Question

Solve the following inequality:
${\log }_{1/3}x+2{\log }_{1/9}(x-1)⩽{\log }_{1/3}6$

Answer 1

${\log }_{1/3}x+2{\log }_{1/9}(x-1)\le {\log }_{1/3}6$

$-{\log }_{3}x-{\log }_3(x-1)\le -{\log }_{3}6$

${\log }_{3}x+{\log }_3(x-1)-{\log }_{3}6\ge 6$

${\log }_3\left(\frac{x(x-1)}{6}\right)$

$\frac{x(x-1)}{6}\ge 1\Rightarrow x^2-x-6\ge 0$

$(x-3)(x+2)\ge 0$

$xϵ(-\infty ,-2]\bigcup [3,\infty )$

Also $x>0$ & $x>1$

Taking intersection we get $xϵ[3,\infty )$