If  ${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$ , then $x$ lies in the interval.

${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$

$\Rightarrow {\log }_{0.3}(x-1)<{\log }_{{\left(0.3\right)}^2}(x-1)$

$\Rightarrow {\log }_{0.3}(x-1)<\frac{1}{2}{\log }_{0.3}(x-1)$

$\Rightarrow {\log }_{0.3}(x-1)<{\log }_{0.3}\sqrt{x-1}$

$x-1>\sqrt{x-1}$

${(x-1)}^2>(x-1)$

${(x-1)}^2-(x-1)>0$

$(x-1)(x-1-1)>0$

$(x-1)(x-2)>0$

So $x>2$ or   $x<1$

Hence $x\in (2,\infty )$  

$(\text{if }x<1{\text{ then log}}_{0.3}(x-1)\text{ not defined})$