a
(a) \({\log _{0.3}}(x - 1) < {\log _{{{(0.3)}^2}}}(x - 1) = {1 \over 2}{\log _{0.3}}(x - 1)\)

\(\therefore {1 \over 2}{\log _{0.3}}(x - 1) < 0\)

or \({\log _{0.3}}(x - 1) < \,0 = \log 1\) or \((x - 1) > 1\) or \(x > 2\)

As base is less than \(1\), therefore the inequality is reversed, now \(x > 2\) \( \Rightarrow \)\(x\) lies in \((2,\infty )\).