ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 781

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\[1)\log_{3}{27\sqrt{a^{2}b}} = \log_{3}27 +\]

\[+ \log_{3}\sqrt{a^{2}} + \log_{3}\sqrt{b} =\]

\[= \log_{3}3^{3} + \log_{3}a + \log_{3}b^{\frac{1}{2}} =\]

\[= 3 + \log_{3}a + \frac{1}{2}\log_{3}\text{b.}\]

\[2)\ \log_{3}\frac{\sqrt[3]{a}}{9b^{2}} = \log_{3}\sqrt[3]{a} - \log_{3}9 -\]

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

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

\[= \frac{1}{3}\log_{3}a - 2 - 2\log_{3}\text{b.}\]

\[3)\ \log_{3}\frac{81\sqrt{a^{3}}}{\sqrt[3]{b^{2}}} = \log_{3}81 +\]

\[+ \log_{3}\sqrt{a^{3}} - \log_{3}\ \sqrt[3]{b^{2}} =\]

\[= \log_{3}3^{4} - \log_{3}a^{\frac{3}{2}} - \log_{3}b^{\frac{2}{3}} =\]

\[= 4 + \frac{3}{2}\log_{3}a - \frac{2}{3}\log_{3}\text{b.}\]

\[4)\ \log_{3}\frac{\sqrt[4]{ab^{2}}}{\sqrt[3]{a^{5}b}} = \log_{3}\sqrt[4]{a} +\]

\[+ \log_{3}\sqrt[4]{b^{2}} - \log_{3}\sqrt[3]{a^{5}} -\]

\[- \log_{3}\sqrt[3]{b} =\]

\[= \frac{1}{4}\log_{3}a + \frac{1}{2}\log_{3}b - \frac{5}{3}\log_{3}a -\]

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

\[= - \frac{17}{12}\log_{3}a + \frac{1}{6}\log_{3}\text{b.}\]