To find the solutions to the equation [tex]\(\log_4 x + \log_4 (x+6) = 2\)[/tex], follow these steps:
1. Combine the logarithms: Use the property of logarithms that states [tex]\(\log_b a + \log_b c = \log_b (a \cdot c)\)[/tex]. This simplifies the given equation:
[tex]\[
\log_4 x + \log_4 (x+6) = \log_4 [x \cdot (x + 6)]
\][/tex]
So, the equation becomes:
[tex]\[
\log_4 [x \cdot (x + 6)] = 2
\][/tex]
2. Convert the logarithmic equation to an exponential equation: Recall that if [tex]\(\log_b y = z\)[/tex], then [tex]\(y = b^z\)[/tex]. Applying this to the equation [tex]\(\log_4 [x \cdot (x + 6)] = 2\)[/tex], we get:
[tex]\[
x \cdot (x + 6) = 4^2
\][/tex]
Simplify [tex]\(4^2\)[/tex]:
[tex]\[
x \cdot (x + 6) = 16
\][/tex]
3. Form a quadratic equation: Distribute [tex]\(x\)[/tex] in the equation [tex]\(x \cdot (x + 6) = 16\)[/tex]:
[tex]\[
x^2 + 6x = 16
\][/tex]
Move 16 to the left side of the equation to set it to zero:
[tex]\[
x^2 + 6x - 16 = 0
\][/tex]
4. Solve the quadratic equation: Factor the quadratic equation [tex]\(x^2 + 6x - 16 = 0\)[/tex]. We look for two numbers that multiply to [tex]\(-16\)[/tex] and add to [tex]\(6\)[/tex]. These numbers are [tex]\(8\)[/tex] and [tex]\(-2\)[/tex].
So, we can factor the quadratic equation as:
[tex]\[
(x + 8)(x - 2) = 0
\][/tex]
5. Find the solutions: Set each factor equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
x + 8 = 0 \quad \Rightarrow \quad x = -8
\][/tex]
[tex]\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\][/tex]
6. Consider the domain of the logarithmic function: Logarithmic functions are only defined for positive arguments. Therefore, we must check that [tex]\(x\)[/tex] and [tex]\(x + 6\)[/tex] are positive:
- For [tex]\(x = 2\)[/tex]:
[tex]\[
\log_4 2 + \log_4 (2 + 6) = \log_4 2 + \log_4 8 = 2
\][/tex]
(This is valid because [tex]\(2 > 0\)[/tex] and [tex]\(2 + 6 > 0\)[/tex]).
- For [tex]\(x = -8\)[/tex]:
[tex]\[
\log_4 (-8) + \log_4 (-8 + 6) = \log_4 (-8) + \log_4 (-2) = \text{undefined}
\][/tex]
(This is invalid because [tex]\(-8 < 0\)[/tex] and [tex]\(-8 + 6 < 0\)[/tex]).
Therefore, the only valid solution is:
[tex]\[
x = 2
\][/tex]
The answer is [tex]\( \boxed{x = 2 \text{, } x = -8} \)[/tex]. However, only [tex]\(x = 2\)[/tex] is valid, considering the domain of the logarithmic function.
