Question

Solve the following inequality:
${\log }_{0.5}(4-x)⩾{\log }_{0.5}2-{\log }_{0.5}(x-1)$

Answer 1

$-{\log }_2(4-x)\ge -{\log }_{2}2+{\log }_2(x-1)$

${\log }_{2}2-{\log }_2(4-x)\ge {\log }_2(x-1)$

Taking antilog,we get,

$\frac{2}{4-x}\ge x-1$

$\frac{2-(4-x)(x-1)}{4-x}\ge 0$

$\frac{2-(4x-4-x^2+x)}{4x}\ge 0$

$\frac{2-5x+x^2+4}{x-4}\ge 0$
$\frac{x^2-5x+6}{x-4}\le 0$

$\frac{(x-3)(x-2)}{(x-4)}\le 0therefore,$

$x\epsilon (-\infty ,2]\cup [3,4)$

$Now,asx<4andx>1$

$x\epsilon (1,4)$

$so,wecansaythat,$

$x\epsilon (1,2]\cup [3,4)$