Question

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

• A. $(2,\infty )$
• B. $(-2,-1)$
• C. $(1,2)$
• D. $(1.3,\infty )$

Answer 1

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

We have,
${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$
Clearly, it is defined for $x>1$
Now,
${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$
$\Rightarrow {\log }_{0.3}(x-1)<{\log }_{(0.3{)}^2}(x-1)$
$\Rightarrow {\log }_{0.3}(x-1)<\frac{1}{2}{\log }_{0.3}(x-1)$
$\Rightarrow {\log }_{0.3}(x-1)<0\Rightarrow x-1>(0.3{)}^0$

${\log }_{a}x$ is decreasing function if $0<a<1$, that's why the inequality reverses

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