Question 192786
    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:
*[invoke quadratic "x", 1, -5, -3 ]