Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 29

\[\boxed{\mathbf{29.}}\]

\[\textbf{а)}\log_{3}x = 4 - 3\log_{3}x\]

\[x > 0;\ \ x \neq 1;\]

\[M = (0;1) \cup (1; + \infty).\]

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

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

\[t = 4 - \frac{3}{t}\]

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

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

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

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

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

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

\[x = 3.\]

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

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

\[x = 27.\]

\[Ответ:x = 3;x = 27.\]

\[\textbf{б)}\log_{4}x + 2 = 3\log_{x}4\]

\[x > 0;\ \ x \neq 1;\]

\[M = (0;1) \cup (1; + \infty).\]

\[\log_{4}x + 2 = \frac{3}{\log_{4}x}\]

\[t = \log_{4}x:\]

\[t + 2 = \frac{3}{t}\]

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

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

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

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

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

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

\[x = 4.\]

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

\[x = \frac{1}{4^{3}}\]

\[x = \frac{1}{64}.\]

\[Ответ:x = \frac{1}{64};\ \ x = 4.\]

\[\textbf{в)}\log_{3}x - 2 = 3\log_{x}3\]

\[x > 0;\ \ x \neq 1;\]

\[M = (0;1) \cup (1; + \infty).\]

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

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

\[t - 2 - \frac{3}{t} = 0\]

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

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

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

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

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

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

\[x = 27.\]

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

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

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

\[Ответ:x = \frac{1}{3};\ \ x = 27.\]

\[\textbf{г)}\log_{2}x + 6\log_{x}2 = 5\]

\[x > 0;\ \ x \neq 1;\]

\[M = (0;1) \cup (1; + \infty).\]

\[\log_{2}x + \frac{6}{\log_{2}x} = 5\]

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

\[t + \frac{6}{t} - 5 = 0\]

\[t - 5t + 6 = 0\]

\[t_{1} + t_{2} = 5;\ \ t_{1} \cdot t_{2} = 6\]

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

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

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

\[x = 4.\]

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

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

\[x = 8.\]

\[Ответ:x = 4;x = 8.\]