Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 18

\[\boxed{\mathbf{18.}}\]

\[\textbf{а)}\log_{2}\left( 4^{x} - 2^{x + 1} + 2 \right) = \log_{2}2^{x}\]

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

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

\[\left( 2^{x} \right)^{2} - 3 \cdot 2^{x} + 2 = 0\]

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

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

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

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

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

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

\[x = 0.\]

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

\[x = 1.\]

\[Проверка:\]

\[\log_{2}\left( 4^{0} - 2^{1} + 2 \right) = \log_{2}2^{0}\]

\[\log_{2}1 = \log_{2}1\]

\[x = 0 - корень.\]

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

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

\[x = 1 - корень.\]

\[Ответ:x = 0;x = 1.\]

\[\textbf{б)}\log_{3}\left( 9^{x} - 3^{x + 1} + 3 \right) = \log_{3}3^{x}\]

\[9^{x} - 3^{x + 1} + 3 = 3^{x}\]

\[\left( 3^{x} \right)^{2} - 3 \cdot 3^{x} - 3^{x} + 3 = 0\]

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

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

\[t^{2} - 4t + 3 = 0\]

\[D_{1} = 4 - 3 = 1\]

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

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

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

\[3^{x} = 3^{0}\]

\[x = 0.\]

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

\[x = 1.\]

\[Проверка:\]

\[\log_{3}\left( 9^{0} - 3^{1} + 3 \right) = \log_{3}3^{0}\]

\[\log_{3}1 = \log_{3}1\]

\[x = 0 - корень.\]

\[\log_{3}\left( 9^{1} - 3^{2} + 3 \right) = \log_{3}3^{1}\]

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

\[x = 1 - корень.\]

\[Ответ:x = 0;x = 1.\]

\[\textbf{в)}\log_{2}\left( 4^{x} + 2^{x + 1} - 8 \right) =\]

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

\[4^{x} + 2^{x + 1} - 8 = 2^{x + 2}\]

\[\left( 2^{2} \right)^{x} + 2 \cdot 2^{x} - 2^{2} \cdot 2^{x} - 8 = 0\]

\[\left( 2^{x} \right)^{2} - 2 \cdot 2^{x} - 8 = 0\]

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

\[t^{2} - 2t - 8 = 0\]

\[D_{1} = 1 + 8 = 9\]

\[t_{1} = 1 + 3 = 4;\]

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

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

\[нет\ корней.\]

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

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

\[x = 2.\]

\[Проверка:\]

\[\log_{2}\left( 4^{2} + 2^{3} - 8 \right) = \log_{2}2^{4}\]

\[\log_{2}16 = \log_{2}16\]

\[x = 2 - корень.\]

\[Ответ:x = 2.\]

\[\textbf{г)}\log_{5}\left( 25^{x} + 5^{x} - 5 \right) = \log_{5}5^{x + 1}\]

\[25^{x} + 5^{x} - 5 = 5^{x + 1}\]

\[\left( 5^{2} \right)^{x} + 5^{x} - 5 - 5^{x + 1} = 0\]

\[\left( 5^{x} \right)^{2} + 5^{x} - 5 \cdot 5^{x} - 5 = 0\]

\[\left( 5^{x} \right)^{2} - 4 \cdot 5^{x} - 5 = 0\]

\[t = 5^{x}:\]

\[t^{2} - 4t - 5 = 0\]

\[D_{1} = 4 + 5 = 9\]

\[t_{1} = 2 + 3 = 5;\]

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

\[1)\ 5^{x} = - 1\]

\[нет\ корней.\]

\[2)\ 5^{x} = 5\]

\[x = 1.\]

\[Проверка:\]

\[\log_{5}(25 + 5 - 5) = \log_{5}5^{2}\]

\[\log_{5}25 = \log_{5}25\]

\[x = 1 - корень.\]