Question

If ${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$, then $x$ lies in the interval.

• A. $(2,\infty )$
• B. $(1,2)$
• C. $(-2,-1)$
• D. None of these

Answer 1

The correct option is A $(2,\infty )$

We have, ${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$,

$\Rightarrow {\log }_{0.3}(x-1)<{\log }_{(0.3{)}^2}(x-1)$

$\Rightarrow {\log }_{0.3}(x-1)<\frac{1}{2}{\log }_{0.3}(x-1)$, using power rule

$\Rightarrow 2{\log }_{0.3}(x-1)<{\log }_{0.3}(x-1)$

$\Rightarrow 2{\log }_{0.3}(x-1)-{\log }_{0.3}(x-1)<0$

$\Rightarrow {\log }_{0.3}(x-1)<0$

$\Rightarrow x-1>{0.3}^0=1$, since base is less that 1, so inequality reversed

$\Rightarrow x>2$

$\Rightarrow x\in (2,\infty )$