Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 390

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\[1)\ 3^{4x} = 10\]

\[\log_{3}3^{4x} = \log_{3}10\]

\[4x = \log_{3}10\]

\[x = \frac{\log_{3}10}{4} = \frac{\log_{3}10}{\log_{3}3^{4}} =\]

\[= \frac{\log_{3}10}{\log_{3}81} = \log_{81}10\]

\[Ответ:\ \ x = \log_{81}10.\]

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

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

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

\[x = \frac{\log_{2}3}{3} = \frac{\log_{2}3}{\log_{2}2^{3}} = \frac{\log_{2}3}{\log_{2}8} =\]

\[= \log_{8}3\]

\[Ответ:\ \ x = \log_{8}3.\]

\[3)\ {1,3}^{3x - 2} = 3\]

\[\log_{1,3}{1,3}^{3x - 2} = \log_{1,3}3\]

\[3x - 2 = \log_{1,3}3\]

\[3x = \log_{1,3}3 + 2\]

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

\[Ответ:\ \ x = \frac{1}{3}\left( \log_{1,3}3 + 2 \right).\]

\[4)\ \left( \frac{1}{3} \right)^{5 + 4x} = 1,5\]

\[\log_{\frac{1}{3}}\left( \frac{1}{3} \right)^{5 + 4x} = \log_{\frac{1}{3}}{1,5}\]

\[5 + 4x = \log_{\frac{1}{3}}{1,5}\]

\[4x = \log_{\frac{1}{3}}{1,5} - 5\]

\[x = \frac{1}{4}\left( \log_{\frac{1}{3}}{1,5} - 5 \right)\]

\[Ответ:\ \ x = \frac{1}{4}\left( \log_{\frac{1}{3}}{1,5} - 5 \right).\]

\[5)\ 16^{x} - 4^{x + 1} - 14 = 0\]

\[4^{2x} - 4 \bullet 4^{x} - 14 = 0\]

\[Пусть\ y = 4^{x}:\]

\[y^{2} - 4y - 14 = 0\]

\[D = 4^{2} + 4 \bullet 14 = 16 + 56 = 72\]

\[y = \frac{4 \pm \sqrt{72}}{2} = \frac{4 \pm 6\sqrt{2}}{2} =\]

\[= 2 \pm 3\sqrt{2}.\]

\[1)\ 4^{x} = 2 - 3\sqrt{2}\]

\[нет\ корней.\]

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

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

\[x = \log_{4}\left( 2 + 3\sqrt{2} \right).\]

\[Ответ:\ \ x = \log_{4}\left( 2 + 3\sqrt{2} \right).\]

\[6)\ 25^{x} + 2 \bullet 5^{x} - 15 = 0\]

\[5^{2x} + 2 \bullet 5^{x} - 15 = 0\]

\[Пусть\ y = 5^{x}:\]

\[y^{2} + 2y - 15 = 0\]

\[D = 2^{2} + 4 \bullet 15 = 4 + 60 = 64\]

\[y_{1} = \frac{- 2 - 8}{2} = - 5;\text{\ \ }\]

\[y_{2} = \frac{- 2 + 8}{2} = 3.\]

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

\[нет\ корней.\]

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

\[\log_{5}5^{x} = \log_{5}3\]

\[x = \log_{5}3.\]

\[Ответ:\ \ x = \log_{5}3.\]