What is x for the logarithm exist log(x+2) (15 + 2x -x^2) ?

To determine x for which log(x+2)(15+2x-x^2) is
defined.

We know that log a exists, if a > 0. And log a is
does not exist for x< 0.

Therefore log(x+2)(15+2x-x^2) exists
if (x+2)(15+2x-x^2) > 0.

Or (x+2)(x^2-2x-15) < 0
........(1), we multiplied  by (-1). So the inequality reversed.

x^2
-2x-15 = (x+3)(x-5).

Substituting in (1) , we get: (x+2)(x+3)(x-5)
< 0.

Or  f(x) = (x+3)(x+2)(x-5) <
0.

Clearly for x> 5, f(x) > 0, as all factors are
positive.

For   -2 < x <
5, f(x) < 0 as (x+3)(x+2) positive and x-5 <
0.

For -3<x<-2, f(x) > 0, as x+3 >0 and
(x+2) <0 and (x-5) < 0.

For x<
-3, f(x) < 0 as all 3 factors are
negative.

Therefore log(x+2)(15+2x-x^2) exits  only if  (-2<
x <5).