Вариант II. Вычислить (1-9):

Решение:

1. \( \log_{15} 5 + \log_{15} 3 = \log_{15} (5 \cdot 3) = \log_{15} 15 = 1 \)
2. \( \log_{0,1} 5 + \log_{0,1} 2 = \log_{0,1} (5 \cdot 2) = \log_{0,1} 10 \)
3. \( \log_{5} 50 - \log_{5} 2 = \log_{5} \frac{50}{2} = \log_{5} 25 = 2 \)
4. \( \log_{3} \frac{1}{9} - \log_{3} 40,5 = \log_{3} \frac{1}{9} - \log_{3} \frac{81}{2} = \log_{3} \frac{1}{9} \cdot \frac{2}{81} = \log_{3} \frac{2}{729} \)
5. \( \log_{2} 8^{7} = 7 \log_{2} 8 = 7 \cdot 3 = 21 \)
6. \( \log_{13} \sqrt[3]{169} = \log_{13} \sqrt[3]{13^2} = \log_{13} 13^{\frac{2}{3}} = \frac{2}{3} \)
7. \( 2 \log_{10} 3 - \frac{1}{2} \log_{10} 0,81 = \log_{10} 3^2 - \log_{10} \sqrt{0,81} = \log_{10} 9 - \log_{10} 0,9 = \log_{10} \frac{9}{0,9} = \log_{10} 10 = 1 \)
8. \( \log_{3} 3,6 - \log_{3} 1,4 + \log_{3} 1 \frac{7}{9} = \log_{3} \frac{3,6}{1,4} + \log_{3} \frac{16}{9} = \log_{3} \frac{36}{14} + \log_{3} \frac{16}{9} = \log_{3} \frac{18}{7} + \log_{3} \frac{16}{9} = \log_{3} (\frac{18}{7} \cdot \frac{16}{9}) = \log_{3} (\frac{2 \cdot 16}{7}) = \log_{3} \frac{32}{7} \)
9. \( \frac{1}{3} \log_{2} \sqrt[8]{8} - 3 \log_{2} 3 + \frac{1}{2} \log_{2} 36 = \log_{2} \sqrt[3]{\sqrt[8]{8}} - \log_{2} 3^3 + \log_{2} \sqrt{36} = \log_{2} 8^{\frac{1}{24}} - \log_{2} 27 + \log_{2} 6 = \log_{2} (2^3)^{\frac{1}{24}} - \log_{2} 27 + \log_{2} 6 = \log_{2} 2^{\frac{1}{8}} - \log_{2} 27 + \log_{2} 6 = \log_{2} (\frac{2^{\frac{1}{8}} \cdot 6}{27}) \)

Ответ: 1. 1; 2. \( \log_{0,1} 10 \); 3. 2; 4. \( \log_{3} \frac{2}{729} \); 5. 21; 6. \( \frac{2}{3} \); 7. 1; 8. \( \log_{3} \frac{32}{7} \); 9. \( \log_{2} (\frac{2^{\frac{1}{8}} \cdot 6}{27}) \).