Question

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

Answer 1

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