1. Вычислить! a) $$log_2 2\sqrt{2}; log_4 64; log_{\sqrt{17}} \frac{1}{7}; log_5 \frac{1}{\sqrt{5}}$$ б) $$log_3 8 - 2log_3 2 + log_3 \frac{9}{2}; log_{\sqrt{22}} - log_{\sqrt{11}} - log_{\sqrt{10}};$$ $$2 log_7 32 - log_7 256 + 2 log_7 14; log_2 7 - log_2 63 + log_2 36$$ в) $$8^{4log_{16}3}; \sqrt{5}^{2log_53}; 2^{log_2 125} \cdot 6^{log_3 \frac{1}{6}}$$ г) $$log_2 log_5 \sqrt{5}; log_3 log_3 27; log_3^2 log_5 \frac{1}{125}$$

a) $$log_2 2\sqrt{2}; log_4 64; log_{\sqrt{17}} \frac{1}{7}; log_5 \frac{1}{\sqrt{5}}$$

1. $$log_2 2\sqrt{2} = log_2 2^{1} \cdot 2^{\frac{1}{2}} = log_2 2^{\frac{3}{2}} = \frac{3}{2}$$
2. $$log_4 64 = log_4 4^3 = 3$$
3. $$log_{\sqrt{17}} \frac{1}{7} = log_{17^{\frac{1}{2}}} 7^{-1} = -1 \cdot log_{17^{\frac{1}{2}}} 7 = -1 \cdot 2 log_{17} 7 = -2 log_{17} 7 = -2 \cdot \frac{log_7 7}{log_7 17} = -2 \cdot \frac{1}{log_7 17}$$
4. $$log_5 \frac{1}{\sqrt{5}} = log_5 5^{-\frac{1}{2}} = -\frac{1}{2}$$

Ответ: $$\frac{3}{2}; 3; -2 \cdot \frac{1}{log_7 17}; -\frac{1}{2}$$

б) $$log_3 8 - 2log_3 2 + log_3 \frac{9}{2}; log_{\sqrt{22}} - log_{\sqrt{11}} - log_{\sqrt{10}};$$ $$2 log_7 32 - log_7 256 + 2 log_7 14; log_2 7 - log_2 63 + log_2 36$$

1. $$log_3 8 - 2log_3 2 + log_3 \frac{9}{2} = log_3 8 - log_3 2^2 + log_3 \frac{9}{2} = log_3 8 - log_3 4 + log_3 \frac{9}{2} = log_3 \frac{8 \cdot \frac{9}{2}}{4} = log_3 \frac{36}{4} = log_3 9 = log_3 3^2 = 2$$
2. $$log_{\sqrt{22}} - log_{\sqrt{11}} - log_{\sqrt{10}} = \frac{1}{2}log_{22} - \frac{1}{2}log_{11} - \frac{1}{2}log_{10} = \frac{1}{2} (log_{22} - log_{11} - log_{10}) = \frac{1}{2}log \frac{22}{11 \cdot 10} = \frac{1}{2}log \frac{2}{10} = \frac{1}{2}log \frac{1}{5} = \frac{1}{2}log 5^{-1} = -\frac{1}{2}log 5$$
3. $$2 log_7 32 - log_7 256 + 2 log_7 14 = log_7 32^2 - log_7 256 + log_7 14^2 = log_7 1024 - log_7 256 + log_7 196 = log_7 \frac{1024 \cdot 196}{256} = log_7 4 \cdot 196 = log_7 784 = log_7 7^2 \cdot 16 = log_7 7^2 + log_7 16 = 2 + log_7 16$$
4. $$log_2 7 - log_2 63 + log_2 36 = log_2 \frac{7 \cdot 36}{63} = log_2 \frac{7 \cdot 4}{7} = log_2 4 = log_2 2^2 = 2$$

Ответ: $$2; -\frac{1}{2}log 5; 2 + log_7 16; 2$$

в) $$8^{4log_{16}3}; \sqrt{5}^{2log_53}; 2^{log_2 125} \cdot 6^{log_3 \frac{1}{6}}$$

1. $$8^{4log_{16}3} = 8^{4log_{4^2}3} = 8^{4 \cdot \frac{1}{2}log_{4}3} = 8^{2log_{4}3} = (2^3)^{2log_{2^2}3} = 2^{6 \cdot \frac{1}{2}log_{2}3} = 2^{3log_{2}3} = 2^{log_{2}3^3} = 3^3 = 27$$
2. $$\sqrt{5}^{2log_53} = (5^{\frac{1}{2}})^{2log_53} = 5^{\frac{1}{2} \cdot 2log_53} = 5^{log_53} = 3$$
3. $$2^{log_2 125} \cdot 6^{log_3 \frac{1}{6}} = 125 \cdot 6^{log_3 6^{-1}} = 125 \cdot 6^{-log_3 6} = 125 \cdot \frac{1}{6^{log_3 6}} = 125 \cdot \frac{1}{6^{log_3 2 + log_3 3}} = 125 \cdot \frac{1}{6^{log_3 2 + 1}} = 125 \cdot \frac{1}{6^{log_3 2} \cdot 6^1} = \frac{125}{6 \cdot 6^{log_3 2}}$$

Ответ: $$27; 3; \frac{125}{6 \cdot 6^{log_3 2}}$$

г) $$log_2 log_5 \sqrt{5}; log_3 log_3 27; log_3^2 log_5 \frac{1}{125}$$

1. $$log_2 log_5 \sqrt{5} = log_2 log_5 5^{\frac{1}{2}} = log_2 \frac{1}{2} = log_2 2^{-1} = -1$$
2. $$log_3 log_3 27 = log_3 log_3 3^3 = log_3 3 = 1$$
3. $$log_3^2 log_5 \frac{1}{125} = (log_3 log_5 5^{-3})^2 = (log_3 (-3))^2 = (log_3 (-3))^2$$

Ответ: $$-1; 1; (log_3 (-3))^2$$