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Question 1104616: I just wanted see if oen of my answeres is correct here. At first there were 20 incidents. Three months later there were 560 incidents. How many incidents would there be at the end of 9 months? I solved this two ways: the first way I simply did 560/40 equals 14, and then for the growth over 9 months 40(14)(14)(14) equals 109,760. Since my textbook never validated this process, I used the At=A0e^kt formula and got 110071 (rounded up to get the decimal out). The two answers are in the same ball park, but they can't both be correct. Can you help? Thanks!
Found 2 solutions by greenestamps, Theo:
Answer by greenestamps(13334)   (Show Source):
You can put this solution on YOUR website!
You are working the same problem by two different methods; if your methods are right, you should be getting the same answer both ways.
Your problem is that you need to keep more decimal places in the value of k if you use the At=A0e^kt formula.
The value of k corresponding to a growth factor of 14 over each 3 month period is ln(14) = 2.6390573 (to 7 decimal places).
When I solve the problem using that many decimal places, I get the same 109760 answer.
When I use 2.64 for ln(14) (rounded to only 2 decimal places), I get your 110071 answer.
In general, because of the rapid growth nature of exponential growth, you need to keep a large number of digits in your value of k to get several digits of accuracy in your answer.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! not sure there were 20 incidents at first or 40.
i'll work with 40.
you started with 40 and then 3 months later there were 560.
the rate of increase for the 3 months was 560/40 = 14.
you then said, at the same rate of increase, you would multiply 40 * 14 * 14 * 14 to get 109760.
what you were doing was applying the discrete compounding formula of:
f = p * (1 + r) ^ n
f is the future value
p is the presentvalue
r is the interest rate per time period.
n is the number of time periods.
your time periods were every 3 months.
your formula would have become 560 = 40 * (1 + r) ^ 1
n was 1 because you were dealing with one 3 month period.
to find r, you would have done the following.
divide both sides of the equation by 40 to get 560/40 = (1 + r) ^ 1
since (1 + r) ^ 1 is the same as 1 + r, the equation becomes 560/40 = 1 + r.
you would then have subtracted 1 from both sides of the equation to get 560/40 - 1 = r
you would then have solved for r to get r = 13.
your formula would have become 560 = 40 * (1 + 13) ^ 1.
when n became three 3 month periods, the formula would have become f = 40 * (1 + 13) ^ 3 which would have become f = 40 * 14 ^ 3 which would have given you f = 109760.
if you used the continuous compounding formula of f = p * e^(rn) formula, then the procedure would be the same except you would have needed to do something different to get r.
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods.
the formula for your problem would have become 560 = 40 * e^(r*1)
this is because you were dealing with one 3 month period and you were looking to find the growth rate which you could then apply to multiple three 3 month periods.
you would have solved for r as follows:
formula becomes 560 = 40 * e^(r).
divide both sides of the equaion by 40 to get 560/40 = e^r.
take the natural log of both sides of the equation to get ln(560/40) = ln(e^r).
since ln(e^r) is equal to r*ln(e) which is equal to r, you get ln(560/40) = r.
you would solve for r to get r = ln(560/40) = 2.63905733.
when n = 3,the formula would becoomes f = 40 * e^(2.63905733 * 3) which would result in f = 109760.
the continous compounding formula and the discrete compounding formula are two diferent animals and you needed to solve for the growth rate in a different way.
the formula you gave me of At=A0 * e^kt is the same as the continuous compounding formula i showed you of f = p * e^(rn).
your At was my f
your A0 was my p
your k was my r
your t was my n
different nomenclature but same formula.
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