Question

lg(x^2-2x)-lg(2x+12)=0

Answer 1

$lg(x^2-2x)-lg(2x+12)=0\\ODZ:; ; left { {{x^2-2x textgreater 0} atop {2x+12 textgreater 0}} right. ; left { {{x(x-2) textgreater 0} atop {2(x+6) textgreater 0}} right. ; left { {{xin (-infty ,0)cup (2,+infty )} atop {xin (-6,+infty )}} right. ; to \\xin (-6,0)cup (2,+infty )\\lg(x^2-2x)=lg(2x+12)\\x^2-2x=2x+12\\x^2-4x-12=0\\x_1=-2; ,; ; x_2=6; ; (teorema; Vieta)\\Otvet:; ; x_1=-2; ,; ; x_2=6; .$