(A) Given the inequality: $\log _{0.3}(x - 1) 0$,so $x > 1$.We can rewrite the bas

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(A) Given the inequality: $\log _{0.3}(x - 1) First,note that the domain requires $x - 1 > 0$,so $x > 1$.We can rewrite the base $0.09$ as $(0.3)^2$. Thus,$\log _{0.09}(x - 1) = \log _{(0.3)^2}(x - 1) = \frac{1}{2} \log _{0.3}(x - 1)$.The inequality becomes $\log _{0.3}(x - 1) Subtracting $\frac{1}{2} \log _{0.3}(x - 1)$ from both sides,we get $\frac{1}{2} \log _{0.3}(x - 1) Since the base $0.3 1$.Therefore,$x > 2$.Thus,$x$ lies in the interval $(2, \infty)$.