Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 38

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\[\textbf{а)}\log_{2}x = 2\log_{2}{(x - 3)} + 2\]

\[x > 0;\]

\[x - 3 > 0\]

\[x > 3.\]

\[M = (3; + \infty).\]

\[\log_{2}x = \log_{2}(x - 3)^{2} + \log_{2}2^{2}\]

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

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

\[x = 4 \cdot \left( x^{2} - 6x + 9 \right)\]

\[x = 4x^{2} - 24x + 36\]

\[4x^{2} - 25x + 36 = 0\]

\[4x^{2} - 9x - 16x + 36 = 0\]

\[x(4x - 9) - 4(4x - 9) = 0\]

\[(4x - 9)(x - 4) = 0\]

\[4\left( x - \frac{9}{4} \right)(x - 4) = 0\]

\[x = \frac{9}{4} = 2,25 < 3;\]

\[x = 4 > 3.\]

\[Ответ:x = 4.\]

\[\textbf{б)}\log_{5}x = 2\log_{5}{(6 - x)} + 1\]

\[x > 0;\]

\[6 - x > 0\]

\[x < 6.\]

\[M = (0;6).\]

\[\log_{5}x = \log_{5}(6 - x)^{2} + \log_{5}5\]

\[\log_{5}x = \log_{5}{5(6 - x)^{2}}\]

\[x = 5 \cdot (6 - x)^{2}\]

\[x = 5 \cdot \left( 36 - 12x + x^{2} \right)\]

\[x = 180 - 60x + 5x^{2}\]

\[5x^{2} - 61x + 180 = 0\]

\[5x^{2} - 25x - 36x + 180 = 0\]

\[5x(x - 5) - 36(x - 5) = 0\]

\[(x - 5)(5x - 36) = 0\]

\[5(x - 5)(x - 7,2) = 0\]

\[x = 5;\]

\[x = 7,2 > 6.\]

\[Ответ:x = 5.\]

\[\textbf{в)}\log_{3}{(x - 2)} - 3\log_{x - 2}9 = 1\]

\[x - 2 > 0\]

\[x > 2.\]

\[x - 2 \neq 1\]

\[x \neq 3.\]

\[M = (2;3) \cup (3; + \infty).\]

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

\[= 3 \cdot 2\log_{x - 2}3 = 6\log_{x - 2}3 =\]

\[= \frac{6}{\log_{3}{(x - 2)}}\]

\[\log_{3}{(x - 2)} - \frac{6}{\log_{3}(x - 2)} = 1\]

\[\log_{3}(x - 2) = t:\]

\[t - \frac{6}{t} - 1 = 0\]

\[t^{2} - t - 6 = 0\]

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

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

\[\log_{3}(x - 2) = 3:\]

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

\[x - 2 = 27\]

\[x = 29 > 3.\]

\[\log_{3}(x - 2) = - 2:\]

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

\[x - 2 = \frac{1}{9}\]

\[x = 2\frac{1}{9} < 3.\]

\[Ответ:x = 2\frac{1}{9};x = 29.\]

\[\textbf{г)}\log_{2}{(x - 3)} + 3\log_{x - 3}4 = 5\]

\[x - 3 > 0\]

\[x > 3.\]

\[x - 3 \neq 1\]

\[x \neq 4.\]

\[M = (3;4) \cup (4; + \infty).\]

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

\[= 3 \cdot 2\log_{x - 3}2 = 6\log_{x - 3}2 =\]

\[= \frac{6}{\log_{2}(x - 3)};\]

\[\log_{2}(x - 3) + \frac{6}{\log_{2}(x - 3)} = 5\]

\[\log_{2}(x - 3) = t:\]

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

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

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

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

\[\log_{2}(x - 3) = 2:\]

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

\[x - 3 = 4\]

\[x = 7 > 4.\]

\[\log_{2}(x - 3) = 3:\]

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

\[x - 3 = 8\]

\[x = 11 > 4.\]

\[Ответ:x = 7;x = 11.\]