Показательные уравнения повышенной сложности. Задача 5

ОТВЕТ: \(-1;\;\;d\frac{{3 + 2{{\log }_5}7}}{{{{\log }_5}7-1}}.\)

\({5^{{x^2} + 4x + 3}} = {7^{{x^2}-x-2}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{5^{\left( {x + 1} \right)\left( {x + 3} \right)}} = {7^{\left( {x + 1} \right)\left( {x-2} \right)}}.\)

Прологарифмируем обе части последнего уравнения по основанию 5:

\({\log _5}{5^{\left( {x + 1} \right)\left( {x + 3} \right)}} = {\log _5}{7^{\left( {x + 1} \right)\left( {x-2} \right)}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left( {x + 1} \right)\left( {x + 3} \right){\log _5}5 = \left( {x + 1} \right)\left( {x-2} \right){\log _5}7\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left( {x + 1} \right)\left( {x + 3-\left( {x-2} \right){{\log }_5}7} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x + 1 = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x + 3-x{{\log }_5}7 + 2{{\log }_5}7 = 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x\left( {{{\log }_5}7-1} \right) = 3 + 2{{\log }_5}7}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.\left[ {\begin{array}{*{20}{c}}{x = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{{3 + 2{{\log }_5}7}}{{{{\log }_5}7-1}}.}\end{array}} \right.\)

Ответ:  \(-1;\,\,\,\,\,\,\dfrac{{3 + 2{{\log }_5}7}}{{{{\log }_5}7-1}}.\)