Задача 2. Решите уравнение \({\lg ^2}\left( {8x-9} \right) = {\lg ^2}\left( {6x-4} \right)\)
ОТВЕТ: \(\dfrac{5}{2};\;\;\dfrac{7}{6}.\)
\({\lg ^2}\left( {8x-9} \right) = {\lg ^2}\left( {6x-4} \right).\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{8x-9 > 0,}\\{6x-4 > 0\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x > \dfrac{9}{8},}\\{x > \dfrac{2}{3}\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x \in \left( {\dfrac{9}{8};\infty } \right).\) \({\lg ^2}\left( {8x-9} \right) = {\lg ^2}\left( {6x-4} \right)\;\;\;\; \Leftrightarrow \;\;\;\;{\lg ^2}\left( {8x-9} \right)-{\lg ^2}\left( {6x-4} \right) = 0\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left( {\lg \left( {8x-9} \right)-\lg \left( {6x-4} \right)} \right)\left( {\lg \left( {8x-9} \right) + \lg \left( {6x-4} \right)} \right) = 0\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\lg \left( {8x-9} \right)-\lg \left( {6x-4} \right) = 0,\,\,}\\{\lg \left( {8x-9} \right) + \lg \left( {6x-4} \right) = \lg 1\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\lg \left( {8x-9} \right) = \lg \left( {6x-4} \right),}\\{\left( {8x-9} \right)\left( {6x-4} \right) = 1\,\,\,\,\,\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{8x-9 = 6x-4,\;\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;\;\;\;\,\;}\\{48{x^2}-32x-54x + 36-1 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{2x-5 = 0,\;\;\;\;\;\;\;\;\;\;\;\;\,\;}\\{48{x^2}-86x + 35 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{x = \dfrac{5}{2},\;\;\;\;\;\;\;\;\;\;\;\,\,}\\{x = \dfrac{7}{6},\;\;\;\;\;\,\,\;\;\;\;\;\;}\end{array}}\\{x = \dfrac{5}{8} \notin \left( {\dfrac{9}{8};\infty } \right)}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = \dfrac{5}{2},}\\{x = \dfrac{7}{6}.\,}\end{array}} \right.\) Ответ: \(\dfrac{5}{2};\;\;\;\dfrac{7}{6}.\)