Логарифмические уравнения повышенной сложности. Задача 21

ОТВЕТ: \(\dfrac{5}{4};\;\;5;\;\;6.\)

\({\left( {x-4} \right)^2}{\log _4}\left( {x-1} \right)-2{\log _4}{\left( {x-1} \right)^2} = {\left( {x-4} \right)^2}{\log _{x-1}}4-2{\log _{x-1}}16.\)

Запишем ОДЗ:

\(\left\{ {\begin{array}{*{20}{c}}{x-1 > 0,\,\,\,\,\,\,}\\{{{\left( {x-1} \right)}^2} > 0,}\\{x-1 \ne 1\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > 1,}\\{x \ne 1,}\\{x \ne 2}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {1;2} \right) \cup \left( {2;\infty } \right).} \right.} \right.\)

\({\left( {x-4} \right)^2}\left( {{{\log }_4}\left( {x-1} \right)-{{\log }_{x-1}}4} \right)-4{\log _4}\left| {x-1} \right| + 4{\log _{x-1}}4 = 0\)

Так как ОДЗ  \(x\, \in \,\left( {1;2} \right) \cup \left( {2;\infty } \right)\), то \(\left| {x-1} \right| = x-1\).

\({\left( {x-4} \right)^2}\left( {{{\log }_4}\left( {x-1} \right)-\dfrac{1}{{{{\log }_4}\left( {x-1} \right)}}} \right)-4\left( {{{\log }_4}\left( {x-1} \right)-\dfrac{1}{{{{\log }_4}\left( {x-1} \right)}}} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left( {{{\left( {x-4} \right)}^2}-4} \right)\left( {{{\log }_4}\left( {x-1} \right)-\dfrac{1}{{{{\log }_4}\left( {x-1} \right)}}} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {x-4} \right)}^2}-4 = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\log }_4}\left( {x-1} \right)-\dfrac{1}{{{{\log }_4}\left( {x-1} \right)}} = 0}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-4 = 2,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x-4 = -2,\,\,\,\,\,\,\,\,}\\{\log _4^2\left( {x-1} \right) = 1}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\)

\( \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = 2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\log }_4}\left( {x-1} \right) = 1,}\\{{{\log }_4}\left( {x-1} \right) = -1}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 6,}\\{x = 2,}\\{x = 5,}\\{x = \dfrac{5}{4}.}\end{array}} \right.\)

Корень \(x = 2\) не удовлетворяет ОДЗ.

Ответ:  \(\dfrac{5}{4};\,\,\,5;\,\,6\).