We have to solve log (2x+2) = 1 - log 3x

log (2x+2) = 1 - log 3x

=> log (2x+2) + log 3x = 1

use the property log a + log b = log a*b

=> log (2x + 2)*3x = 1

=> (2x + 2)*3x = 10^1

=> 6x^2 + 6x - 10 = 0

=> 3x^2 + 3x - 5 = 0

x1 = -3/6 + sqrt (69)/6

x1 = -1/2 + sqrt 69 / 6

x2 = -1/2 - sqrt 69 / 6

We see that log 3x is not defined for x = -1/2 - sqrt 69 / 6, so we eliminate it.

**The solution of the equation is -1/2 + (sqrt 69)/6**

We'll impose the constraints of existence of logarithms:

2x+ 2 >0

2x>-2

x>-1

3x>0

x>0

The interval of admissible solutions for the given equation is: (0 ; +infinite).

Now, we'll sove the equation, adding log 3x both sides:

log (2x+2) + log 3x = 1

We'll apply the product rule:

log [6x(x+1)] = 1

We'll take antilogarithm:

6x(x+1) = 10

We'll remove the brackets:

6x^2 + 6x - 10 = 0

We'll divide by 2:

3x^2 + 3x - 5 = 0

We'll apply the quadratic formula:

x1 = [-3+sqrt(9 + 60)]/6

x1 = [-3+sqrt(69)]/6

x2 = [-3-sqrt(69)]/6

Since the 2nd value is negative, it doesn't belong to the range of admissible values, so we'll reject it.

**The only valid solution of the equation is: **

**x1 = [-3+sqrt(69)]/6**