Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 286

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\[1)\ 7^{2x} + 7^{x} - 12 = 0\]

\[Пусть\ y = 7^{x}:\]

\[y^{2} + y - 12 = 0\]

\[D = 1^{2} + 4 \bullet 12 = 1 + 48 = 49\]

\[y_{1} = \frac{- 1 - 7}{2} = - 4;\text{\ \ }\]

\[y_{2} = \frac{- 1 + 7}{2} = 3.\]

\[1)\ 7^{x} = - 4\]

\[нет\ корней.\]

\[2)\ 7^{x} = 3\]

\[\log_{7}7^{x} = \log_{7}3\]

\[x = \log_{7}3.\]

\[Ответ:\ \ x = \log_{7}3.\]

\[2)\ 9^{x} - 3^{x} - 12 = 0\]

\[3^{2x} - 3^{x} - 12 = 0\]

\[Пусть\ y = 3^{x}:\]

\[y^{2} - y - 12 = 0\]

\[D = 1^{2} + 4 \bullet 12 = 1 + 48 = 49\]

\[y_{1} = \frac{1 - 7}{2} = - 3;\text{\ \ }\]

\[y_{2} = \frac{1 + 7}{2} = 4.\]

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

\[нет\ корней.\]

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

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

\[x = \log_{3}{4.}\]

\[Ответ:\ \ x = \log_{3}4.\]

\[3)\ 8^{x + 1} - 8^{2x - 1} = 30\]

\[8 \bullet 8^{x} - 8^{- 1} \bullet 8^{2x} - 30 = 0\]

\[8 \bullet 8^{x} - \frac{8^{2x}}{8} - 30 = 0\ \ \ \ \ | \bullet ( - 8)\]

\[8^{2x} - 64 \bullet 8^{x} + 240 = 0\]

\[Пусть\ y = 8^{x}:\]

\[y^{2} - 64y + 240 = 0\]

\[D = 64^{2} - 4 \bullet 240 =\]

\[= 4096 - 960 = 3136\]

\[y_{1} = \frac{64 - 56}{2} = 4;\text{\ \ }\]

\[= y_{2} = \frac{64 + 56}{2} = 60.\]

\[1)\ 8^{x} = 4\]

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

\[3x = 2\]

\[x = \frac{2}{3}.\]

\[2)\ 8^{x} = 60\]

\[\log_{8}8^{x} = \log_{8}60\]

\[x = \log_{8}60.\]

\[Ответ:\ \ x_{1} = \frac{2}{3};\ \ x_{2} = \log_{8}60.\]

\[4)\left( \frac{1}{9} \right)^{x} - 5\left( \frac{1}{3} \right)^{x} + 6 = 0\]

\[\left( \frac{1}{3} \right)^{2x} - 5\left( \frac{1}{3} \right)^{x} + 6 = 0\]

\[Пусть\ y = \left( \frac{1}{3} \right)^{x}:\]

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

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

\[y_{1} = \frac{5 - 1}{2} = 2;\text{\ \ }y_{2} = \frac{5 + 1}{2} = 3.\]

\[1)\ \left( \frac{1}{3} \right)^{x} = 2\]

\[\log_{\frac{1}{3}}\left( \frac{1}{3} \right)^{x} = \log_{\frac{1}{3}}2\]

\[x = \log_{\frac{1}{3}}2.\]

\[2)\ \left( \frac{1}{3} \right)^{x} = 3\]

\[\left( \frac{1}{3} \right)^{x} = \left( \frac{1}{3} \right)^{- 1}\ \]

\[x = - 1\]

\[Ответ:\ \ x_{1} = \log_{\frac{1}{3}}2;\ \ x_{2} = - 1.\]