Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 860

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\[1)\ x^{\lg 9} + 9^{\lg x} = 6\ \]

\[Область\ определения:\]

\[x - 1 > 0\]

\[x > 1.\]

\[x^{\frac{\log_{x}9}{\log_{x}10}} + 9^{\lg x} = 6\ \]

\[\left( x^{\log_{x}9} \right)^{\frac{1}{\log_{x}10}} + 9^{\lg x} = 6\ \]

\[9^{\frac{1}{\log_{x}10}} + 9^{\lg x} = 6\ \]

\[9^{\lg x} + 9^{\lg x} = 6\]

\[2 \cdot 9^{\lg x} = 6\]

\[9^{\lg x} = 3\]

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

\[2\lg x = 3\]

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

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

\[x = \sqrt{10}.\]

\[Ответ:x = \sqrt{10}.\]

\[2)\ x^{\log_{2}\frac{x}{98}} \cdot 14^{\log_{2}7} = 1\ \]

\[\log_{2}{(x^{\log_{2}\frac{x}{98}} \cdot}14^{\log_{2}7}) = \log_{2}1\]

\[\log_{2}x^{\log_{2}\frac{x}{98}} + \log_{2}14^{\log_{2}7} = 0\]

\[\log_{2}{\left( \log_{2}x - \log_{2}98 \right) + \log_{2}14 \cdot \log_{2}7} = 0\]

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

\[t^{2} - \log_{2}98t +\]

\[+ \log_{2}14 \cdot \log_{2}7 = 0\]

\[D = \log_{2}^{2}98 -\]

\[- 4 \cdot \log_{2}14 \cdot \log_{2}7 = 1\]

\[t_{1} = \frac{\log_{2}98 - 1}{2} =\]

\[= \frac{2\log_{2}7}{2} = \log_{2}7;\]

\[t_{2} = \frac{\log_{2}98 + 1}{2} =\]

\[= \frac{2 + 2\log_{2}7}{2} = 1 + \log_{2}7.\]

\[\textbf{а)}\log_{2}x = \log_{2}7\]

\[x = 7.\]

\[\textbf{б)}\ \log_{2}x = 1 + \log_{2}7\]

\[x = 2^{1 + \log_{2}7}\]

\[x = 2 \cdot 7 = 14.\]

\[Ответ:x = 7;\ \ x = 14.\]