Asked by Megan
In 1960, the population was 291000. In 1970, the population was 480000. Assuming exponential growth, what is the annual percent rate? and doubling time?
The answers are 5.13% and 14 years. I do not know how to get this though :( Please help!
The answers are 5.13% and 14 years. I do not know how to get this though :( Please help!
Answers
Answered by
Megan
Please help! test tomorrow
Answered by
Megan
thanks:)
Answered by
Reiny
Let's use the equation
number = a e^kt, where a is the initial value, k is the rate of growth and t is the number of years
so let 1960 correspond to a time of t = 0
then 1970 ----> t = 10
also a = 291000
480000 = 291000 e^10k
1.64948 = e^10k
take ln of both sides
ln 1.64948 = ln e^10k = 10k
k = .50046/10 = .050046 = 5.005 %
(no idea how they got their answer, it is not correct)
check:
291000 e^(10(.050046)) = 479999
pretty close to 480000
their answer:
291000 e^(10(.0513)) = 486056 , too big
for doubling time:
2 = 1(e^.050046t)
ln 2 = .050046t
t = ln 2/.050046 = 13.8 years.
number = a e^kt, where a is the initial value, k is the rate of growth and t is the number of years
so let 1960 correspond to a time of t = 0
then 1970 ----> t = 10
also a = 291000
480000 = 291000 e^10k
1.64948 = e^10k
take ln of both sides
ln 1.64948 = ln e^10k = 10k
k = .50046/10 = .050046 = 5.005 %
(no idea how they got their answer, it is not correct)
check:
291000 e^(10(.050046)) = 479999
pretty close to 480000
their answer:
291000 e^(10(.0513)) = 486056 , too big
for doubling time:
2 = 1(e^.050046t)
ln 2 = .050046t
t = ln 2/.050046 = 13.8 years.
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