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

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) (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 (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=29 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 4.19258240356725, -1.19258240356725. Here's your graph:

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