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

(2log_{5}(x+1) = 2)

(x = -6)
(x = -4)
(x = -2)
(x = -1)

Question

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

(2log_{5}(x+1) = 2)

(x = -6)
  (x = -4)
  (x = -2)
  (x = -1)

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", "Extraneous Solutions of Logarithmic Equations" ], "new_concepts": [], "current_concepts": [ "Logarithmic Equations", "Extraneous Solutions of Logarithmic Equations", "Logarithmic Domain" ] } ## Solve the logarithmic equation algebraically $$ \begin{aligned} 2\log_5(x+1) &= 2 \ \log_5(x+1) &= 1 \ x+1 &= 5^1 \ x &= 4 \end{aligned} $$ ## Determine the domain restriction of the logarithmic function $$ \begin{aligned} x+1 &> 0 \ x &> -1 \end{aligned} $$ ## Identify the extraneous solution from the given options $$ \begin{aligned} &\text{Candidates from options: } x = -6, -4, -2, -1 \ &\text{Since } x > -1 \text{ is required for the logarithm to be defined, any value } x \le -1 \text{ is extraneous.} \ &\text{An extraneous solution arises when applying the power property: } \log_5(x+1)^2 = 2 \implies (x+1)^2 = 25 \ &x+1 = \pm 5 \implies x = 4 \text{ or } x = -6 \ &\text{Since } x = -6 \text{ makes the original argument } x+1 = -5 < 0\text{, } x = -6 \text{ is the extraneous solution.} \end{aligned} $$

(A) $x = -6$

(B) $x = -4$

(C) $x = -2$

(D) $x = -1$

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

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QUESTION IMAGE

Question

Explanation:

Solve the logarithmic equation algebraically

Determine the domain restriction of the logarithmic function

Identify the extraneous solution from the given options

Answer: