what are the solutions to $\log_{6}(x^2 + 8) = 1 + log_{6}(x)$?

$x = -2$ and $x = -4$
$x = -1$ and $x = 8$
$x = 1$ and $x = -8$
$x = 2$ and $x = 4$

Question

what are the solutions to $\log_{6}(x^2 + 8) = 1 + log_{6}(x)$?

$x = -2$ and $x = -4$
$x = -1$ and $x = 8$
$x = 1$ and $x = -8$
$x = 2$ and $x = 4$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Logarithmic Equations", "Logarithmic Properties", "Extraneous Solutions of Logarithmic Equations" ], "new_concepts": [], "current_concepts": [ "Logarithmic Equations", "Logarithmic Properties", "Extraneous Solutions of Logarithmic Equations" ] } ## Express the constant as a logarithm and apply properties $$ \begin{aligned} &\log_6(x^2 + 8) = 1 + \log_6(x) \ &\log_6(x^2 + 8) = \log_6(6) + \log_6(x) \ &\log_6(x^2 + 8) = \log_6(6x) \end{aligned} $$ ## Solve the resulting quadratic equation $$ \begin{aligned} &x^2 + 8 = 6x \ &x^2 - 6x + 8 = 0 \ &(x - 2)(x - 4) = 0 \ &x = 2 \quad \text{or} \quad x = 4 \end{aligned} $$ ## Verify the solutions against domain restrictions $$ \begin{aligned} &\text{Domain restriction: } x > 0 \ &x = 2 > 0 \quad (\text{valid}) \ &x = 4 > 0 \quad (\text{valid}) \end{aligned} $$

$x = -2$ and $x = -4$

$x = -1$ and $x = 8$

$x = 1$ and $x = -8$

$x = 2$ and $x = 4$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Logarithmic Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Express the constant as a logarithm and apply properties

Solve the resulting quadratic equation

Verify the solutions against domain restrictions

Answer: