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

Question 281573: how do you solve
ln (2x+3)+ln(x-6)-2 ln x =0
Thank you soooo much

Answer by nerdybill(7384) (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 (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=153 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 10.6846584384265, -1.68465843842649. Here's your graph:

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