If $\log_{0.1} (x - 3) < \log_{0.01} (x - 3)$ then x lies in

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$(4, \infty)$

$(3, \infty)$

[3,4]

(3,4)

answer is D.

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Detailed Solution

$\begin{array}{l} \log_{0.1} (x - 3) < \log_{(0.1)^{2}} (x - 3) \\ \Leftrightarrow 2 \log_{0.1} (x - 3) < \log_{(0.1)} (x - 3) \\ \Leftrightarrow \log_{0.1} (x - 3) < 0 \\ \Leftrightarrow x - 3 > 1 \Leftrightarrow x > 4 \end{array}$

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