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

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

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

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

$\Rightarrow \frac{1}{2}{\log }_{0.3}(x-1)<0$

$(x-1)>1$  as base<1

$\Rightarrow x>2$  i.e., $x\in (2,\infty )$ .