Solve ln(3x+8)=ln(2x+2)+ln(x-2) for x.
We'll impose the constraints of existence of logarithms:
The interval of admissible values for x is (2, +infinite).
Now, we'll solve the equation:
Since the logarithms have the matching bases, we'll apply the product rule to the right side:
ln(3x+8) = ln[(2x+2)*(x-2)]
Since the bases are matching, we'll apply one to one rule:
3x + 8 = [(2x+2)*(x-2)]
We'll remove the brackets from the right side:
3x + 8 = 2x^2 - 4x + 2x - 4
We'll move all terms to the right side and we'll use symmetric property:
2x^2 - 4x + 2x - 4 - 3x - 8 = 0
2x^2 - 5x - 12 = 0
We'll apply quadratic formula;
x1 = [5 + sqrt(25 + 96)]/4
x1 = (5+11)/4
x1 = 16/4
x1 = 4
x2 = -6/4
x2 = -3/2
Since the second solution is not in the interval of admissible values, we'll reject it. The only valid solution is x = 4.