Question

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

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

Answer 1

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

Given,

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

Upon applying log rules, we get,

${\log }_{0.3}(x-1)<{\log }_{0.3}\left(\sqrt{x-1}\right)$

$\Rightarrow x-1>\sqrt{x-1}$ and

$\sqrt{x-1}>0$

Now,

$x-1>\sqrt{x-1}$

$(\sqrt{x-1}{)}^2<(x-1{)}^2$

$x-1<x^2-2x+1$

$x<1orx>2$

Now,

$x-1\ge 0:x\ge 1$

Overlapping all the intervals, we get,

$x>2$