Question

The complete set of values of $x$ which satisfies the inequation ${\log }_{\left(1.5\right)}\left(\frac{2x-8}{x-2}\right)>0$ is

• A. $x\in (-\infty ,2)\cup (4,\infty )$
• B. $x\in (2,6)$
• C. $x\in (-\infty ,2)\cup (6,\infty )$
• D. $x\in (6,\infty )$

Answer 1

The correct option is C $x\in (-\infty ,2)\cup (6,\infty )$

${\log }_{\left(1.5\right)}\left(\frac{2x-8}{x-2}\right)>0$
For ${\log }_{\left(1.5\right)}\left(\frac{2x-8}{x-2}\right)$ to be defined if
$\frac{2x-8}{x-2}>0$
$\Rightarrow \frac{x-4}{x-2}>0$
$\Rightarrow x\in (-\infty ,2)\cup (4,\infty )...\left(1\right)$

Now, $\frac{2x-8}{x-2}>1$
$\Rightarrow \frac{2x-8}{x-2}-1>0$
$\Rightarrow \frac{2x-8-x+2}{x-2}>0$
$\Rightarrow \frac{x-6}{x-2}>0$
By using the wavy curve method, we have
$x\in (-\infty ,2)\cup (6,\infty )...\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we have
$x\in (-\infty ,2)\cup (6,\infty )$