Question

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

• A. $x\in (0,\infty )$
• B. $x\in (1,\infty )$
• C. $x\in (2,\infty )$
• D. $x\in (3,\infty )$

Answer 1

The correct option is C $x\in (2,\infty )$

${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$
$\Rightarrow {\log }_{0.3}(x-1)-\frac{1}{2}{\log }_{0.3}(x-1)<0$
$\Rightarrow {\log }_{0.3}(x-1)<0$
$x-1>1$, as $0<$ base $<1$
$x>2$