解不等式log0.5(3-x)>=-2

log(0.5)(3-x)≥-2
log(0.5)(3-x)≥log(0.5)(0.5⁻²)
log(0.5)(3-x)≥log(0.5)4
因为f(x)=log(0.5)x是减函数，所以有不等式
3-x≤4，x≥-1
又因为真数要大于0，所以3-x>0，x<3
不等式的解是：-1≤x<3

log0.5(x)是单调减函数
=>
log0.5(3-x) >= log0.5(4)
=>
0 < 3-x <= 4
=>
-1<=x<3

log0.5(3-x)>=-2
3-x>=（1/2）^-2
3-x>=4
x<=-1

上面都解了吗