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 $(2,\infty )$

$\begin{array}{c}{\log }_{0.3}\left(x-1\right)<{\log }_{0.09}\left(x-1\right)\\ {\log }_{0.3}{\left(x-1\right)}^2<{\log }_{0.3}\left(x-1\right)\\ {\left(x-1\right)}^2>\left(x-1\right)x-1>0\\ x>0---\left(1\right)\\ x^2+1-2x-x+1>0\\ x^2-3x+2>0\\ \left(x-2\right)\left(x-1\right)>0\\ x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)\\ x\in \left(2,\infty \right)----\left(2\right)\\ From\left(i\right)and\left(2\right)weget\\ option\left(A\right)isthecorrectanswer.\end{array}$