Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 753

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\[\log_{3}3 = \log_{3}3^{1} = 1\]

\[\log_{3}9 = \log_{3}3^{2} = 2\]

\[\log_{3}27 = \log_{3}3^{3} = 3\]

\[\log_{3}81 = \log_{3}3^{4} = 4\]

\[\log_{3}1 = \log_{3}3^{0} = 0\]

\[\log_{3}\frac{1}{3} = \log_{3}3^{- 1} = - 1\]

\[\log_{3}\frac{1}{9} = \log_{3}\frac{1}{3^{2}} =\]

\[= \log_{3}3^{- 2} = - 2\]

\[\log_{3}\frac{1}{243} = \log_{3}\frac{1}{3^{5}} =\]

\[= \log_{3}3^{- 5} = - 5\]

\[\log_{3}\sqrt[3]{3} = \log_{3}3^{\frac{1}{3}} = \frac{1}{3}\]

\[\log_{3}\frac{1}{3\sqrt{3}} = \log_{3}\frac{1}{3^{1} \bullet 3^{\frac{1}{2}}} =\]

\[= \log_{3}\frac{1}{3^{1\frac{1}{2}}} = \log_{3}3^{- 1\frac{1}{2}} =\]

\[= - 1\frac{1}{2} = - 1,5\]

\[\log_{3}{9\sqrt[4]{3}} = \log_{3}{3^{2} \bullet 3^{\frac{1}{4}}} =\]

\[= \log_{3}3^{2\frac{1}{4}} = 2\frac{1}{4} = 2,25\]