Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (boxman61):

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)

OpenStudy (anonymous):

wanna do it the snap way?

OpenStudy (boxman61):

rater not do it at all lol... but have to show work.

OpenStudy (anonymous):

oh ok then you have to do it the hard way ready?

OpenStudy (anonymous):

we start counting in 1965, so we will make that year zero, and so\(P_0=55,000\)

OpenStudy (anonymous):

then we know that our formula will look like \[A=55,000e^{kt}\] and we need to find \(k\)

OpenStudy (anonymous):

how did you get \(k\)?

OpenStudy (boxman61):

sorry wrong problem, my mistake

OpenStudy (anonymous):

wrong problem as in the question is the wrong problem or as in you solved the wrong problem?

OpenStudy (boxman61):

solved the wrong problem..... to find k i found the percentage growth of 55,000 to 89,500 which is 63%

OpenStudy (anonymous):

ok if you are going to do that, then it is a different formula

OpenStudy (boxman61):

\[89500e ^{.63(15)}\]\[89500e ^{9.45}\] = 1137380791

OpenStudy (anonymous):

you would use \[55,000\times (1.63)^{\frac{t}{30}}\]

OpenStudy (anonymous):

if you are using \(e\) as the base, you have to solve for k

OpenStudy (boxman61):

thought i was doing something wrong....

OpenStudy (anonymous):

\[A=55,000e^{kt}\] and you know \(A=89,500\) when \(t=30\) so you have to solve \[89,500=55,000e^{30k}\] for k

OpenStudy (anonymous):

divide by 55,000 get \[\frac{179}{110}=e^{30k}\] \[30k=\ln(\frac{179}{110})\] \[k=\frac{1}{30}\ln(\frac{179}{110})\]

OpenStudy (anonymous):

i get \(.02623\) rounded and now you have your \(k\) and therefore your formula

OpenStudy (boxman61):

im getting k=0.016

OpenStudy (anonymous):

your way was the snap way, but it doesn't use \[A=P_0e^{dt}\] as i said you would use \[A=55,000\times \left(\frac{89,500}{55,000}\right)^{\frac{t}{30}}\]

OpenStudy (anonymous):

you are right, i copied wrong

OpenStudy (anonymous):

.01623

OpenStudy (boxman61):

so it becomes \[A=89500e ^{0.16(15)}\]\[A=895e ^{2.4}=986574.29\] correct?

OpenStudy (anonymous):

from 1995 to 2005 is ten years, not 15

OpenStudy (anonymous):

\[A=89500e^{.1623}\]

OpenStudy (anonymous):

i get 105,271

OpenStudy (boxman61):

mixing up problems again (cant post my actualy problem so crafter a similar one)... thank you

OpenStudy (anonymous):

why can't you post the actual one?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!