\(\log _3^2\left( {x + 2} \right) \cdot \left( {{{\log }_4}\left( {x + 3} \right)-1} \right) \ge 0.\)
ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x + 2 > 0,}\\{x + 3 > 0\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > -2,\,}\\{x > -3}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {-2;\infty } \right).} \right.} \right.\)
Решим данное неравенство методом интервалов:
\(\log _3^2\left( {x + 2} \right) \cdot \left( {{{\log }_4}\left( {x + 3} \right)-1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\log _3^2\left( {x + 2} \right) = 0,\,\,\,\,\,}\\{{{\log }_4}\left( {x + 3} \right)-1 = 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_3}\left( {x + 2} \right) = 0,}\\{{{\log }_4}\left( {x + 3} \right) = 1\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = 1.\,\,\,}\end{array}} \right.} \right.\)
Таким образом, решением исходного неравенства является: \(x \in \,\left\{ {-1} \right\} \cup \left[ {1;\infty } \right).\)
Ответ: \(\,\left\{ {-1} \right\} \cup \left[ {1;\infty } \right).\)