Question 898387
your equation is:


ln(2x+3) + ln(x-6) - 2*ln(x) = 0



since ln(2x+3) + ln(x-6) = ln((2x+3) * (x-6)), your equation becomes:


ln((2x+3)*(x-6)) - 2*ln(x) = 0


since 2*ln(x) = ln(x^2) and since ln((2x+3)*(x-6)) - ln(x^2) = ln((2x+3)*(x-6)/x^2), your equation becomes:


ln((2x+3)*(x-6)/x^2) = 0


this is true if and only if:


e^0 = (2x+3)*(x-6) / x^2


since e^0 = 1, your equation becomes:


1 = (2x+3)*(x-6) / x^2


multiply both sides of this equation by x^2 and simplify to get:


x^2 = 2x^2 - 12x + 3x - 18 = 2x^2 - 9x - 18


subtract x^2 from both sides of this equation to get:


0 = x^2 - 9x - 18


factor this equation using the quadratic formula to get:


x = 10.68465... or x = -1.68465...


since x has to be positive, the solution is x = 10.68465... which rounds to 10.68.