To solve the equation [tex]\(\log_4 x + \log_4 (x + 6) = 2\)[/tex], we start by using properties of logarithms. The logarithmic sum [tex]\(\log_a b + \log_a c\)[/tex] can be combined to [tex]\(\log_a (bc)\)[/tex]. Thus, we can rewrite the given equation as:
[tex]\[
\log_4 (x(x + 6)) = 2
\][/tex]
This implies that:
[tex]\[
\log_4 (x^2 + 6x) = 2
\][/tex]
To simplify further, we can rewrite the logarithmic equation using exponential form. Recall that [tex]\(\log_a b = c\)[/tex] is equivalent to [tex]\(a^c = b\)[/tex]. Thus, we have:
[tex]\[
4^2 = x^2 + 6x
\][/tex]
Which simplifies to:
[tex]\[
16 = x^2 + 6x
\][/tex]
This is a quadratic equation. To solve it, we bring all terms to one side of the equation:
[tex]\[
x^2 + 6x - 16 = 0
\][/tex]
Next, we solve this quadratic equation using the quadratic formula [tex]\(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -16\)[/tex]:
[tex]\[
x = \frac{{-6 \pm \sqrt{{6^2 - 4 \cdot 1 \cdot (-16)}}}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{{-6 \pm \sqrt{{36 + 64}}}}{2}
\][/tex]
[tex]\[
x = \frac{{-6 \pm \sqrt{100}}}{2}
\][/tex]
[tex]\[
x = \frac{{-6 \pm 10}}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{{-6 + 10}}{2} = \frac{4}{2} = 2
\][/tex]
[tex]\[
x = \frac{{-6 - 10}}{2} = \frac{-16}{2} = -8
\][/tex]
We have [tex]\(x = 2\)[/tex] and [tex]\(x = -8\)[/tex] as the potential solutions. However, not all potential solutions are necessarily valid in the context of logarithms. The arguments of the logarithms must be positive. Therefore, we need:
[tex]\[
x > 0 \quad \text{and} \quad x + 6 > 0
\][/tex]
From [tex]\(x > 0\)[/tex], it is clear that [tex]\(x = -8\)[/tex] is not valid because it does not satisfy the requirement. So, we discard [tex]\(x = -8\)[/tex].
The only valid solution is:
[tex]\[
x = 2
\][/tex]
Thus, the correct answer is:
[tex]\[
x = 2 \quad \text{and} \quad x = -8 \quad (\text{Potential, but only } x = 2 \text{ is} \quad \text{valid})
\][/tex]
So, in the multiple choice context, the correct answer is:
[tex]\[
x = 2 \quad \text{and} \quad x = -8
\][/tex]
