Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 812

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\[1)\log_{2}^{2}x - 9\log_{8}x = 4\]

\[\log_{2}^{2}x - \frac{9}{3}\log_{2}x - 4 = 0\]

\[\log_{2}^{2}x - 3\log_{2}x - 4 = 0\]

\[\log_{2}x = t:\]

\[t^{2} - 3t - 4 = 0\]

\[D = 9 + 16 = 25\]

\[t_{1} = \frac{3 + 5}{2} = 4;\ \ \]

\[\ t_{2} = \frac{3 - 5}{2} = - 1.\]

\[1)\ \log_{2}x = 4\]

\[x = 16.\]

\[2)\ \log_{2}x = - 1\]

\[x = \frac{1}{2}.\]

\[Ответ:x = 16;\ \ x = \frac{1}{2}.\]

\[2)\ 16\log_{16}^{2}x + 3\log_{4}x - 1 = 0\]

\[4\log_{4}^{2}x + 3\log_{4}x - 1 = 0\]

\[\log_{4} = t:\]

\[4t^{2} + 3t - 1 = 0\]

\[D = 9 + 16 = 25\]

\[t_{1} = \frac{- 3 + 5}{8} = \frac{1}{4};\ \]

\[\text{\ \ \ }t_{2} = \frac{- 3 - 5}{8} = - 1.\]

\[1)\ \log_{4}x = \frac{1}{4}\]

\[x = \sqrt{2}.\]

\[2)\ \log_{4}x = - 1\]

\[x = \frac{1}{4}.\]

\[Ответ:x = \sqrt{2};\ \ x = \frac{1}{4}.\]

\[3)\ \log_{3}^{2}x + 5\log_{9}x - 1,5 = 0\]

\[\log_{3}^{2}x + \frac{5}{2}\log_{3}x - 1,5 = 0\ \ | \cdot 2\]

\[2\log_{3}^{2}x + 5\log_{3}x - 3 = 0\]

\[\log_{3}x = t:\]

\[2t^{2} + 5t - 3 = 0\]

\[D = 25 + 24 = 49\]

\[t_{1} = \frac{- 5 + 7}{4} = \frac{1}{2};\ \ \ \]

\[\ t_{2} = \frac{- 5 - 7}{4} = - 3.\]

\[1)\ \log_{3}x = \frac{1}{2}\]

\[x = \sqrt{3}.\]

\[2)\ \log_{3}x = - 3\]

\[x = \frac{1}{27}.\]

\[Ответ:x = \sqrt{3};\ \ x = \frac{1}{27}.\]

\[4)\log_{3}^{2}x - 15\log_{27}x + 6 = 0\]

\[\log_{3}^{2}x - \frac{15}{3}\log_{3}x + 6 = 0\]

\[\log_{3}^{2}x - 5\log_{3}x + 6 = 0\]

\[\log_{3}x = t:\]

\[t^{2} - 5t + 6 = 0\]

\[t_{1} + t_{2} = 5;\ \ \ t_{1} \cdot t_{2} = 6\]

\[t_{1} = 2;\ \ \ t_{2} = 3.\]

\[1)\ \log_{3}x = 2\]

\[x = 3^{2} = 9.\]

\[2)\ \log_{3}x = 3\]

\[x = 3^{3} = 27.\]

\[Ответ:x = 9;\ \ x = 27.\]