Question

Solve the following equation:
$2{\log }_3(x-2)+{\log }_3(x-4{)}^2=0$

Answer 1

${2log}_3(x-2)+{log}_3{(x-4)}^2=0\Rightarrow {log}_3{(x-2)}^2+{log}_3({x-4)}^2=0\Rightarrow [(x-2)(x-4){]}^2=1\Rightarrow (x-2)(x-4)=\pm 1\Rightarrow x^2-6x+8=\pm 1$

$x^2-6x+9=0$ or $x^2-6x+7=0$

Therefore, $x=3$ or $x=\frac{6\pm \sqrt{8}}{2}=3\pm \sqrt{2}$

at $x=3-\sqrt{2},x-2$ is negative.

Therefore, $x=3,3+\sqrt{2}$ are only solutions