Задача 18. Решите уравнение \(3 \cdot {2^{{{\log }_x}\left( {3x-2} \right)}} + 2 \cdot {3^{{{\log }_x}\left( {3x-2} \right)}} = 5 \cdot {6^{{{\log }_{{x^2}}}\left( {3x-2} \right)}}\)
ОТВЕТ: 2.
\(3 \cdot {2^{{{\log }_x}\left( {3x-2} \right)}} + 2 \cdot {3^{{{\log }_x}\left( {3x-2} \right)}} = 5 \cdot {6^{{{\log }_{{x^2}}}\left( {3x-2} \right)}}.\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{3x-2 > 0,}\\{x > 0,\,\,\,\,\,\,\,\,\,\,\,}\\{x \ne 1,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} > 0,\,\,\,\,\,\,\,\,}\\{{x^2} \ne 1\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > \dfrac{2}{3},\,}\\{x > 0,\,\,\,}\\{x \ne 1,\,\,\,}\\{x \ne -1\,}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {\dfrac{2}{3};1} \right) \cup \left( {1;\infty } \right).} \right.} \right.\) \(3 \cdot {2^{{{\log }_x}\left( {3x-2} \right)}} + 2 \cdot {3^{{{\log }_x}\left( {3x-2} \right)}} = 5 \cdot {6^{{{\log }_{{x^2}}}\left( {3x-2} \right)}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3 \cdot {2^{{{\log }_x}\left( {3x-2} \right)}} + 2 \cdot {3^{{{\log }_x}\left( {3x-2} \right)}} = 5 \cdot {6^{\frac{1}{2}{{\log }_x}\left( {3x-2} \right)}}.\) Пусть \(\dfrac{1}{2}{\log _x}\left( {3x-2} \right) = t\). Тогда уравнение примет вид: \(3 \cdot {2^{2t}} + 2 \cdot {3^{2t}} = 5 \cdot {6^t}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3 \cdot {4^t}-5 \cdot {6^t} + 2 \cdot {9^t} = 0\left| {:{9^t} \ne 0\,\,\,\,\,\,\, \Leftrightarrow } \right.\) \( \Leftrightarrow \,\,\,\,\,\,\,3 \cdot {\left( {\dfrac{4}{9}} \right)^t}-5 \cdot {\left( {\dfrac{6}{9}} \right)^t} + 2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3 \cdot {\left( {\dfrac{2}{3}} \right)^{2t}}-5 \cdot {\left( {\dfrac{2}{3}} \right)^t} + 2 = 0.\) Пусть \({\left( {\dfrac{2}{3}} \right)^t} = y\). Тогда: \(3{y^2}-5y + 2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{y = 1,\,\,}\\{y = \dfrac{2}{3}.}\end{array}} \right.\) Вернёмся к переменной y: \(\left[ {\begin{array}{*{20}{c}}{{{\left( {\dfrac{2}{3}} \right)}^t} = 1,}\\{{{\left( {\dfrac{2}{3}} \right)}^t} = \dfrac{2}{3}}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 0,}\\{t = 1.}\end{array}} \right.} \right.\) Вернёмся к переменной x: \(\left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}{{\log }_x}\left( {3x-2} \right) = 0}\\{\dfrac{1}{2}{{\log }_x}\left( {3x-2} \right) = 1}\end{array}\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_x}\left( {3x-2} \right) = 0}\\{{{\log }_x}\left( {3x-2} \right) = 2}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{3x-2 = 1,\,}\\{3x-2 = {x^2}}\end{array}\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1,}\\{x = 2.}\end{array}} \right.} \right.\) Корень \(x = 1\) не удовлетворяет ОДЗ. Ответ: 2.