Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 851

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

\[\ x > 0;\ \ \ x \neq 1.\ \ \]

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

\[t - \frac{2}{t} = - 1\]

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

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

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

\[\textbf{а)}\ \log_{2}x = - 2\]

\[x = 2^{- 2} = \frac{1}{4}.\]

\[\textbf{б)}\ \log_{2}x = 1\]

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

\[Ответ:x = \frac{1}{4};\ \ x = 2.\]

\[2)\log_{2}x + \log_{x}2 = 2,5;\ \ \]

\[\ x > 0;\ \ x \neq 1\]

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

\[t + \frac{1}{t} = 2,5\ \ \ \ \ \ \ | \cdot 2t\]

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

\[D = 25 - 16 = 9\]

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

\[\textbf{а)}\ \log_{2}x = 2\]

\[x = 2^{2} = 4.\]

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

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

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

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

\[\ x > 0;\ \ x \neq 1\]

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

\[t + \frac{2}{t} = 3\]

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

\[t_{1} + t_{2} = 3;\ \ \ t_{1} \cdot t_{2} = 2\]

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

\[\textbf{а)}\ \log_{3}x = 1\]

\[x = 3^{1} = 3.\]

\[\textbf{б)}\ \log_{3}x = 2\]

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

\[Ответ:x = 3;\ \ x = 9.\]

\[4)\log_{3}x - 6\log_{x}3 = 1;\ \ \ \]

\[\ x > 0;\ \ x \neq 1\ \]

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

\[t - \frac{6}{t} = 1\]

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

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

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

\[\textbf{а)}\ \log_{3}x = 3\]

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

\[\textbf{б)}\ \log_{3}x = - 2\]

\[x = 3^{- 2} = \frac{1}{9}.\]

\[Ответ:x = - \frac{1}{9};\ \ x = 27.\]