log(x+1) (x^2+3x-14)=2

Supposing that x + 1 is the base of logarithm, we'll
impose the constraints of existence of logarithms:

x+ 1
>0

x + 1different from
1

x different from
0

x>-1

x^2+3x-14
> 0

x^2+3x-14 = 0

x1 =
[-3+sqrt(9 + 56)]/2

x1 = (-3 +
sqrt65)/2

The range of admissible values of x for the
logarithms to exists, is: ((-3 + sqrt65)/2 ;
+infinitte).

We'll solve the
equation:

x^2+3x-14 =
(x+1)^2

We'll expand the square from the right
side:

x^2+3x-14 = x^2 + 2x +
1

We'll eliminate x^2:

3x-14 =
2x + 1

We'll subtract 2x + 1 both
side:

3x - 2x - 14 - 1 = 0

x -
15 = 0

x =
15

Since the value of x belongs to the
interval of admissible values, we'll accept as solution of the equation x =
15.