Question

which of the following shows the true solution to the logarithmic equation \\(3\log_{2}(2x) = 3\\)?

\\(x = -1\\)
\\(x = 1\\)
\\(x = -1\\) and \\(x = 1\\)
\\(x = 0\\), \\(x = -1\\), and \\(x = 1\\)

Explanation:

<pre_analysis>
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<reasoning>

Isolate the logarithmic term

\[

$$\begin{aligned} 3\log_{2}(2x) &= 3 \\ \log_{2}(2x) &= 1 \end{aligned}$$

\]

Convert to exponential form

\[

$$\begin{aligned} 2x &= 2^1 \\ 2x &= 2 \end{aligned}$$

\]

Solve for x and check domain

\[

$$\begin{aligned} x &= 1 \\ \text{Domain check: } 2x > 0 &\implies 2(1) = 2 > 0 \quad (\text{Valid}) \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>\(x = -1\)</mcq-option>
<mcq-correct>\(x = 1\)</mcq-correct>
<mcq-option>\(x = -1\) and \(x = 1\)</mcq-option>
<mcq-option>\(x = 0\), \(x = -1\), and \(x = 1\)</mcq-option>
</answer>

<post_analysis>
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"subject": "Mathematics",
"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Algebra",
"Logarithmic Equations"
]
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</post_analysis>

Answer:

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</pre_analysis>

<reasoning>

Isolate the logarithmic term

\[

$$\begin{aligned} 3\log_{2}(2x) &= 3 \\ \log_{2}(2x) &= 1 \end{aligned}$$

\]

Convert to exponential form

\[

$$\begin{aligned} 2x &= 2^1 \\ 2x &= 2 \end{aligned}$$

\]

Solve for x and check domain

\[

$$\begin{aligned} x &= 1 \\ \text{Domain check: } 2x > 0 &\implies 2(1) = 2 > 0 \quad (\text{Valid}) \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>\(x = -1\)</mcq-option>
<mcq-correct>\(x = 1\)</mcq-correct>
<mcq-option>\(x = -1\) and \(x = 1\)</mcq-option>
<mcq-option>\(x = 0\), \(x = -1\), and \(x = 1\)</mcq-option>
</answer>

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