We have to find x given that lg(2x+2)=lg(8+3x)/(x-2)

Now log (a/b) = log a - log b

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

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

=> lg (2x + 2) - lg ( 8 + 3x) + lg ( x -2 ) = 0

=> lg [( 2x + 2)( x - 2) / ( 8 + 3x)] = 0

=> ( 2x + 2)( x - 2) / ( 8 + 3x) = 1

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

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

=> 2x^2 + 2x - 4x - 3x - 4 - 8 =0

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

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

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

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

Therefore x = -2 and 4.

We can eliminate x = -2 as lg(8+3x)/(x-2) would not be defined.

**Therefore x = 4.**

Find x x if lg(2x+2)=lg(8+3x)/(x-2).

lg(2x+2)=lg(8+3x)/(x-2).

We take the anti log of both sides:

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

We multiply by (x-2).

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

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

2x^2-2x-4-8-3x = 0

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

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

2x+3 = 0. Or x-4 = 0.

x = -3/2 ). Or 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.**