Question

If $\sqrt{{\log }_2\left(\frac{2x-3}{x-1}\right)}<1,$ then $x\in$

• A. $(2,\infty )$
• B. $[2,\infty )$
• C. $(1,\infty )$
• D. $[1,\infty )$

Answer 1

The correct option is B $[2,\infty )$

${\log }_2\left(\frac{2x-3}{x-1}\right)\ge 0$ and $\frac{2x-3}{x-1}>0$
$\Rightarrow \left(\frac{2x-3}{x-1}\right)\ge 1$
$\Rightarrow \left(\frac{2x-3-x+1}{x-1}\right)\ge 0$
$\Rightarrow \frac{x-2}{x-1}\ge 0$
$\Rightarrow x\in (-\infty ,1)\cup [2,\infty )\cdots \left(1\right)$

Now, $\sqrt{{\log }_2\left(\frac{2x-3}{x-1}\right)}<1$
$\Rightarrow {\log }_2\left(\frac{2x-3}{x-1}\right)<1$
$\Rightarrow \frac{2x-3}{x-1}<2$
$\Rightarrow \frac{-1}{x-1}<0$
$\Rightarrow x-1>0$
$\Rightarrow x>1\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$x\in [2,\infty )$