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!
3 answers
Please help! test tomorrow
thanks:)
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.