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

*print*Print*list*Cite

### 2 Answers

We are given that log(x+1) (x^2+3x-14)=2.

We know that log(a) b = c => b = c^a

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

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

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

=> x = 15

**The required value of x is 15.**

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.**