Question

what are the potential solutions of \\(\log_{6}x + \log_{6}(x + 5) = 2\\)?

-12
-9
4
32
36

Explanation:

Apply the product rule of logarithms

Using the Logarithmic Product Rule knowledge point

$$
LATEXBLOCK0
$$

Convert to exponential form

Using the Solving Logarithmic Equations knowledge point

$$
LATEXBLOCK1
$$

Solve the quadratic equation

Using the Polynomial Factoring knowledge point

$$
LATEXBLOCK2
$$

Check for extraneous solutions

Using the Extraneous Solutions knowledge point

$$
LATEXBLOCK3
$$

Thus, the potential algebraic solutions are \(x = -9\) and \(x = 4\), but only \(x = 4\) is a valid solution. The question asks for "potential solutions" (which typically refers to the algebraic solutions obtained before checking the domain restrictions, or the options are checkboxes to select all candidate values). Looking at the options, both \(-9\) and \(4\) are listed. Let's check the wording "potential solutions" which usually means the solutions to the algebraic equation derived, or the valid solutions. Since it is a multiple-choice question with checkboxes, the algebraic solutions derived are \(-9\) and \(4\).

Answer:

• -12
• -9 (Correct answer)
• 4 (Correct answer)
• 32
• 36