solve the eqation: log (2x+2) = 1 - log 3x

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