Logarithmic Function

2015-05-24 7:18 am
1. Given that log4=x and log24=y, express the following in terms of x and y.
(a) log0.06 (b)log3

2. Given that log2=a and log3=b, express log0.75 in terms of a and b.

3. Solve the following logarithmic equations.
(a) log(4-3x)=-1
(b) log(x-6)-log(x-8)=log2
(c) 5^(x-1)=3(x+1)
(d) log[x+(1/2)] + log8=log(2x+1)+1

回答 (1)

2015-05-24 8:09 am
✔ 最佳答案

1.
(a)
log(0.06)
= log(24 ÷ 4 ÷ 100)
= log(24) ‒ log(4) ‒ log(100)
= x ‒ y ‒ 2

(b)
log(3)
= log(24 ÷ 8)
= log(24 ÷ 2³)
= log(24) ‒ log(4³´²)
= log(24) ‒ (3/2)log(4)
= a ‒ (3/2)b
= (2a ‒ 3b)/2


====
2.
log(0.75)
= log(3 / 4)
= log(3 / 2²)
= log(3) ‒ log(2²)
= log(3) ‒ 2log(2)
= b ‒ 2a


====
3.
(a)
log(4 ‒ 3x) = 1
log(4 ‒ 3x) = log(10)
4 ‒ 3x = 10
3x = ‒6
x = ‒2

(b)
log(x ‒ 6) ‒ log(x ‒ 8) = log(2)
log[(x ‒ 6) / (x ‒ 8)] = log(2)
(x ‒ 6) / (x ‒ 8) = 2
x ‒ 6 = 2x ‒ 16
x = 10

(c)
5ˣ⁻¹ = 3ˣ⁺¹
log(5ˣ⁻¹) = log(3ˣ⁺¹)
(x ‒ 1) log(5) = (x + 1) log(3)
x log(5) ‒ log(5) = x log(3) + log(3)
x [log(5) ‒ log(3)] = log(5) + log(3)
x = [log(5) + log(3)] / [log(5) ‒log(3)]
x ≈ 5.301

(d)
log[x + (1/2)] + log8 = log(2x + 1) + 1
log[x + (1/2)] + log8 = log(2x + 1) + log(10)
log{8[x + (1/2)]} = log[10(2x + 1)]
8[x + (1/2)] = 10(2x + 1)
8x + 4 = 20x + 10
12x = ‒6
x = ‒1/2 (rejected for log[(‒1/2) + (1/2)] is undefined.)
Hence, there is no solution.

2015-05-24 00:10:45 補充:
3(d) The last two lines should be :

x = ‒1/2 (rejected for log[(‒1/2) + (1/2)] = log(0) is undefined.)
Hence, there is no solution.


收錄日期: 2021-04-15 20:16:40
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