Verified answer

ОДЗ: x²-3x>0

x∈(-∞;0)∪(3;+∞)

Решение:

log2(x²-3x)≤2

log2(x²-3x)≤log2(4)

x²-3x≤4

x²-4x+x-4≤0

x(x-4)+(x-4)≤0

(x-4)(x+1)≤0

x∈[-1;4]

Ответ: x∈[-1;0)∪(3;4]