If log0.5 (x−1)<log0.25 (x−1), then x lies in the interval.
(2,∞)
(3,∞)
(−∞,0)
(0, 3)
log0.5 (x−1)<log0.25 (x−1)⇔log0.5 (x−1)<log(0.5)2 (x−1)=log0.5 (x−1)log0.5 (0.5)2 =12log0.5 (x−1)⇔log0.5 (x−1)<0⇔x−1>1⇔xϵ(2,∞)