You need to use the information that logarithmic function is inverse to exponential function such that:

`a^x = b => log_a (a^x) = log_a b => x = log_a b`

Since `a^x > 0` and `a^x = b => b > 0` , hence, reasoning by analogy, yields `15 + 2x -x^2 > 0` .

You need to attach the quadratic equation, such that:

`15 + 2x -x^2 = 0 => x^2 - 2x - 15 = 0`

Completing the square ` x^2 - 2x` yields:

`x^2 - 2x + 1 = 15 + 1 => (x - 1)^2 = 16 => x - 1 = +-4`

`x - 1 = 4 => x = 5`

`x - 1 = -4 => x = -3`

Graphing the parabola `y = 15 + 2x -x^2` yields:

You need to notice that the parabola is above x axis, hence `y = 15 + 2x -x^2 > 0` for `x in (-3,5).`

You also need to consider the condition `x + 2 > 0 => x > -2 => x in (-2,oo).`

**Hence, the logarithm `log_(x+2) (15 + 2x -x^2)` is valid if **`-2<x<5.`

We'll impose the constraints of existence of logarithms:

1) x+2 > 0

2) x + 2 different from 1.

3) 15 + 2x -x^2 > 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) 15 + 2x -x^2 > 0

We'll multiply by -1:

x^2 - 2x - 15 < 0

We'll calculate the roots of the quadratic:

x1 = [2 + sqrt(4+60)]/2

x1 = (2+8)/2

x1 = 5

x2 = (2-8)/2

x2 = -3

The expression is negative if x is in the interval (-3 ; 5)

From this interval, we'll reject the value -1.

The interval (-3 ; 5) - {-1} will be intersected by the interval (-2 ; +infinity).

The interval of admissible values for the logarithm to be defined is (-2;5).