Log(2)(6-x)*log(3)(2x+4)+3=log(2)(6-x)+3log(2x+4) *** первый числа в скобках - это основание

[latex]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[/latex] [latex] \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[/latex]