You need first to test in what conditions the given logarithmic equation holds, snce both base and argument depend on variable x.

You need to remember that the base of all logarithms has the following properties:

`x + 2 > 0 => x > -2`

`x + 2 != 1 => x != -1`

You also need to remember that the argument of logarithm needs to fulfil the following condition, such that:

`3x + 6 > 0 => x > -6/3 => x > -2`

Hence, the equation holds for all real values `x in (-2,oo)-{-1}.`

You may solve the equation using the following property of logarithms, such that:

`log_(x+2) (3x + 6) = 2 => 3x + 6 = (x+2)^2`

Expanding the square yields:

`3x + 6 = x^2 + 4x + 4 => x^2 + x - 2 = 0`

Using quadratic formula yields:

`x_(1,2) = (-1+-sqrt(1+8))/2 => x_(1,2) = (-1+-3)/2`

`x_1 = 1, x_2 = -2`

**You should notice that only one value belongs to the interval (-2,oo)-{-1}, hence, the equation has the solution `x = 1` .**