which equation is equivalent to $\log 4 + \log(x + 2) = 1$?

$x + 2 = 10^1$
$4x + 8 = 10^1$
$4x + 8 = 1$
$\frac{4}{x + 2} = 10^1$

the solution to the equation is $x = $

Question

which equation is equivalent to $\log 4 + \log(x + 2) = 1$?

$x + 2 = 10^1$
$4x + 8 = 10^1$
$4x + 8 = 1$
$\frac{4}{x + 2} = 10^1$

the solution to the equation is $x = $

Accepted Answer

{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Logarithmic Product Rule", "Solving Logarithmic Equations" ], "new_concepts": [], "current_concepts": [ "Logarithmic Product Rule", "Solving Logarithmic Equations" ] } ## Apply the product rule of logarithms Using the *Logarithmic Product Rule* knowledge point $$ \begin{aligned} &\log(4) + \log(x + 2) = 1 \ &\log(4(x + 2)) = 1 \ &\log(4x + 8) = 1 \end{aligned} $$ ## Convert to exponential form Using the *Solving Logarithmic Equations* knowledge point $$ \begin{aligned} &10^1 = 4x + 8 \ &4x + 8 = 10^1 \end{aligned} $$ ## Solve for x Using the *Solving Logarithmic Equations* knowledge point $$ \begin{aligned} &4x + 8 = 10 \ &4x = 2 \ &x = \frac{1}{2} = 0.5 \end{aligned} $$ ## Question 1 (A) $x + 2 = 10^1$

(B) $4x + 8 = 10^1$

(C) $4x + 8 = 1$

(D) $\frac{4}{x+2} = 10^1$ ## Question 2 The solution to the equation is $x =$ $0.5$ { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Algebra", "Solving Logarithmic Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Apply the product rule of logarithms

Convert to exponential form

Solve for x

Answer:

Question 1

Question 2