Задача 7А. Решите неравенство \({\log _2}\left( {3x + 1} \right) \le {\log _2}\left( {x + 2} \right)\)
Ответ
ОТВЕТ: \(\,\left( {-\dfrac{1}{3};\dfrac{1}{2}} \right].\)
Решение
\({\log _2}\left( {3x + 1} \right) \le {\log _2}\left( {x + 2} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{3x + 1 \le x + 2}\\{3x + 1 > 0\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le \dfrac{1}{2},\,}\\{x > -\dfrac{1}{3}}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-\dfrac{1}{3};\dfrac{1}{2}} \right].} \right.\)
Ответ: \(\,\left( {-\dfrac{1}{3};\dfrac{1}{2}} \right].\)