Профиль №15. Логарифмические неравенства. Задача 39А

ОТВЕТ: \(\left( {0;\dfrac{1}{6}} \right) \cup \left[ {6;\infty } \right).\)

\(\dfrac{2}{{{{\log }_6}x + 1}} \le 1.\)

Пусть \({\log _6}x = t\). Тогда:

\(\dfrac{2}{{t + 1}} \le 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{2}{{t + 1}}-1 \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{1-t}}{{t + 1}} \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t < -1,\,}\\{t \ge 1.\,\,\,\,\,}\end{array}} \right.\)

Вернёмся к прежней переменной:

\(\left[ {\begin{array}{*{20}{c}}{{{\log }_6}x < -1,}\\{{{\log }_6}x \ge 1\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_6}x < {{\log }_6}\dfrac{1}{6},}\\{{{\log }_6}x \ge {{\log }_6}6\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{0 < x < \dfrac{1}{6},}\\{x \ge 6\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.} \right.} \right.x \in \,\left( {0;\dfrac{1}{6}} \right) \cup \left[ {6;\infty } \right).\)

Ответ:  \(\left( {0;\dfrac{1}{6}} \right) \cup \left[ {6;\infty } \right).\)