Posted by Megan on Sunday, October 14, 2012 at 9:29pm.
Please help! test tomorrow
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)
291000 e^(10(.050046)) = 479999
pretty close to 480000
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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