數學題PLZ

2011-11-13 6:29 am
(1+y)^20=1.47
求y
400(1-R)^24=400/1.47
求R

要埋方法PLZ

回答 (2)

2011-11-13 8:13 am
✔ 最佳答案
(1 + y)^20 = 1.47
log(1 + y)^20 = log1.47
20 * log(1 + y) = log1.47
log(1 + y) = (1/20) * log1.47
log(1 + y) = log(1.47)^(1/20)
1 + y = 1.47^(1/20)
y = [1.47^(1/20)] - 1
y ≈ 0.01945 (to 4 sig. fig.)


=====
400(1 - R)^24 = 400/1.47
(1 - R)^24 = 400/(1.47 * 400)
(1 - R)^24 = 1/1.47
(1 - R)^24 = 1.47^(-1)
log(1 - R)^24 = log(1.47)^(-1)
24 * log(1 - R) = log(1.47)^(-1)
log(1 - R) = (1/24) * log(1.47)^(-1)
log(1 - R) = log(1.47)^(-1/24)
1 - R = 1.47^(-1/24)
R = 1 - [1.47^(-1/24)]
R = 0.01592 (to 4 sig. fig.)
參考: andrew
2011-11-13 11:03 am
(1+y)^20=1.47
1+y = (1.47)^1/20
y= 1.47^1/20-1
y=0.019449851

400(1-R)^24=400/1.47
1-R= (1/1.47)^(1/24)
R= (1/1.47)^(1/24)+1
R=1.9841


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