ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 848

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\[1)\ 2^{3\lg x} \cdot 5^{\lg x} = 1600\]

\[\left( 2^{3} \cdot 5 \right)^{\lg x} = 1600\]

\[40^{\lg x} = 1600\]

\[\lg 40^{\lg x} = \lg 1600\]

\[\lg x \cdot \lg 40 = \lg 1600\]

\[\lg x = \frac{\lg 1600}{\lg 40}\]

\[\lg x = \log_{40}1600\]

\[\lg x = 2\]

\[x = 10^{2}\]

\[x = 100.\]

\[Ответ:x = 100.\]

\[2)\ 2^{\log_{3}x^{2}} \cdot 5^{\log_{3}x} = 400\]

\[\left( 2^{2} \cdot 5 \right)^{\log_{3}x} = 400\]

\[20^{\log_{3}x} = 400\]

\[\lg 20^{\log_{3}x} = \lg 400\]

\[\log_{3}x \cdot \lg 20 = \lg 400\]

\[\log_{3}x = \frac{\lg 400}{\lg 20}\]

\[\log_{3}x = \log_{20}400\]

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

\[x = 3^{2}\]

\[x = 9\]

\[Ответ:x = 9.\ \]

\[3)\ \frac{1}{4 + \lg x} + \frac{2}{2 - \lg x} = 1\]

\[\frac{\log^{2}x + 3\lg x + 2}{\left( 4 + \lg x \right)\left( 2 - \lg x \right)} = 0\]

\[\log^{2}x + 3\lg x + 2 = 0\]

\[\lg x = t:\]

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

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

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

\[\textbf{а)}\ \lg x = - 2\]

\[x = 10^{- 2} = \frac{1}{100}.\]

\[\textbf{б)}\ \lg x = - 1\]

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

\[Ответ:\ \ x = \frac{1}{10};\ \ x = \frac{1}{100}\text{.\ }\]

\[4)\ \frac{1}{5 - \lg x} + \frac{2}{1 + \lg x} = 1\]

\[\log^{2}x - 5\lg x + 6 = 0\]

\[t = \lg x:\]

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

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

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

\[\textbf{а)}\ \lg x = 2\]

\[x = 10^{2}\]

\[x = 100.\]

\[\textbf{б)}\ \lg x = 3\]

\[x = 10^{3}\]

\[x = 1000.\]

\[Ответ:x = 100;x = 1000.\]