Question

Find the range of $x$ for which the inequality ${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$ holds true.

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

Answer 1

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

Given: ${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$
The logarithm function is defined iff $(x-1)>0$
$\Rightarrow x>1\Rightarrow x\in (1,\infty )\cdots \left(A\right)$
Now, Simplifying the given inequality, we get:
${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$
$\Rightarrow {\log }_{0.3}(x-1)<{\log }_{(0.3{)}^2}(x-1)$
$\Rightarrow {\log }_{0.3}(x-1)<\frac{1}{2}{\log }_{0.3}(x-1)$
$\Rightarrow 2{\log }_{0.3}(x-1)<{\log }_{0.3}(x-1)$
$\Rightarrow {\log }_{0.3}(x-1{)}^2<{\log }_{0.3}(x-1)$

Now, using the property that for $0<a<1,{\log }_{a}b>{\log }_{a}c\Rightarrow b<c$
$\Rightarrow (x-1{)}^2>(x-1)$
$\Rightarrow (x-1{)}^2-(x-1)>0$
$\Rightarrow (x-1)\left[\right(x-1)-1]>0$
$\Rightarrow (x-1)(x-2)>0$
$\Rightarrow x\in (-\infty ,1)\cup (2,\infty )\cdots \left(B\right)$

Thus, the final solution set is the intersection of the solutions $A,B$
$\Rightarrow x\in A\cap B$
$\Rightarrow x\in (1,\infty )\cap \left\{\right(-\infty ,1)\cup (2,\infty \left)\right\}$

$\Rightarrow x\in (2,\infty )$