JEE MainMathematicsToughLogarithm

$\log_{0.3}(x - 1) < \log_{0.09}(x - 1)$ then x lies in the
interval

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\log_{0.3}(x - 1) < \log_{0.09}(x - 1)

\Rightarrow \log_{0.3}(x - 1) < \log_{(0.3)^{2}}(x - 1)

\Rightarrow 2\log_{0.3}(x - 1) < \log_{0.3}(x - 1)

\Rightarrow (x - 1)^{2} > x - 1

\Rightarrow (x - 1)(x - 2) > 0

\Rightarrow x \in ( - \infty,1) \cup (2,\infty)

.......(1)
But to define the given relation

x - 1 > 0 \Rightarrow x \in (1,\infty)

......(2)

\therefore(1) \cap (2) \Rightarrow x \in (2,\infty)

Source: Grass-G / M-16 Mathematics Paper 25 Nov First.docx

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