Question 630498: (Pre-calculus) In 1987, the number of AIDS cases in the US was estimated to be 50,000. In 2005, this number was estimated to be 1,000,000. Assume the trend continued. [*leave answers in terms of e and ln, etc*]
a) Set up a function of the form N(t) = N0 x e^kt
b) According to your model, how many cases occurred in 2000?
c) When, according to your model, will there be 5,000,000 cases?
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! (Pre-calculus) In 1987, the number of AIDS cases in the US was estimated to be 50,000. In 2005, this number was estimated to be 1,000,000. Assume the trend continued. [*leave answers in terms of e and ln, etc*]
a) Set up a function of the form N(t) = N0 x e^kt
t: 2005-1987 = 18 years
No: 50000
N(t): 1000000
plug above into:
N(t) = N0 x e^kt
1000000 = 50000 x e^(k*18)
solve for k:
20 = e^(k*18)
ln(20) = 18k
ln(20)/18 = k
Our equation is then:
N(t) = 50000e^(ln(20)/18*t)
.
b) According to your model, how many cases occurred in 2000?
t=2000-1987 = 13 years
N(13) = 50000e^(ln(20)/18*13)
N(13) = 50000e^(2.16358)
N(13) = 50000(8.7023)
N(13) = 435114
.
c) When, according to your model, will there be 5,000,000 cases?
set N(t) to 5000000 and solve for t:
5000000 = 50000e^(ln(20)/18*t)
100 = e^(ln(20)/18*t)
ln(100) = ln(20)/18*t
18*ln(100) = ln(20)*t
18*ln(100)/ln(20) = t
27.7 years = t
.
Year:
1987+28 = 2015
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