What are the potential solutions of [tex]\(\log_4 x + \log_4 (x+6) = 2\)[/tex]?

A. [tex]\(x = -2\)[/tex] and [tex]\(x = -8\)[/tex]
B. [tex]\(x = -2\)[/tex] and [tex]\(x = 8\)[/tex]
C. [tex]\(x = 2\)[/tex] and [tex]\(x = -8\)[/tex]
D. [tex]\(x = 2\)[/tex] and [tex]\(x = 8\)[/tex]



Answer :

To solve the given equation [tex]\(\log_4 x + \log_4 (x + 6) = 2\)[/tex], let's follow the steps:

1. Understand the logarithmic properties:
Recall the property of logarithms that states [tex]\(\log_b(m) + \log_b(n) = \log_b(m \cdot n)\)[/tex].

2. Combine the logarithms:
Using the property, we can combine the logarithms on the left-hand side:
[tex]\[ \log_4 x + \log_4 (x + 6) = \log_4 [x(x + 6)] \][/tex]

3. Substitute and simplify:
Substitute the combined logarithm back into the equation:
[tex]\[ \log_4 [x(x + 6)] = 2 \][/tex]

4. Exponentiate to remove the logarithm:
To isolate [tex]\(x\)[/tex], rewrite the equation in exponential form:
[tex]\[ x(x + 6) = 4^2 \][/tex]
[tex]\[ x(x + 6) = 16 \][/tex]

5. Solve the quadratic equation:
The equation [tex]\(x(x + 6) = 16\)[/tex] simplifies to:
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
To solve this quadratic equation, we use 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].

6. Apply the quadratic formula:
[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]

7. Calculate the solutions:
This gives us two potential solutions:
[tex]\[ x = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8 \][/tex]

8. Check the validity of the solutions:
Since logarithms are only defined for positive real numbers in this context, we must ensure that both [tex]\(x\)[/tex] and [tex]\(x+6\)[/tex] are positive:
- For [tex]\(x = 2\)[/tex]: [tex]\(\log_4 2\)[/tex] and [tex]\(\log_4 8\)[/tex] are valid.
- For [tex]\(x = -8\)[/tex]: [tex]\(\log_4 (-8)\)[/tex] is not valid as logarithms of negative numbers are undefined in the real number system.

Thus, the only valid solution is:
[tex]\[ x = 2 \][/tex]

Therefore, the potential solution among the given choices is:
[tex]\[ x = 2 \][/tex]

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