Question

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

• A. $\left(2,\infty \right)$
• B. $\left(1,2\right)$
• C. $\left(-2,-1\right)$
• D. None of these

Answer 1

The correct option is A

$\left(2,\infty \right)$

Explanation for the correct option.

Step 1: Simplify the given inequality.

$$\begin{gathered} {\log }_{0.3}(x-1)<{\log }_{0.09}(x-1) \\ \Rightarrow {\log }_{0.3}(x-1)<{\log }_{{\left(0.3\right)}^2}(x-1) \\ \Rightarrow {\log }_{0.3}(x-1)<\frac{1}{2}{\log }_{\left(0.3\right)}(x-1) \\ \Rightarrow 2{\log }_{0.3}(x-1)<{\log }_{\left(0.3\right)}(x-1) \\ \Rightarrow {\log }_{0.3}{(x-1)}^2<{\log }_{\left(0.3\right)}(x-1) \end{gathered}$$

Step 2: Apply log rule.

We know that if ${\log }_{b}a<{\log }_{b}c$, where $0<b<1$, then $a>c$. So here we can say that,

$\begin{array}{rcl}{\left(x-1\right)}^2& >& x-1\\ \Rightarrow x^2-2x+1& >& x-1\\ \Rightarrow x^2-3x+2& >& 0\\ & \Rightarrow & \left(x-1\right)\left(x-2\right)>0\end{array}$

Step 3: Find the interval of $x$.

From $\begin{array}{rcl}\left(x-1\right)\left(x-2\right)& >& 0\end{array}$ we get $x\in \left(1,\infty \right)\cup \left(2,\infty \right)$, therefore, $x$ lies in $\left(2,\infty \right)$.

Hence, option A is correct.