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

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.