\(\left( {{{\log }_9}2-{{\log }_5}2} \right) \cdot {\log _3}\left( {x-18} \right) > 0.\)
Так как \({\log _9}2 = \dfrac{1}{{{{\log }_2}9}}\) и \({\log _5}2 = \dfrac{1}{{{{\log }_2}5}}\), при этом:
\({\log _2}9 > {\log _2}5 > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{1}{{{{\log }_2}9}} < \dfrac{1}{{{{\log }_2}5}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _9}2 < {\log _5}2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _9}2-{\log _5}2 < 0,\)
то:
\(\left( {{{\log }_9}2-{{\log }_5}2} \right) \cdot {\log _3}\left( {x-18} \right) > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _3}\left( {x-18} \right) < 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\log _3}\left( {x-18} \right) < {\log _3}1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,0 < x-18 < 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,18 < x < 19\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\,\left( {18;19} \right).\)
Ответ: \(\,\left( {18;19} \right).\)