Запишем ОДЗ: \(x > 0.\)
\(f’\left( x \right) = {\left( {{x^3}\ln x} \right)^\prime } = {\left( {{x^3}} \right)^\prime }\ln x + {x^3} \cdot {\left( {\ln x} \right)^\prime } = 3{x^2}\ln x + {x^3} \cdot \dfrac{1}{x} = 3{x^2}\ln x + {x^2}.\)
Тогда:
\(f’\left( x \right)-\dfrac{2}{x}f\left( x \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3{x^2}\ln x + {x^2}-\dfrac{2}{x} \cdot {x^3}\ln x = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,3{x^2}\ln x + {x^2}-2{x^2}\ln x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{x^2}\ln x + {x^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{x^2} \cdot \left( {\ln x + 1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{x^2} = 0,\,\,\,\,\,\,\,\,\,}\\{\ln x + 1 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,}\\{x = \dfrac{1}{e}.}\end{array}} \right.\)
Корень \(x = 0\) не удовлетворяет ОДЗ.
Тогда: \(5e \cdot x = 5e \cdot \dfrac{1}{e} = 5.\)
Ответ: 5.