Вариант I. Вычислить (1-9):

Решение:

1. \( \log_{12} 3 + \log_{12} 4 = \log_{12} (3 \cdot 4) = \log_{12} 12 = 1 \)
2. \( \log_{4} 4 + \log_{9} 9 = 1 + 1 = 2 \)
3. \( \log_{3} 192 - \log_{3} 4 = \log_{3} \frac{192}{4} = \log_{3} 48 \)
4. \( \log_{2} 13 - \log_{2} 1 \frac{3}{8} = \log_{2} 13 - \log_{2} \frac{11}{8} = \log_{2} \frac{13 \cdot 8}{11} = \log_{2} \frac{104}{11} \)
5. \( \log_{3} 9^{10} = 10 \log_{3} 9 = 10 \cdot 2 = 20 \)
6. \( \log_{15} \sqrt{225} = \log_{15} 15 = 1 \)
7. \( \frac{1}{2} \log_{3} 81 - \frac{1}{3} \log_{3} 27 = \frac{1}{2} \cdot 4 - \frac{1}{3} \cdot 3 = 2 - 1 = 1 \)
8. \( \log_{2} 0,8 - \log_{2} 1 \frac{1}{4} + \log_{2} 22,5 = \log_{2} \frac{0,8}{1,25} + \log_{2} 22,5 = \log_{2} 0,64 + \log_{2} 22,5 = \log_{2} (0,64 \cdot 22,5) = \log_{2} 14,4 \)
9. \( 2 \log_{6} 10 - \log_{6} 28 + \frac{1}{2} \log_{6} \sqrt{49} = \log_{6} 100 - \log_{6} 28 + \log_{6} 7 = \log_{6} \frac{100 \cdot 7}{28} = \log_{6} \frac{700}{28} = \log_{6} 25 \)

Ответ: 1. 1; 2. 2; 3. \( \log_{3} 48 \); 4. \( \log_{2} \frac{104}{11} \); 5. 20; 6. 1; 7. 1; 8. \( \log_{2} 14,4 \); 9. \( \log_{6} 25 \).