Question

Solve the following inequality:
$\sqrt{{\log }_2\frac{3-2x}{1-x}}<1$

Answer 1

${\log }_2\frac{3-2x}{1-x}<1$

$0<\frac{3-2x}{1-x}<2$

$\frac{3-2x}{1-x}>0$

$\frac{x-\frac{3}{2}}{x-1}>0$

$Requiredregionis(1,\frac{3}{2})$

$Nowtaking,\frac{3-2x}{1-x}<2$

$\frac{x-\frac{3}{2}}{x-1}<1$

$\frac{x-\frac{3}{2}-x+1}{x-1}<0$

$\frac{\frac{-1}{2}}{x-1}<0$

$\frac{1}{x-1}>0$

$x>1$

$Therefore,x\epsilon (1,\frac{3}{2})$