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