Вычислить log5 3 – 1/2 log5 12 + log5 50.

Для решения этих задач используем основные свойства логарифмов: 1) $\log_a b + \log_a c = \log_a (b \cdot c)$ 2) $\log_a b - \log_a c = \log_a (b / c)$ 3) $n \cdot \log_a b = \log_a (b^n)$ 4) $\log_{a^k} b^m = \frac{m}{k} \log_a b$ **№ 290** 1) $\log_{10} 5 + \log_{10} 2 = \log_{10} (5 \cdot 2) = \log_{10} 10 = 1$ 2) $\log_{10} 8 + \log_{10} 125 = \log_{10} (8 \cdot 125) = \log_{10} 1000 = 3$ 3) $\log_{12} 2 + \log_{12} 72 = \log_{12} (2 \cdot 72) = \log_{12} 144 = 2$ 4) $\log_3 6 + \log_3 \frac{3}{2} = \log_3 (6 \cdot \frac{3}{2}) = \log_3 9 = 2$ **№ 291** 1) $\log_2 15 - \log_2 \frac{15}{16} = \log_2 (15 : \frac{15}{16}) = \log_2 16 = 4$ 2) $\log_5 75 - \log_5 3 = \log_5 (75 : 3) = \log_5 25 = 2$ 3) $\log_{\frac{1}{3}} 54 - \log_{\frac{1}{3}} 2 = \log_{\frac{1}{3}} (54 : 2) = \log_{\frac{1}{3}} 27 = -3$ (так как $(1/3)^{-3} = 3^3 = 27$) 4) $\log_8 \frac{1}{16} - \log_8 32 = \log_8 (\frac{1}{16} : 32) = \log_8 \frac{1}{512} = -3$ (так как $8^{-3} = 1/512$) **№ 292** 1) $\log_{13} \sqrt[4]{169} = \log_{13} 169^{1/4} = \log_{13} (13^2)^{1/4} = \log_{13} 13^{1/2} = 0.5$ 2) $\log_{11} \sqrt[3]{121} = \log_{11} (11^2)^{1/3} = \log_{11} 11^{2/3} = 2/3$ 3) $\log_{\frac{1}{3}} \sqrt[4]{243} = \log_{3^{-1}} (3^5)^{1/4} = \log_{3^{-1}} 3^{5/4} = \frac{5/4}{-1} = -1.25$ 4) $\log_2 \frac{1}{\sqrt{128}} = \log_2 (128^{-1/2}) = \log_2 ((2^7)^{-1/2}) = \log_2 2^{-3.5} = -3.5$ **№ 293** 1) $\log_8 12 - \log_8 15 + \log_8 20 = \log_8 (12 : 15 \cdot 20) = \log_8 (0.8 \cdot 20) = \log_8 16 = \log_{2^3} 2^4 = 4/3 = 1.333$ 2) $\log_9 15 + \log_9 18 - \log_9 10 = \log_9 (15 \cdot 18 : 10) = \log_9 (270 : 10) = \log_9 27 = \log_{3^2} 3^3 = 3/2 = 1.5$ 3) $\frac{1}{2} \log_7 36 - \log_7 14 - 3 \log_7 \sqrt[3]{21} = \log_7 6 - \log_7 14 - \log_7 21 = \log_7 (6 / 14 / 21) = \log_7 (3 / 7 / 21) = \log_7 (1/49) = -2$ 4) $2 \log_{\frac{1}{3}} 6 - \frac{1}{2} \log_{\frac{1}{3}} 400 + 3 \log_{\frac{1}{3}} \sqrt[3]{45} = \log_{\frac{1}{3}} (6^2) - \log_{\frac{1}{3}} (400^{1/2}) + \log_{\frac{1}{3}} (45) = \log_{\frac{1}{3}} 36 - \log_{\frac{1}{3}} 20 + \log_{\frac{1}{3}} 45 = \log_{\frac{1}{3}} (36 : 20 \cdot 45) = \log_{\frac{1}{3}} 81 = -4$ (так как $(1/3)^{-4} = 3^4 = 81$) **№ 294** 1) $\frac{\log_3 8}{\log_3 16} = \log_{16} 8 = \log_{2^4} 2^3 = 3/4 = 0.75$ 2) $\frac{\log_5 27}{\log_5 9} = \log_9 27 = \log_{3^2} 3^3 = 3/2 = 1.5$ 3) $\frac{\log_5 36 - \log_5 12}{\log_5 9} = \frac{\log_5 3}{\log_5 9} = \log_9 3 = \log_{3^2} 3^1 = 1/2 = 0.5$ 4) $\frac{\log_7 8}{\log_7 15 - \log_7 30} = \frac{\log_7 8}{\log_7 0.5} = \log_{0.5} 8 = \log_{2^{-1}} 2^3 = -3$ **№ 295** Дано: $\log_a b = 3, \log_a c = -2$. 1) $\log_a (a^3 b^2 \sqrt{c}) = \log_a a^3 + \log_a b^2 + \log_a c^{0.5} = 3 + 2 \log_a b + 0.5 \log_a c = 3 + 2(3) + 0.5(-2) = 3 + 6 - 1 = 8$ 2) $\log_a \frac{a^4 \sqrt[3]{b}}{c^3} = \log_a a^4 + \log_a b^{1/3} - \log_a c^3 = 4 + \frac{1}{3} \log_a b - 3 \log_a c = 4 + \frac{1}{3}(3) - 3(-2) = 4 + 1 + 6 = 11$