Log (x+2)(x^2-4) is real iff (x+2)(x^2-4) >0, as logarithme are not real for negative values.

So (x+2)(x^2-4) > 0.

(x+2)(x+2)(x-2) > 0.

For x > 2 all facors are > 0, so LHS is > 0

For x < -2 , LHS is negative.

So x> 2 is the only solution.

We'll impose the constraints of existence of logarithms:

1) x+2 > 0

2) x + 2 different from 1

3) x^2 - 4 > 0

We'll solve the first constraint:

1) x+2 > 0

We'll subtract 2 both sides:

x > -2

2) x + 2 different from 1

We'll subtract 2 both sides:

x different from -2+1 = -1

3) x^2 - 4 > 0

We'll re-write the difference of squares:

x^2 – 4 = (x - 2)(x + 2)

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

x – 2>0 and x + 2>0

The expression is positive if x is in the interval (2 ; infinite)

x – 2<0 and x + 2<0

The expression is positive if x is in the interval (infinite ; -2)

**The interval of admissible values for the logarithm to be defined is**** ****(infinite ; -2) U (2 ; infinite).**