Question

ПОМОГИТЕ, ПОЖАЛУЙСТА, ДАЮ 20 БАЛЛОВ ​

Answer 1

$\displaystyle\bf\\ODZ:\\\\\left \{ {{x+4>0} \atop {2x-12>0}} \right. \\\\\\\left \{ {{x>-4} \atop {x>6}} \right. \ \ \Rightarrow \ x>6\\\\\\\log_{3}(x+4)=\log_{3} (2x-12)\\\\x+4=2x-12\\\\2x-x=4+12\\\\x=16\\\\Otvet:16$