Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1358

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\[1)\ x^{1 + \lg x} = 10x\]

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

\[1 + \lg x = \log_{x}10 + \log_{x}x\]

\[1 + \lg x = \frac{\lg 10}{\lg x} + 1\]

\[\lg x = \frac{1}{\lg x}\ \ \ \ \ | \bullet \lg x\]

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

\[\lg x = \pm 1.\]

\[1)\ \lg x = - 1\]

\[\lg x = \lg 10^{- 1}\]

\[x = 0,1.\]

\[2)\ \lg x = 1\]

\[\lg x = \lg 10\]

\[x = 10.\]

\[Ответ:\ \ x_{1} = 0,1;\ \ x_{2} = 10.\]

\[2)\ x^{\lg x} = 100x\]

\[\log_{x}x^{\lg x} = \log_{x}{100x}\]

\[\lg x = \log_{x}100 + \log_{x}x\]

\[\lg x = \frac{\lg 100}{\lg x} + 1\]

\[\lg x = \frac{2}{\lg x} + 1\ \ \ \ \ | \bullet \lg x\]

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

\[y = \lg x:\]

\[y^{2} = 2 + y\]

\[y^{2} - y - 2 = 0\]

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{1 - 3}{2} = - 1;\]

\[y_{2} = \frac{1 + 3}{2} = 2.\]

\[1)\ \lg x = - 1\]

\[\lg x = \lg 10^{- 1}\]

\[x = 0,1.\]

\[2)\ \lg x = 2\]

\[\lg x = \lg 10^{2}\]

\[x = 100.\]

\[Ответ:\ \ x_{1} = 0,1;\ \ x_{2} = 100.\]

\[3)\log_{2}\left( 17 - 2^{x} \right) + \log_{2}\left( 2^{x} + 15 \right) = 8\]

\[\log_{2}\left( \left( 17 - 2^{x} \right)\left( 2^{x} + 15 \right) \right) = \log_{2}2^{8}\]

\[17 \bullet 2^{x} + 255 - 2^{2x} - 15 \bullet 2^{x} = 256\]

\[2^{2x} - 2 \bullet 2^{x} + 1 = 0\]

\[\left( 2^{x} - 1 \right)^{2} = 0\]

\[2^{x} - 1 = 0\]

\[2^{x} = 1\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]

\[4)\log_{2}\left( 3 + 2^{x} \right) + \log_{2}\left( 5 - 2^{x} \right) = 4\]

\[\log_{2}\left( \left( 3 + 2^{x} \right)\left( 5 - 2^{x} \right) \right) = \log_{2}2^{4}\]

\[15 - 3 \bullet 2^{x} + 5 \bullet 2^{x} - 2^{2x} = 16\]

\[2^{2x} - 2 \bullet 2^{x} + 1 = 0\]

\[\left( 2^{x} - 1 \right)^{2} = 0\]

\[2^{x} - 1 = 0\]

\[2^{x} = 1\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]