Question

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

• A. $x>2$
• B. $x<2$
• C. $x>-2$
• D. None of these

Answer 1

The correct option is A $x>2$

$lo{g}_{0.3}(x-1)<lo{g}_{0.09}(x-1)$

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

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

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

Taking antilog

$\Rightarrow (x-1)>\sqrt{x-1}$

$\Rightarrow (x-1{)}^2-(x-2)>0$

$\Rightarrow (x-1)(x-2)>0$

$Asx\ne 1$

$\Rightarrow x>2$