Краткое пояснение: Необходимо вычислить значения логарифмов, используя свойства логарифмов и определение логарифма.
1)
- \(\log_a a = 1\)
- \(\log_a 1 = 0\)
- \(\log_a a^5 = 5\)
- \(\log_a \frac{1}{a} = \log_a a^{-1} = -1\)
- \(\log_a \sqrt{a} = \log_a a^{\frac{1}{2}} = \frac{1}{2}\)
- \(\log_a \sqrt[5]{a^3} = \log_a a^{\frac{3}{5}} = \frac{3}{5}\)
2)
- \(\log_{\frac{1}{2}} \frac{1}{4} = \log_{\frac{1}{2}} (\frac{1}{2})^2 = 2\)
- \(\log_{\frac{1}{2}} 2 = -1\)
- \(\log_{\frac{1}{2}} 1 = 0\)
- \(\log_{\frac{1}{2}} 8 = \log_{\frac{1}{2}} (\frac{1}{2})^{-3} = -3\)
- \(\log_{\frac{1}{2}} \sqrt{2} = \log_{\frac{1}{2}} (\frac{1}{2})^{- \frac{1}{2}} = - \frac{1}{2}\)
- \(\log_{\frac{1}{2}} \frac{\sqrt{2}}{2} = \log_{\frac{1}{2}} \frac{2^{\frac{1}{2}}}{2} = \log_{\frac{1}{2}} 2^{-\frac{1}{2}} = -\frac{1}{2} \cdot (-1) = \frac{1}{2}\)
3)
- \(\log_3 27 = \log_3 3^3 = 3\)
- \(\log_3 \frac{1}{9} = \log_3 3^{-2} = -2\)
- \(\log_9 \frac{1}{27} = \log_{3^2} 3^{-3} = -\frac{3}{2}\)
- \(\log_2 \sqrt{2} = \log_2 2^{\frac{1}{2}} = \frac{1}{2}\)
- \(\log_2 \frac{1}{\sqrt{2}} = \log_2 2^{-\frac{1}{2}} = -\frac{1}{2}\)
- \(\log_{\sqrt{2}} 4 = \log_{2^{\frac{1}{2}}} 2^2 = 4\)
- \(\log_{\sqrt{5}} \sqrt[4]{125} = \log_{5^{\frac{1}{2}}} (5^3)^{\frac{1}{4}} = \log_{5^{\frac{1}{2}}} 5^{\frac{3}{4}} = \frac{\frac{3}{4}}{\frac{1}{2}} = \frac{3}{2}\)