Odpowiedź:
a)
log₃(2x - 2) = 3
założenie:
2x- 2 > 0
2x > 2
x > 2/2
x > 1
D: x ∈ ( 1 , + ∞ )
3³ = 2x - 2
9 = 2x- 2
2x = 9+2= 11
x = 11/2 = 5 1/2
b)
log₂x + 4 = 4
założenie :
x + 4 > 0
x > - 4
D: x ∈ ( - 4 , + ∞ )
2⁴= x+ 4
16 = x + 4
x = 16 - 4 = 12
c)
log₂(x - 1) = 1
x - 1 > 0
2¹ = x - 1
2 = x - 1
x = 2 + 1 = 3
d)
log₂(x² + 7x) = 3
x² + 7x > 0
x(x + 7) > 0
x > 0 ∧ x + 7 > 0 ∨ x < 0 ∧ x + 7 < 0
x > 0 ∧ x > - 7 ∨ x < 0 ∧ x < - 7
x < - 7 ∨ x > 0
D: x ∈ (- ∞ , - 7 ) ∪ ( 0 , + ∞ )
∧ - znaczy "i"
∨ - znaczy "lub"
2³ = x² + 7x
8 = x² + 7x
x² + 7x - 8 =0
a = 1 , b = 7, c = - 8
Δ = b² - 4ac = 7² - 4 * 1 * (- 8) = 49 + 32 = 81
√Δ = √81 = 9
x₁= (- b - √Δ)/2a = ( - 7 - 9)/2 = - 16/2 = - 8
x₂ = ( - b + √Δ)/2a = ( - 7 + 9)/2= 2/2= 1
e)
log₅(x - 2)/5 = 2
(x - 2)/5 > 0
x - 2 > 0
x > 2
D: x ∈ ( 2 , + ∞ )
5² = (x - 2)/5
25 = (x - 2)/5
25*5= x - 2
125 = x - 2
x = 125 + 2 = 127
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Odpowiedź:
a)
log₃(2x - 2) = 3
założenie:
2x- 2 > 0
2x > 2
x > 2/2
x > 1
D: x ∈ ( 1 , + ∞ )
log₃(2x - 2) = 3
3³ = 2x - 2
9 = 2x- 2
2x = 9+2= 11
x = 11/2 = 5 1/2
b)
log₂x + 4 = 4
założenie :
x + 4 > 0
x > - 4
D: x ∈ ( - 4 , + ∞ )
log₂x + 4 = 4
2⁴= x+ 4
16 = x + 4
x = 16 - 4 = 12
c)
log₂(x - 1) = 1
założenie:
x - 1 > 0
x > 1
D: x ∈ ( 1 , + ∞ )
log₂(x - 1) = 1
2¹ = x - 1
2 = x - 1
x = 2 + 1 = 3
d)
log₂(x² + 7x) = 3
założenie:
x² + 7x > 0
x(x + 7) > 0
x > 0 ∧ x + 7 > 0 ∨ x < 0 ∧ x + 7 < 0
x > 0 ∧ x > - 7 ∨ x < 0 ∧ x < - 7
x < - 7 ∨ x > 0
D: x ∈ (- ∞ , - 7 ) ∪ ( 0 , + ∞ )
∧ - znaczy "i"
∨ - znaczy "lub"
log₂(x² + 7x) = 3
2³ = x² + 7x
8 = x² + 7x
x² + 7x - 8 =0
a = 1 , b = 7, c = - 8
Δ = b² - 4ac = 7² - 4 * 1 * (- 8) = 49 + 32 = 81
√Δ = √81 = 9
x₁= (- b - √Δ)/2a = ( - 7 - 9)/2 = - 16/2 = - 8
x₂ = ( - b + √Δ)/2a = ( - 7 + 9)/2= 2/2= 1
e)
log₅(x - 2)/5 = 2
założenie:
(x - 2)/5 > 0
x - 2 > 0
x > 2
D: x ∈ ( 2 , + ∞ )
log₅(x - 2)/5 = 2
5² = (x - 2)/5
25 = (x - 2)/5
25*5= x - 2
125 = x - 2
x = 125 + 2 = 127