# Solve ln(3x+8)=ln(2x+2)+ln(x-2) for x.

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We'll impose the constraints of existence of logarithms:

3x+8>0

x>-8/3

2x+2>0

x>-1

x-2>0

x>2

The interval of admissible values for x is (2, +infinite).

Now, we'll solve the equation:

ln(3x+8)=ln(2x+2)+ln(x-2)

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.**