Показательные уравнения повышенной сложности. Задача 6

ОТВЕТ: \(3;\;\;1 + {\log _2}3.\)

Так как  \(216 = {6^3} = {3^3} \cdot {2^3},\)  то уравнение примет вид:

\({3^x} = 216 \cdot {2^{{x^2}-4x}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{3^x} = {3^3} \cdot {2^3} \cdot {2^{{x^2}-4x}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{3^{x-3}} = {2^{{x^2}-4x + 3}}.\)

Прологарифмируем обе части последнего уравнения по основанию 3:

\({\log _3}{3^{x-3}} = {\log _3}{2^{{x^2}-4x + 3}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left( {x-3} \right){\log _3}3 = \left( {{x^2}-4x + 3} \right){\log _3}2\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,x-3-\left( {x-3} \right)\left( {x-1} \right){\log _3}2 = 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left( {x-3} \right)\left( {1-\left( {x-1} \right){{\log }_3}2} \right) = 0\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-3 = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{1-\left( {x-1} \right){{\log }_3}2 = 0}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x-1 = \dfrac{1}{{{{\log }_3}2}}}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = 1 + {{\log }_2}3.}\end{array}} \right.\)

Ответ:  \(3;\;\;1 + {\log _2}3.\)