Question 1239
 Given ln(3x+8)=ln(2x+2)+ln(x-2)
 Since the right hand side ln(2x+2)+ln(x-2) = ln(2x+2)(x-2) = ln(2x^2 -2x -4),
 So,ln(3x+8) = ln(2x^2 -2x -4) and we have

 3x+8 = 2x^2 -2x -4, or 2x^2 -5x -12  = 0.
 Factor: (2x + 3)(x - 4) = 0.
 So, x = -3/2 or 4 

 But by the definition of ln (log), 3x +8 , 2(x+1) and x-2 must be positive.
 Hence, x > -8/3, x > -1 and x > 2. Therefore, x must be greater than 2 and
 the solution -3/2 is invalid. The only solution is x =4.

 Check,when x = 4, ln(3x+8)=ln(2x+2)+ln(x-2) becomes
 ln 20 = ln 10 + ln 2 = ln 10*2 ,so 4 is a correct solution of the
 given equation.