\({\log _2}{\left( {x + 2} \right)^2} + {\log _2}{\left( {x + 10} \right)^2} = 4{\log _2}3.\)
Запишем ОДЗ:
\(\left\{ {\begin{array}{*{20}{c}}{{{\left( {x + 2} \right)}^2} > 0,\;\,}\\{{{\left( {x + 10} \right)}^2} > 0,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x \ne -2,\,\,}\\{x \ne -10}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x \in \left( {-\infty ;-10} \right) \cup \left( {-10;-2} \right) \cup \left( {-2;\infty } \right).\)
\({\log _2}{\left( {x + 2} \right)^2} + {\log _2}{\left( {x + 10} \right)^2} = 4{\log _2}3\;\;\;\; \Leftrightarrow \;\;\;\;{\log _2}{\left( {\left( {x + 2} \right) \cdot \left( {x + 10} \right)} \right)^2} = {\log _2}{3^4}\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;{\left( {\left( {x + 2} \right) \cdot \left( {x + 10} \right)} \right)^2} = 81\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + 12x + 20 = -9,}\\{{x^2} + 12x + 20 = 9\;\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2} + 12x + 29 = 0,}\\{{x^2} + 12x + 11 = 0\,\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -6 + \sqrt 7 ,}\\{x = -6-\sqrt {7,} }\\{x = -11,\;\;\;\;\;\;}\\{x = -1.\;\;\;\;\;\;\;\;}\end{array}} \right.\)
Ответ: \(-11;\;\;\;-1;\;\;\;-6 \pm \sqrt 7 .\)