Задача 25. Решите уравнение \(\dfrac{3}{2}\log _5^2{\left( {2x-3} \right)^2} + 12\,\,\log _5^2\sqrt x = {\log _5}{\left( {2x-3} \right)^3} \cdot {\log _5}{x^3}\)
ОТВЕТ: \(\dfrac{9}{4};\;\;3.\)
\(\dfrac{3}{2}\log _5^2{\left( {2x-3} \right)^2} + 12\log _5^2\sqrt x = {\log _5}{\left( {2x-3} \right)^3} \cdot {\log _5}{x^3}.\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{{{\left( {2x-3} \right)}^3} > 0,}\\{x > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > \dfrac{3}{2},}\\{x > 0}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {\dfrac{3}{2};\infty } \right).} \right.} \right.\) \(\dfrac{3}{2} \cdot {\left( {2{{\log }_5}\left( {2x-3} \right)} \right)^2} + 12 \cdot {\left( {\dfrac{1}{2}{{\log }_5}x} \right)^2}-9{\log _5}\left( {2x-3} \right) \cdot {\log _5}x = 0\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,6\log _5^2\left( {2x-3} \right) + 3\log _5^2x-9{\log _5}\left( {2x-3} \right) \cdot {\log _5}x = 0.\) Так как \(x = 1\) не является корнем уравнения, то разделим обе части последнего уравнения на: \(3\log _5^2x\). \(2 \cdot {\left( {\dfrac{{{{\log }_5}\left( {2x-3} \right)}}{{{{\log }_5}x}}} \right)^2}-3 \cdot \dfrac{{{{\log }_5}\left( {2x-3} \right)}}{{{{\log }_5}x}} + 1 = 0.\) Пусть \(\dfrac{{{{\log }_5}\left( {2x-3} \right)}}{{{{\log }_5}x}} = t\). Тогда: \(2{t^2}-3t + 1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,}\\{t = \dfrac{1}{2}.}\end{array}} \right.\) Возвращаясь к прежней переменной, получим: \(\left[ {\begin{array}{*{20}{c}}{\dfrac{{{{\log }_5}\left( {2x-3} \right)}}{{{{\log }_5}x}} = 1,}\\{\dfrac{{{{\log }_5}\left( {2x-3} \right)}}{{{{\log }_5}x}} = \dfrac{1}{2}}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_5}\left( {2x-3} \right) = {{\log }_5}x,}\\{2{{\log }_5}\left( {2x-3} \right) = {{\log }_5}x}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\) \( \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x-3 = x,\,\,\,\,}\\{{{\left( {2x-3} \right)}^2} = x}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,\,\,\,}\\{x = 1,\,\,\,}\\{x = \dfrac{9}{4}.\,}\end{array}} \right.\) Корень \(x = 1\) не удовлетворяет ОДЗ. Ответ: \(\dfrac{9}{4};\,\,\,3\).