294 1) \( \frac{\log _{3} 8}{\log _{3} 16} \); 2) \( \frac{\log _{5} 27}{\log _{5} 9} \); 3) \( \frac{\log _{5} 36-\log _{5} 12}{\log _{5} 9} \); 4) \( \frac{\log _{7} 8}{\log _{7} 15-\log _{7} 30} \)

Let’s break these down into manageable pieces and find their values! 1) \( \frac{\log _{3} 8}{\log _{3} 16} = \frac{\log_{3}(2^3)}{\log_{3}(2^4)} = \frac{3 \log_{3} 2}{4 \log_{3} 2} = \frac{3}{4} \) 2) \( \frac{\log _{5} 27}{\log _{5} 9} = \frac{\log_{5}(3^3)}{\log_{5}(3^2)} = \frac{3 \log_{5} 3}{2 \log_{5} 3} = \frac{3}{2} \) 3) \( \frac{\log _{5} 36 - \log _{5} 12}{\log _{5} 9} = \frac{\log_{5}\left(\frac{36}{12}\right)}{\log_{5}(3^2)} = \frac{\log_{5}(3)}{2 \log_{5}(3)} = \frac{1}{2} \) 4) \( \frac{\log _{7} 8}{\log _{7} 15 - \log _{7} 30} = \frac{\log_{7}(2^3)}{\log_{7}\left(\frac{15}{30}\right)} = \frac{3 \log_{7} 2}{\log_{7}(0.5)} \) Using the change of base property: \( \log_{7}(0.5) = -\log_{7}(2) \) Thus, \( \frac{3 \log_{7} 2}{-\log_{7}(2)} = -3 \) In summary: 1) \( \frac{3}{4} \) 2) \( \frac{3}{2} \) 3) \( \frac{1}{2} \) 4) \( -3 \)