SOLUTION: Log(x-3) + log(2x + 1) = 2logx

Question 192786: Log(x-3) + log(2x + 1) = 2logx
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Log(x-3) + log(2x + 1) = 2logx
Log(x-3)(2x + 1) = logx^2
(x-3)(2x + 1) = x^2
2x^2 +x -6x - 3 = x^2
2x^2 - 5x - 3 = x^2
x^2 - 5x - 3 = 0
.
Using the quadratic equation to solve for x yields:
x = {5.541, -0.541}
We can toss out the negative solution leaving:
x = 5.541
.
Details of quadratic:

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=37 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 5.54138126514911, -0.54138126514911. Here's your graph:

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