(A) Given,$\log _{0.3}(x-1) 0$,which implies $x > 1$.We can rewrite the inequ

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(A) Given,$\log _{0.3}(x-1) For the logarithm to be defined,we must have $x-1 > 0$,which implies $x > 1$.We can rewrite the inequality as $\log _{0.3}(x-1) Using the property $\log _{a^n} b = \frac{1}{n} \log_a b$,we get $\log _{0.3}(x-1) Multiplying by $2$,we get $2 \log _{0.3}(x-1) Subtracting $\log _{0.3}(x-1)$ from both sides,we get $\log _{0.3}(x-1) Since the base $0.3 (0.3)^0$.$x-1 > 1$.$x > 2$.Thus,$x$ lies in the interval $(2, \infty)$.