0} \atop {6-x>0}} \right.
\end{array}\right\\
\left[\begin{array}{c}log_{3}(2x+4) = log_{3}(3^1)\\log_{2}(6-x) = log_{2}(2^3)\\
\left \{ {{x>-2} \atop {x<6}} \right.
\end{array}\right" alt="log_{2}(6-x)*log_{3}(2x+4)+3=log_{2}(6-x)+3log_{3}(2x+4);\\
log_{2}(6-x)*(log_{3}(2x+4) - 1) = 3*(log_{3}(2x+4)-1);\\
(log_{3}(2x+4) - 1)*(log_{2}(6-x) - 3) = 0;\\
\left[\begin{array}{c}log_{3}(2x+4) - 1 = 0\\log_{2}(6-x) - 3 = 0\\
\left \{ {{2x+4>0} \atop {6-x>0}} \right.
\end{array}\right\\
\left[\begin{array}{c}log_{3}(2x+4) = log_{3}(3^1)\\log_{2}(6-x) = log_{2}(2^3)\\
\left \{ {{x>-2} \atop {x<6}} \right.
\end{array}\right" align="absmiddle" class="latex-formula">
Answer: x = -0.5" alt=" \left[\begin{array}{c}2x+4 = 3\\6-x = 8\end{array}\right \ x \in (-2; 6)\\
\left[\begin{array}{c}x = -\frac{1}{2}\\x = -2\end{array}\right \ x \in (-2; 6)\\
=> Answer: x = -0.5" align="absmiddle" class="latex-formula">