Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 914

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\[1)\ 5x^{\log_{3}2} + 2^{\log_{3}x} = 24\]

\[5 \cdot 2^{\log_{3}x} + 2^{\log_{3}x} = 24\]

\[2^{\log_{3}x}(5 + 1) = 24\]

\[2^{\log_{3}x} \cdot 6 = 24\]

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

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

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

\[3^{2} = x\]

\[x = 9.\]

\[Ответ:x = 9.\]

\[2)\ x^{3\lg^{3}x - \frac{2}{3}\lg x} = 100\sqrt[3]{10};\ \ \]

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

\[x^{3\lg^{3}x - \frac{2}{3}\lg x} = 10^{\frac{7}{3}}\]

\[\lg\left( x^{3\lg^{3}x - \frac{2}{3}\lg x} \right) = \lg\left( 10^{\frac{7}{3}} \right)\]

\[\left( 3\lg^{3}x - \frac{2}{3}\lg x \right)\lg x = \frac{7}{3}\lg 10\]

\[\left( 3\lg^{3}x - \frac{2}{3}\lg x \right)\lg x = \frac{7}{3}\]

\[t = \lg x:\]

\[\left( 3t^{3} - \frac{2}{3}t \right) \cdot t = \frac{7}{3}\]

\[3t^{4} - \frac{2}{3}t^{2} - \frac{7}{3} = 0\ \ \ \ \ \ \ \ \ \ \ | \cdot 3\]

\[9t^{4} - 2t^{2} - 7 = 0\]

\[t^{2} = a \geq 0:\]

\[9a^{2} - 2a - 7 = 0\]

\[D_{1} = 1 + 63 = 64\]

\[a_{1} = \frac{1 + 8}{9} = 1;\ \ \ \ \]

\[\ a_{2} = \frac{1 - 8}{9} =\]

\[= - \frac{7}{9}\ (не\ подходит).\]

\[t^{2} = 1\]

\[t = \pm 1.\]

\[\textbf{а)}\ \lg x = 1\]

\[x = 10^{1} = 10.\ \ \]

\[\textbf{б)}\ \lg x = - 1\]

\[x = 10^{- 1} = \frac{1}{10}.\]

\[Ответ:x = 1;\ \ x = \frac{1}{10}\text{.\ }\]