Find, without using logarithm tables, the value of \(\frac{log_3 27 - log_{\frac{1}{4}} 64}{log_3 \frac{1}{81}}\)
A. \(\frac{-3}{9}\)
B. \(\frac{-3}{2}\)
C. \(\frac{6}{11}\)
D. \(\frac{43}{78}\)
Correct Answer: B
Explanation
\(\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}\)
\(\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3\)
\(\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3\)
\(\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4\)
\(\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}\)
= \(\frac{6}{-4} = \frac{-3}{2}\)
