what is the true solution to $3 \ln 2 + \ln 8 = 2 \ln(4x)$?

$x = 1$
$x = 2$
$x = 4$
$x = 8$

Question

what is the true solution to $3 \ln 2 + \ln 8 = 2 \ln(4x)$?

$x = 1$
$x = 2$
$x = 4$
$x = 8$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Logarithmic Power Rule", "Logarithmic Product Rule", "Solving Logarithmic Equations" ], "new_concepts": [], "current_concepts": [ "Logarithmic Power Rule", "Logarithmic Product Rule", "Solving Logarithmic Equations" ] } ## Apply logarithmic power rules Using the *Logarithmic Power Rule* knowledge point $$ \begin{aligned} 3 \ln 2 &= \ln(2^3) = \ln 8 \ 2 \ln(4x) &= \ln((4x)^2) = \ln(16x^2) \quad (x > 0) \end{aligned} $$ ## Combine terms using product rule Using the *Logarithmic Product Rule* knowledge point $$ \begin{aligned} \ln 8 + \ln 8 &= \ln(8 \cdot 8) = \ln 64 \end{aligned} $$ ## Equate arguments and solve Using the *Solving Logarithmic Equations* knowledge point $$ \begin{aligned} \ln 64 &= \ln(16x^2) \ 64 &= 16x^2 \ x^2 &= 4 \ x &= 2 \quad (\text{since } x > 0) \end{aligned} $$

(A) $x = 1$

(B) $x = 2$

(C) $x = 4$

(D) $x = 8$

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

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QUESTION IMAGE

Question

Explanation:

Apply logarithmic power rules

Combine terms using product rule

Equate arguments and solve

Answer: