3. N43.3. a) log₃ 7 - log₃ ⅘ b) log₁⁄₂ 128 - log₁⁄₂ 7 б) log₂ 15 - log₂ 30 г) log₀.₂ 40 - log₀.₂ 0,28

Решение:

• а) \( \log_{3} 7 - \log_{3} \frac{4}{9} = \log_{3} (7 \div \frac{4}{9}) = \log_{3} (7 \times \frac{9}{4}) = \log_{3} \frac{63}{4} \)
• б) \( \log_{\frac{1}{2}} 128 - \log_{\frac{1}{2}} 7 = \log_{\frac{1}{2}} (128 \div 7) = \log_{\frac{1}{2}} \frac{128}{7} \)
• б) \( \log_{2} 15 - \log_{2} 30 = \log_{2} (15 \div 30) = \log_{2} \frac{1}{2} = -1 \)
• г) \( \log_{0.2} 40 - \log_{0.2} 0.28 = \log_{0.2} (40 \div 0.28) = \log_{0.2} (40 \div \frac{28}{100}) = \log_{0.2} (40 \times \frac{100}{28}) = \log_{0.2} (40 \times \frac{25}{7}) = \log_{0.2} \frac{1000}{7} \)

Финальный ответ:

• а) \( \log_{3} \frac{63}{4} \)
• б) \( \log_{\frac{1}{2}} \frac{128}{7} \)
• б) \( -1 \)
• г) \( \log_{0.2} \frac{1000}{7} \)