Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 313

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\[\log_{a^{p}}b = \frac{1}{p}\log_{a}b\]

\[1)\log_{2}^{2}x - 9\log_{8}x = 4\]

\[\log_{2}^{2}x - 9\log_{2^{3}}x - 4 = 0\]

\[\log_{2}^{2}x - 3\log_{2}x - 4 = 0\]

\[Пусть\ y = \log_{2}x:\]

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

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

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

\[y_{2} = \frac{3 + 5}{2} = 4.\]

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

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

\[x = 2^{- 1} = \frac{1}{2} = 0,5.\]

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

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

\[x = 2^{4} = 16.\]

\[Ответ:\ \ x_{1} = 0,5;\ \ x_{2} = 16.\]

\[2)\ 16\log_{16}^{2}x + 3\log_{4}x - 1 = 0\]

\[16\log_{4^{2}}^{2}x + 3\log_{4}x - 1 = 0\]

\[4\log_{4}^{2}x + 3\log_{4}x - 1 = 0\]

\[Пусть\ y = \log_{4}x:\]

\[4y^{2} + 3y - 1 = 0\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

\[y_{1} = \frac{- 3 - 5}{2 \bullet 4} = - \frac{8}{8} = - 1;\]

\[y_{2} = \frac{- 3 + 5}{2 \bullet 4} = \frac{2}{8} = \frac{1}{4}.\]

\[1)\ \log_{4}x = - 1\]

\[\log_{4}x = \log_{4}4^{- 1}\]

\[x = 4^{- 1} = \frac{1}{4} = 0,25.\]

\[2)\ \log_{4}x = \frac{1}{4}\]

\[\log_{4}x = 4^{\frac{1}{4}}\]

\[x = 4^{\frac{1}{4}} = 2^{\frac{1}{2}} = \sqrt{2}.\]

\[Ответ:\ \ x_{1} = 0,25;\ \ x_{2} = \sqrt{2}.\]

\[3)\log_{3}^{2}x + 5\log_{9}x - 1,5 = 0\]

\[\log_{3}^{2}x + 5\log_{3^{2}}x - 1,5 = 0\]

\[\log_{3}^{2}x + \frac{5}{2}\log_{3}x - 1,5 = 0\]

\[Пусть\ y = \log_{3}x:\]

\[y^{2} + \frac{5}{2}y - 1,5 = 0\]

\[2y^{2} + 5y - 3 = 0\]

\[D = 5^{2} + 4 \bullet 2 \bullet 3 =\]

\[= 25 + 24 = 49\]

\[y_{1} = \frac{- 5 - 7}{2 \bullet 2} = - \frac{12}{4} = - 3;\]

\[y_{2} = \frac{- 5 + 7}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2}.\]

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

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

\[x = 3^{- 3} = \frac{1}{3^{3}} = \frac{1}{27}.\]

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

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

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

\[Ответ:\ \ x_{1} = \frac{1}{27};\ \ x_{2} = \sqrt{3}.\]

\[4)\log_{3}^{2}x - 15\log_{27}x + 6 = 0\]

\[\log_{3}^{2}x - 15\log_{3^{3}}x + 6 = 0\]

\[\log_{3}^{2}x - 5\log_{3}x + 6 = 0\]

\[Пусть\ y = \log_{3}x:\]

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

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

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

\[1)\ \log_{3}x = 2\]

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

\[x = 3^{2} = 9.\]

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

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

\[x = 3^{3} = 27.\]

\[Ответ:\ \ x_{1} = 9;\ \ x_{2} = 27.\]