Решение
log²₁/₂ x + log₁/₂ x - 2 ≤ 0
ОДЗ: x > 0
log₁/₂ x = t
t² + t - 2 ≤ 0
t₁ = 1
t₂ = - 2
- 2 ≤ t ≤ 1
1) log₁/₂ x ≤ 1
log₁/₂ x ≤ log₁/₂ (1/2)
так как 0 < 1/2 < 1, то
x ≥ 1/2
2) log₁/₂ x ≥ - 2
log₁/₂ x ≥ - 2log₁/₂ (1/2)
log₁/₂ x ≥ log₁/₂ (1/2)⁻²
так как 0 < 1/2 < 1, то
x ≤ (1/2)⁻²
x ≤ 4
x ∈ [1/2; 4]