which of the following shows the extraneous solution to the logarithmic equation?

$\log_{4}(x)+\log_{4}(x-3)=\log_{4}(-7x+21)$

$x=-7$
$x=-3$
$x=3$ and $x=-7$
$x=7$ and $x=-3$

Question

which of the following shows the extraneous solution to the logarithmic equation?

$\log_{4}(x)+\log_{4}(x-3)=\log_{4}(-7x+21)$

$x=-7$
$x=-3$
$x=3$ and $x=-7$
$x=7$ and $x=-3$

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" ], "new_concepts": [ "Extraneous Solutions of Logarithmic Equations" ], "current_concepts": [ "Logarithmic Equations", "Extraneous Solutions of Logarithmic Equations" ] } ## Solve the logarithmic equation $$ \begin{aligned} \log_4(x) + \log_4(x-3) &= \log_4(-7x+21) \ \log_4(x(x-3)) &= \log_4(-7x+21) \ x(x-3) &= -7x+21 \ x^2 - 3x &= -7x+21 \ x^2 + 4x - 21 &= 0 \ (x+7)(x-3) &= 0 \ x = -7 \quad &\text{or} \quad x = 3 \end{aligned} $$ ## Check for extraneous solutions $$ \begin{aligned} &\text{For } x = -7: \ &\log_4(-7) \text{ is undefined (argument } \le 0\text{)} \ \ &\text{For } x = 3: \ &\log_4(3-3) = \log_4(0) \text{ is undefined (argument } \le 0\text{)} \end{aligned} $$ ## Identify the extraneous solutions $$ x = 3 \text{ and } x = -7 $$

$x = -7$

$x = -3$

$x = 3 \text{ and } x = -7$

$x = 7 \text{ and } x = -3$

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

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QUESTION IMAGE

Question

Explanation:

Solve the logarithmic equation

Check for extraneous solutions

Identify the extraneous solutions

Answer: