Решение заданий на логарифмы

1) log₃ 6+log₃ 18-log₃ 4 = log₃ (6 × 18 / 4) = log₃ (27) = log₃ (3³) = 3.

2) $$log_{2}(sin \frac{\pi}{8})+log_{2}(2cos \frac{\pi}{8})=log_{2}(2sin \frac{\pi}{8}cos \frac{\pi}{8})=log_{2}(sin \frac{\pi}{4})=log_{2}(\frac{\sqrt{2}}{2})=log_{2}(2^{-\frac{1}{2}})=-\frac{1}{2}$$

3) $$\frac{log_{3}^{2} 6 - log_{3}^{2} 2}{log_{3} 12}=\frac{(log_{3} 6 - log_{3} 2)(log_{3} 6 + log_{3} 2)}{log_{3} 12}=\frac{log_{3}(\frac{6}{2}) \cdot log_{3}(6 \cdot 2)}{log_{3} 12}=\frac{log_{3} 3 \cdot log_{3} 12}{log_{3} 12}=log_{3} 3 = 1$$

4) $$9^{0.5-log_{3} 2} - log_{3} log_{2} 8 = (3^{2})^{0.5-log_{3} 2} - log_{3} 3 = 3^{1-2log_{3} 2} - 1 = \frac{3}{3^{2log_{3} 2}} - 1 = \frac{3}{3^{log_{3} 4}} - 1 = \frac{3}{4} - 1 = -\frac{1}{4}$$

5) $$lg 5 \cdot lg 20 + lg^{2} 2 = lg 5 \cdot lg (2 \cdot 10) + lg^{2} 2 = lg 5 \cdot (lg 2 + lg 10) + lg^{2} 2 = lg 5 \cdot (lg 2 + 1) + lg^{2} 2 = lg 5 \cdot lg 2 + lg 5 + lg^{2} 2 = lg 5 \cdot lg 2 + lg 5 + lg 2 \cdot lg 2 = lg 5 \cdot lg 2 + lg 5 + lg 2 \cdot lg 2 = lg 5 \cdot lg 2 + lg 2 \cdot lg 2 + lg 5 = lg 2 \cdot (lg 5 + lg 2) + lg 5 = lg 2 \cdot lg 10 + lg 5 = lg 2 \cdot 1 + lg 5 = lg 2 + lg 5 = lg (2 \cdot 5) = lg 10 = 1$$

6)

$$ log_{\sqrt{7}} 2 - log_{4}{5} \cdot log_{125}{49} = log_{7^{1/2}} 2 - log_{2^2}{5} \cdot log_{5^3}{7^2} = 2log_{7} 2 - \frac{1}{2}log_{2}{5} \cdot \frac{2}{3}log_{5}{7} = 2log_{7} 2 - \frac{1}{3}log_{2}{5} \cdot log_{5}{7} = 2log_{7} 2 - \frac{1}{3}log_{2}{7} = \frac{2}{log_{2} 7} - \frac{1}{3}log_{2}{7} = \frac{6 - log_{2}^2 7}{3log_{2} 7} $$