Ответ:
Корни уравнения logₐx = b:
x = aᵇ
Решение неравенства logₐx > logₐc:
Решить уравнения:
log₂(x - 15) = 4:
x - 15 = 2⁴
x - 15 = 16
x = 16 + 15
x = 31
ln²(x - 2) = 4:
ln(x - 2) = ±2
x = e² + 2 или x = e⁻² + 2
lg²x + 2lg x = 8:
lg²x + 2lg x - 8 = 0
Пусть y = lg x, тогда y² + 2y - 8 = 0
D = 2² - 4 ⋅ 1 ⋅ (-8) = 4 + 32 = 36
y₁ = (-2 + √36) / 2 = (-2 + 6) / 2 = 4 / 2 = 2
y₂ = (-2 - √36) / 2 = (-2 - 6) / 2 = -8 / 2 = -4
lg x = 2 → x = 10² = 100
lg x = -4 → x = 10⁻⁴ = 0.0001
x = 100 или x = 0.0001
lg(x² - 2x - 4) = lg 11:
x² - 2x - 4 = 11
x² - 2x - 15 = 0
D = (-2)² - 4 ⋅ 1 ⋅ (-15) = 4 + 60 = 64
x₁ = (2 + √64) / 2 = (2 + 8) / 2 = 10 / 2 = 5
x₂ = (2 - √64) / 2 = (2 - 8) / 2 = -6 / 2 = -3
x = 5 или x = -3
Решить неравенства:
log₀.₆x > 2:
x < (0.6)²
x < 0.36
x ∈ (0, 0.36)
lg x ≤ -2:
x ≤ 10⁻²
x ≤ 0.01
x ∈ (0, 0.01]
ln x ≥ -3:
x ≥ e⁻³
x ≥ 1 / e³
x ∈ [1/e³, ∞)
log₇x < 1:
x < 7¹
x < 7
x ∈ (0, 7)