# Solve the equation : log 3 (x^2 + 4x + 12) =2

neela | High School Teacher | (Level 3) Valedictorian

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log3 (x^2+4x+12) = 2. To solve this we proceed as below:

We know if loga(b) = x, then  b = a^x.

Therefore x^2+4x+12 = 3^2

x^2+4x+12 = 9

x^2+4x+12-9 = 0.

x^2+4x+3 = 0

We proceed to factorise the left:

x^2+3x+x+3 = 0

x(x+3) +1(x+3) = 0

(x+3)(x+1) = 0

Equate each factor to zero, as right side is zero.

x+3 = 0 or x+1 = 0

x +3 = 0 gives x = -3.

x + 1 = 0 gives x = -1.

Therefore x = -1, or x= -3 are the solution to the given logarithmatic equation.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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First, we'll verify if the argument of the logarithm is positive. For this reason, we'll calculate the discriminant of the quadratic.

If the discriminant is negative and the coefficient of x^2 is positive, then the expression x^2 + 4x + 12 is positive for any value of x.

delta = b^2 - 4ac

We'll identify the coefficients a,b,c:

a = 1

b = 4

c = 12

delta = 16 - 4*12

delta = 16 - 48

delta = -32

Since delta is negative and a is positive, the expression x^2 + 4x + 12 > 0.

Now, we'll solve the equation. We'll take anti-logarithm:

x^2 + 4x + 12 = 3^2

x^2 + 4x + 12 = 9

We'll subtract 9 both sides:

x^2 + 4x + 12 - 9 = 0

We'll combine like terms:

x^2 + 4x + 3 = 0

x1 = [-4 +/- sqrt(16-12)]/2

x1 = (-4+2)/2

x1 = -1

x2 = -3

Since all values of x are admissible, we'll not reject either of resulted roots.

The solutions of the equation are: {-3 ; -1}.