which of the following shows the true solution to the logarithmic equation solved below?

$\log_{2}(x) + \log_{2}(x+7) = 3$
$\log_{2}x(x+7) = 3$
$x(x+7) = 2^{3}$
$x^{2} + 7x - 8 = 0$
$(x+8)(x-1) = 0$
$x = -8, 1$

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

Question

which of the following shows the true solution to the logarithmic equation solved below?

$\log_{2}(x) + \log_{2}(x+7) = 3$
$\log_{2}x(x+7) = 3$
$x(x+7) = 2^{3}$
$x^{2} + 7x - 8 = 0$
$(x+8)(x-1) = 0$
$x = -8, 1$

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

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" ] } ## Identify potential solutions from the algebraic steps $$ x = -8, \quad x = 1 $$ ## Check domain restrictions of the original logarithmic equation $$ \begin{aligned} &\text{For } \log_2(x) \text{ to be defined: } x > 0 \ &\text{For } \log_2(x+7) \text{ to be defined: } x + 7 > 0 \implies x > -7 \end{aligned} $$ ## Verify each candidate solution against the domain $$ \begin{aligned} &x = -8: \text{ invalid because } -8 gtr 0 \quad (\text{extraneous}) \ &x = 1: \text{ valid because } 1 > 0 \text{ and } 1 > -7 \end{aligned} $$

$x = -8$

$x = 1$

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

$x = 1$ and $x = 8$

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

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Question

Explanation:

Identify potential solutions from the algebraic steps

Check domain restrictions of the original logarithmic equation

Verify each candidate solution against the domain

Answer: