simplify; 2log\(_{3}\) 6 + log\(_{3}\) 12 – log\(_{3}\) 16
- 2
-
3
- 2 - 2log32
- 3 - log32
- 4 - log32
The correct answer is: B
Explanation
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