Реши примеры с логарифмами: 1) log8 12-log8 15 + log8 20

1) $\log_8 12 - \log_8 15 + \log_8 20 = \log_8 \frac{12 \cdot 20}{15} = \log_8 \frac{12 \cdot 4}{3} = \log_8 (4 \cdot 4) = \log_8 16 = \log_8 8^{\frac{4}{3}} = \frac{4}{3}$ *Перевод:* $\log_8 12 - \log_8 15 + \log_8 20 = \log_8 \frac{12 \cdot 20}{15} = \log_8 \frac{12 \cdot 4}{3} = \log_8 (4 \cdot 4) = \log_8 16 = \log_8 8^{\frac{4}{3}} = \frac{4}{3}$ 2) $\log_9 15 + \log_9 18 - \log_9 10 = \log_9 \frac{15 \cdot 18}{10} = \log_9 \frac{3 \cdot 18}{2} = \log_9 (3 \cdot 9) = \log_9 27 = \log_9 9^{\frac{3}{2}} = \frac{3}{2}$ *Перевод:* $\log_9 15 + \log_9 18 - \log_9 10 = \log_9 \frac{15 \cdot 18}{10} = \log_9 \frac{3 \cdot 18}{2} = \log_9 (3 \cdot 9) = \log_9 27 = \log_9 9^{\frac{3}{2}} = \frac{3}{2}$ 3) $\frac{1}{2} \log_7 36 - \log_7 14 - 3 \log_7 \sqrt[3]{21} = \log_7 36^{\frac{1}{2}} - \log_7 14 - \log_7 (\sqrt[3]{21})^3 = \log_7 6 - \log_7 14 - \log_7 21 = \log_7 \frac{6}{14 \cdot 21} = \log_7 \frac{3}{7 \cdot 21} = \log_7 \frac{1}{7 \cdot 7} = \log_7 \frac{1}{49} = \log_7 7^{-2} = -2$ *Перевод:* $\frac{1}{2} \log_7 36 - \log_7 14 - 3 \log_7 \sqrt[3]{21} = \log_7 36^{\frac{1}{2}} - \log_7 14 - \log_7 (\sqrt[3]{21})^3 = \log_7 6 - \log_7 14 - \log_7 21 = \log_7 \frac{6}{14 \cdot 21} = \log_7 \frac{3}{7 \cdot 21} = \log_7 \frac{1}{7 \cdot 7} = \log_7 \frac{1}{49} = \log_7 7^{-2} = -2$ 4) $2 \log_{\frac{1}{3}} 6 - \frac{1}{2} \log_{\frac{1}{3}} 400 + 3 \log_{\frac{1}{3}} \sqrt[3]{45} = \log_{\frac{1}{3}} 6^2 - \log_{\frac{1}{3}} 400^{\frac{1}{2}} + \log_{\frac{1}{3}} (\sqrt[3]{45})^3 = \log_{\frac{1}{3}} 36 - \log_{\frac{1}{3}} 20 + \log_{\frac{1}{3}} 45 = \log_{\frac{1}{3}} \frac{36 \cdot 45}{20} = \log_{\frac{1}{3}} \frac{9 \cdot 45}{5} = \log_{\frac{1}{3}} (9 \cdot 9) = \log_{\frac{1}{3}} 81 = \log_{\frac{1}{3}} (\frac{1}{3})^{-4} = -4$ *Перевод:* $2 \log_{\frac{1}{3}} 6 - \frac{1}{2} \log_{\frac{1}{3}} 400 + 3 \log_{\frac{1}{3}} \sqrt[3]{45} = \log_{\frac{1}{3}} 6^2 - \log_{\frac{1}{3}} 400^{\frac{1}{2}} + \log_{\frac{1}{3}} (\sqrt[3]{45})^3 = \log_{\frac{1}{3}} 36 - \log_{\frac{1}{3}} 20 + \log_{\frac{1}{3}} 45 = \log_{\frac{1}{3}} \frac{36 \cdot 45}{20} = \log_{\frac{1}{3}} \frac{9 \cdot 45}{5} = \log_{\frac{1}{3}} (9 \cdot 9) = \log_{\frac{1}{3}} 81 = \log_{\frac{1}{3}} (\frac{1}{3})^{-4} = -4$