Question

The range of values of $x$ which satisfy the inequation $lo{g}_{\left(1.5\right)}\left(\frac{2x-8}{x-2}\right)>0$ is

• A. $x\in [2,6]$
• B. $x\in (2,6)$
• C. $x\in (-\infty ,2)\cup (6,\infty )$
• D. None of these

Answer 1

Answer: (C)

$x\in (-\infty ,2)\cup (6,\infty )$

The base of the logarithm is $1.5$ which is greater than $1$

So, in order for the inequation to be greater than $0$

$\left(\frac{2x-8}{x-2}\right)$ has to be greater than $1$

$\therefore \frac{2x-8}{x-2}>1\Rightarrow \frac{2x-8}{x-2}-1>0$

$\Rightarrow \frac{2x-8-x+2}{x-2}-1>0$

$\Rightarrow \frac{x-6}{x-2}>0$

$\Rightarrow (x>6,x>2)$ or $(x<6,x<2)$

Now, for the region refer to figure

So, $x>2andx<6$

$\therefore [2,6]$ region is avoided

$\therefore x>6$ or $x<2$.

Hence the solution is
$(x>6orx<2)$

i.e., $x\in (-\infty ,2)\cup (6,\infty )$