\(\frac{-3}{9}\)
\(\frac{-3}{2}\)
\(\frac{6}{11}\)
\(\frac{43}{78}\)
Correct answer is B
\(\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}\)
\(\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3\)
\(\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3\)
\(\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4\)
\(\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}\)
= \(\frac{6}{-4} = \frac{-3}{2}\)