To solve the equation [tex]\(\log_3 (x^2 + 6x) = \log_3 (2x + 12)\)[/tex], let's follow a systematic approach.
### Step 1: Use properties of logarithms
Given that the logs are to the same base, we can set the arguments equal to each other:
[tex]\[ x^2 + 6x = 2x + 12 \][/tex]
### Step 2: Rearrange the equation
Move all terms to one side to form a quadratic equation:
[tex]\[ x^2 + 6x - 2x - 12 = 0 \][/tex]
[tex]\[ x^2 + 4x - 12 = 0 \][/tex]
### Step 3: Solve the quadratic equation
We solve the quadratic equation [tex]\(x^2 + 4x - 12 = 0\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this equation, [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = -12\)[/tex]. Substituting these values into the quadratic formula gives:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot -12}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 48}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 8}{2} \][/tex]
### Step 4: Calculate the potential solutions
This produces two potential solutions:
[tex]\[ x = \frac{-4 + 8}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-4 - 8}{2} = \frac{-12}{2} = -6 \][/tex]
### Step 5: Check domain restrictions
Since the original equation involves logarithms, we must ensure the arguments of the logarithms are positive. Therefore, both [tex]\(x^2 + 6x\)[/tex] and [tex]\(2x + 12\)[/tex] should be positive for the solutions to be valid.
- For [tex]\(x = 2\)[/tex]:
[tex]\[ x^2 + 6x = 2^2 + 6 \cdot 2 = 4 + 12 = 16 \][/tex]
[tex]\[ 2x + 12 = 2 \cdot 2 + 12 = 4 + 12 = 16 \][/tex]
Both expressions are positive.
- For [tex]\(x = -6\)[/tex]:
[tex]\[ x^2 + 6x = (-6)^2 + 6 \cdot (-6) = 36 - 36 = 0 \][/tex]
[tex]\[ 2x + 12 = 2 \cdot (-6) + 12 = -12 + 12 = 0 \][/tex]
Both expressions become zero, which is not within the logarithm's domain (logarithm of zero is undefined).
Hence, the valid solution is:
[tex]\[ \boxed{x = 2} \][/tex]
