\(\log _{\frac{1}{5}}^2{x^2}-31 \cdot {\log _{\frac{1}{5}}}x-8 < 0.\)
ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{{x^2} > 0,\,\,\,\,}\\{x > 0\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ne 0,}\\{x > 0}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {0;\infty } \right).} \right.} \right.\)
\(\log _{\frac{1}{5}}^2{x^2}-31 \cdot {\log _{\frac{1}{5}}}x-8 < 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{\left( {-2{{\log }_5}x} \right)^2} + 31 \cdot {\log _5}x-8 < 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,4\log _5^2x + 31 \cdot {\log _5}x-8 < 0.\)
Пусть \({\log _5}x = t\). Тогда:
\(4{t^2} + 31t-8 < 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-8 \le t \le \dfrac{1}{4}.\)
Вернёмся к прежней переменной:
\(-8 < {\log _5}x < \dfrac{1}{4}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _5}{5^{-8}} < {\log _5}x < {\log _5}{5^{\frac{1}{4}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{1}{{{5^8}}} < x < \sqrt[4]{5}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {\dfrac{1}{{{5^8}}};\sqrt[4]{5}} \right).\)
Ответ: \(\left( {\dfrac{1}{{{5^8}}};\sqrt[4]{5}} \right)\).