Question

If ${\log }_{0.04}(x-1)\ge {\log }_{0.2}(x-1),$ then

• A. $x\in (1,\infty )$
• B. $x\in (-\infty ,1]\cup [2,\infty )$
• C. $x\in [2,\infty )$
• D. $x\in [1,2]$

Answer 1

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

For the inequality to be defined,
$x-1>0\Rightarrow x>1$

Now, ${\log }_{0.04}(x-1)\ge {\log }_{0.2}(x-1)$
$\Rightarrow {\log }_{{0.2}^2}(x-1)\ge {\log }_{0.2}(x-1)\Rightarrow \frac{1}{2}{\log }_{0.2}(x-1)\ge {\log }_{0.2}(x-1)\Rightarrow {\log }_{0.2}(x-1)\ge 2{\log }_{0.2}(x-1)\Rightarrow {\log }_{0.2}(x-1)\le 0$
$\Rightarrow (x-1)\ge 1[\because {\log }_{a}x\text{is decreasing for}0<a<1]\Rightarrow x\ge 2$

Hence, $x\ge 2$