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mysheva
@mysheva
September 2021
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Пожалуйста,решите что сможете,очень нужно
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nKrynka
Решение
1) log₂ (3 - x) - log₂ (1 - x) = 3
ОДЗ: 3 - x > 0, x < 3
1 - x > 0, x < 1
x ∈ (- ∞; 1)
log₂ (3 - x) = log₂ (1 - x) + 3log
₂ 2
log₂ (3 - x) = log₂ (1 - x) + log
₂ 2³
log₂ (3 - x) = log₂ [(1 - x)*8]
3 - x = 8 - 8x
8x - x = 8 - 3
7x = 5
x = 5/7
Ответ:
x = 5/7
2) log₂ x + log₂ (x - 2) ≤ 3
ОДЗ: x > 0
x - 2 > 0, x > 2
x ∈ (2; + ∞)
log₂ x + log₂ (x - 2) ≤ 3log₂ 2
log₂ [x*(x - 2)] ≤ log₂ 2³
так как основание логарифма 2 > 1, то
x*(x - 2) ≤ 8
x² - 2x - 8
≤ 0
x₁= - 2
x₂ = 4
x ∈ [- 2; 4]
С учётом ОДЗ
x ∈ (2; 4]
Ответ:
x ∈ (2; 4]
2 votes
Thanks 2
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Answers & Comments
1) log₂ (3 - x) - log₂ (1 - x) = 3
ОДЗ: 3 - x > 0, x < 3
1 - x > 0, x < 1
x ∈ (- ∞; 1)
log₂ (3 - x) = log₂ (1 - x) + 3log₂ 2
log₂ (3 - x) = log₂ (1 - x) + log₂ 2³
log₂ (3 - x) = log₂ [(1 - x)*8]
3 - x = 8 - 8x
8x - x = 8 - 3
7x = 5
x = 5/7
Ответ: x = 5/7
2) log₂ x + log₂ (x - 2) ≤ 3
ОДЗ: x > 0
x - 2 > 0, x > 2
x ∈ (2; + ∞)
log₂ x + log₂ (x - 2) ≤ 3log₂ 2
log₂ [x*(x - 2)] ≤ log₂ 2³
так как основание логарифма 2 > 1, то
x*(x - 2) ≤ 8
x² - 2x - 8 ≤ 0
x₁= - 2
x₂ = 4
x ∈ [- 2; 4]
С учётом ОДЗ
x ∈ (2; 4]
Ответ: x ∈ (2; 4]