SOLUTION: how do you solve ln (2x+3)+ln(x-6)-2 ln x =0 Thank you soooo much

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Question 281573: how do you solve
ln (2x+3)+ln(x-6)-2 ln x =0
Thank you soooo much
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Applying "log rules":
ln (2x+3)+ln(x-6)-2 ln x =0
ln (2x+3)+ln(x-6)- ln x^2 =0
ln (2x+3)(x-6)- ln x^2 =0
ln [(2x+3)(x-6)]/x^2 =0
[(2x+3)(x-6)]/x^2 = e^0
[(2x+3)(x-6)]/x^2 = 1
(2x+3)(x-6) = x^2
2x^2-12x+3x-18 = x^2
2x^2-9x-18 = x^2
x^2-9x-18 = 0
Solve by applying the quadratic equation. This results in:
x = {10.685, -1.685}
But, we can toss out the negative solution since this results in taking a ln of a negative number. This is an extraneous solution -- toss it out. Leaving our solution as:
x = 10.685
.
Details of quadratic equation follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-9x%2B-18+=+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=%28-9%29%5E2-4%2A1%2A-18=153.

Discriminant d=153 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--9%2B-sqrt%28+153+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-9%29%2Bsqrt%28+153+%29%29%2F2%5C1+=+10.6846584384265
x%5B2%5D+=+%28-%28-9%29-sqrt%28+153+%29%29%2F2%5C1+=+-1.68465843842649

Quadratic expression 1x%5E2%2B-9x%2B-18 can be factored:
1x%5E2%2B-9x%2B-18+=+1%28x-10.6846584384265%29%2A%28x--1.68465843842649%29
Again, the answer is: 10.6846584384265, -1.68465843842649. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-9%2Ax%2B-18+%29