Iflog_0.3(x-1)​<log_0.09(x-1) then x lies in the interval

I can understand up to ---​2log_0.3(x-1)<log_0.3(x-1)
(x-1)²>(x-1) --- next step sign changed from < to >. Can you please explain sir

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Note} \mathrm{that} \\ {\log }_{\mathrm{a}}{\mathrm{x}}_1>{\log }_{\mathrm{a}}{\mathrm{x}}_2 \mathrm{gives} \\ \left\{\begin{array}{ll}{\mathrm{x}}_1>{\mathrm{x}}_2& \mathrm{if} \mathrm{a}>1\\ {\mathrm{x}}_1<{\mathrm{x}}_2& \mathrm{if} 0<\mathrm{a}<1\end{array}\right. \\ \mathrm{i}.\mathrm{e}. \mathrm{sign} \mathrm{of} \mathrm{inequality} \mathrm{reverses} \mathrm{if} \mathrm{base} \mathrm{lies} \mathrm{between} 0 \mathrm{and} 1 \\ \mathrm{Also} {\mathrm{x}}_1>0 \mathrm{and} {\mathrm{x}}_2>0 \\ \mathrm{Now} \\ {\log }_{0.3} \left(\mathrm{x}-1\right)<{\log }_{0.09}\left(\mathrm{x}-1\right) \\ \mathrm{Quantity} \mathrm{inside} \log \mathrm{is} \mathrm{always} \mathrm{positive}, \mathrm{hence} \\ \mathrm{x}-1>0 \\ \Rightarrow \mathrm{x}\in \left(1,\infty \right)...\left(\mathrm{i}\right) \\ {\log }_{0.3} \left(\mathrm{x}-1\right)<{\log }_{0.09}\left(\mathrm{x}-1\right) \\ \Rightarrow {\log }_{0.3} \left(\mathrm{x}-1\right)<{\log }_{{\left(0.3\right)}^2}\left(\mathrm{x}-1\right) \\ \Rightarrow {\log }_{0.3} \left(\mathrm{x}-1\right)<\frac{1}{2}{\log }_{\left(0.3\right)}\left(\mathrm{x}-1\right) \left[\mathrm{As} {\log }_{{\mathrm{a}}^{\mathrm{n}}}\mathrm{b}=\frac{1}{\mathrm{n}} {\log }_{\mathrm{a}}\mathrm{b}\right] \\ \Rightarrow 2{\log }_{0.3} \left(\mathrm{x}-1\right)<{\log }_{\left(0.3\right)}\left(\mathrm{x}-1\right) \\ \Rightarrow {\log }_{0.3} {\left(\mathrm{x}-1\right)}^2<{\log }_{0.3}\left(\mathrm{x}-1\right) \left[\mathrm{As} {\log }_{\mathrm{a}}{\mathrm{b}}^{\mathrm{m}}=\mathrm{m} {\log }_{\mathrm{a}}\mathrm{b}\right] \\ \mathrm{Now} \mathrm{using} \left\{\begin{array}{ll}{\mathrm{x}}_1>{\mathrm{x}}_2& \mathrm{if} \mathrm{a}>1\\ {\mathrm{x}}_1<{\mathrm{x}}_2& \mathrm{if} 0<\mathrm{a}<1\end{array}\right., \mathrm{we} \mathrm{get} \\ {\left(\mathrm{x}-1\right)}^2>\left(\mathrm{x}-1\right) \\ \Rightarrow {\left(\mathrm{x}-1\right)}^2-\left(\mathrm{x}-1\right)>0 \\ \Rightarrow \left(\mathrm{x}-1\right)\left[\mathrm{x}-1-1\right]>0 \\ \Rightarrow \left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)>0 \\ \mathrm{Now} \mathrm{using} \mathrm{method} \mathrm{of} \mathrm{interval} \\ -\infty +++++1----2+++++\infty \\ \mathrm{Hence} \mathrm{x}\in \left(-\infty ,1\right)\cup \left(2,\infty \right)...\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)\cap \left(\mathrm{ii}\right) \mathrm{gives} \\ \mathrm{x}\in \left(2,\infty \right) \end{gathered}$$

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