We have to solve lg(2x+2)=lg[(8+3x)/(x-2)]

Equate 2x + 2 = (8+3x)/(x-2)

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

=> 2x^2 + 2x - 4x - 4 = 8 + 3x

=> 2x^2 - 5x - 12 = 0

=> 2x^2 - 8x + 3x - 12 = 0

=> 2x(x - 4) + 3(x - 4) = 0

=> (2x + 3) = 0 and (x - 4) = 0

=> x = -3/2 and x = 4.

For x = -3/2, the given logarithms become logs of negative numbers and are undefined.

**The solution of the equation is x = 4**

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:

lg(2x+2)=lg [(8+3x)/(x-2)]

Since the logarithms have the matching bases, we'll apply the one to one rule:

2x + 2 = [(8+3x)/(x-2)]

We'll cross multiply and we'll get:

[(2x+2)*(x-2)] = 8 + 3x

We'll remove the brackets from the left side:

2x^2 - 4x + 2x - 4 = 8 + 3x

We'll move all terms to the left 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 does not belong to the range of admissible values, we'll reject it. **

**The only admissible solution is x = 4.**