2
3
2 - 2log32
3 - log32
4 - log32
Correct answer is B
2log\(_{3}\) 6 + log\(_{3}\) 12 - log\(_{3}\) 16
= \(\log_3 6^2 + \log_3 12 - \log_3 16\)
= \(\log_{3} (\frac{36 \times 12}{16})\)
= \(\log_{3} (27)\)
= \(\log_{3} 3^3\)
= \(3 \log_{3} 3\)
= 3
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