in 1965 a town had a population of 55,000. in 1995 a town had a population of 89,500. assuming exponential growth, estimate the population in the year 2005? use A=P(o)e^(kt)
wanna do it the snap way?
rater not do it at all lol... but have to show work.
oh ok then you have to do it the hard way ready?
we start counting in 1965, so we will make that year zero, and so\(P_0=55,000\)
then we know that our formula will look like \[A=55,000e^{kt}\] and we need to find \(k\)
how did you get \(k\)?
sorry wrong problem, my mistake
wrong problem as in the question is the wrong problem or as in you solved the wrong problem?
solved the wrong problem..... to find k i found the percentage growth of 55,000 to 89,500 which is 63%
ok if you are going to do that, then it is a different formula
\[89500e ^{.63(15)}\]\[89500e ^{9.45}\] = 1137380791
you would use \[55,000\times (1.63)^{\frac{t}{30}}\]
if you are using \(e\) as the base, you have to solve for k
thought i was doing something wrong....
\[A=55,000e^{kt}\] and you know \(A=89,500\) when \(t=30\) so you have to solve \[89,500=55,000e^{30k}\] for k
divide by 55,000 get \[\frac{179}{110}=e^{30k}\] \[30k=\ln(\frac{179}{110})\] \[k=\frac{1}{30}\ln(\frac{179}{110})\]
i get \(.02623\) rounded and now you have your \(k\) and therefore your formula
im getting k=0.016
your way was the snap way, but it doesn't use \[A=P_0e^{dt}\] as i said you would use \[A=55,000\times \left(\frac{89,500}{55,000}\right)^{\frac{t}{30}}\]
you are right, i copied wrong
.01623
so it becomes \[A=89500e ^{0.16(15)}\]\[A=895e ^{2.4}=986574.29\] correct?
from 1995 to 2005 is ten years, not 15
\[A=89500e^{.1623}\]
i get 105,271
mixing up problems again (cant post my actualy problem so crafter a similar one)... thank you
why can't you post the actual one?
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