Question
If $${\log _{0.3}}\left( {x - 1} \right) < {\log _{0.09}}\left( {x - 1} \right),$$ then $$x$$ lies in the interval-
A.
$$\left( {2,\infty } \right)$$
B.
$$\left( {1,2} \right)$$
C.
$$\left( { - 2, - 1} \right)$$
D.
none of these
Answer :
$$\left( {2,\infty } \right)$$
Solution :
First of all for $$\log \left( {x - 1} \right)$$ to be defined, $$x - 1 > 0$$
$$\eqalign{
& \Rightarrow \,x > 1\,\,\,\,\,\,\,.....\left( 1 \right) \cr
& {\text{Now, }}{\log _{0.3}}\left( {x - 1} \right) < {\log _{0.09}}\left( {x - 1} \right) \cr
& \Rightarrow \,\,{\log _{0.3}}\left( {x - 1} \right) < {\log _{\left( {0.3} \right)}}^2\left( {x - 1} \right) \cr
& \Rightarrow \,\,{\log _{0.3}}\left( {x - 1} \right) < \frac{1}{2}{\log _{0.3}}\left( {x - 1} \right) \cr
& \Rightarrow \,\,2{\log _{0.3}}\left( {x - 1} \right) < {\log _{0.3}}\left( {x - 1} \right) \cr
& \Rightarrow \,\,{\log _{0.3}}{\left( {x - 1} \right)^2} < {\log _{0.3}}\left( {x - 1} \right) \cr
& \Rightarrow \,\,{\left( {x - 1} \right)^2} > \left( {x - 1} \right)\,\,\,\,\,\,\,\,{\bf{NOTE}}\,\,{\bf{THIS}}\,\,{\bf{STEP}} \cr
& \,\,\,\,\,\,\left[ {{\text{The inequality is reversed since base lies between 0 and 1}}} \right] \cr
& \Rightarrow \,{\left( {x - 1} \right)^2} - \left( {x - 1} \right) > 0 \cr
& \Rightarrow \,\left( {x - 1} \right)\left( {x - 2} \right) > 0\,\,\,\,\,\,\,.....\left( 2 \right) \cr
& \,\,\,\,\,{\text{Combining}}\left( 1 \right){\text{and}}\left( 2 \right){\text{we get}}\,\,x > 2 \cr
& \therefore \,\,x \in \left( {2,\infty } \right) \cr} $$