Найдите значение выражения: 1) log2 128 + log3 3 - log12 1; 2) $$\frac{1}{16}$$log4 16 - $$\frac{1}{6}$$log6 216 + log2 32; 3) $$\sqrt{10^{\lg39} + 9^{\log_9 45} + 6^{\log_6 16}}$$

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Accepted Answer

$$log_{2}128 + log_{3}3 - log_{12}1 = log_{2}2^{7} + log_{3}3 - log_{12}1 = 7 + 1 - 0 = 8$$
$$\frac{1}{16}log_{4}16 - \frac{1}{6}log_{6}216 + log_{2}32 = \frac{1}{16}log_{4}4^{2} - \frac{1}{6}log_{6}6^{3} + log_{2}2^{5} = \frac{1}{16} \cdot 2 - \frac{1}{6} \cdot 3 + 5 = \frac{1}{8} - \frac{1}{2} + 5 = \frac{1 - 4 + 40}{8} = \frac{37}{8} = 4,625$$
$$\sqrt{10^{\lg 39} + 9^{\log_{9}45} + 6^{\log_{6}16}} = \sqrt{39 + 45 + 16} = \sqrt{100} = 10$$