5+=+1%28x-4.19258240356725%29%2A%28x--1.19258240356725%29
Again, the answer is: 4.19258240356725, -1.19258240356725. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-3%2Ax%2B-5+%29