Verified answer

1) log₈(4-2x)>=log₈(8²)

4-2x>0      x<2

4-2x>=64  x<=-30  x∈(-∞;-30]

2) log₁/₃(x-1)>=log₁/₃(1/3)⁻²

x-1>0 x>1

x-1<=9( знак поменяли т.к. основание  <1)

x<=10  x∈(1;10]

3) log₂/₃(2-5x)<log₂/₃(2/3)⁻²

2-5x>0   x<0.4

2-5x>9/4   x<-0.05

x∈(-∞;-0.05)

Verified answer

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