Question

log(2)(6-x)*log(3)(2x+4)+3=log(2)(6-x)+3log(2x+4)

*** первый числа в скобках - это основание

Answer 1

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