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Mathematics 7 Online
OpenStudy (anonymous):

assume that A(1)=$100.00 and A(2)=$200.00. use A=Pe^rt to find a) the rate b) the principal

OpenStudy (anonymous):

i know that to find the first part i am suppose have the equation of 100=Pe^r(1) which is just 100=Pe^r

OpenStudy (agent0smith):

\[\Large A(t)=P e^{rt}\] A(1) means when t=1, so \[\Large A(1)=$100.00 = P e ^{r}\]then A(2) means t=2 \[\Large A(2)=$200.00 = P e ^{2r}\] so we have \[\Large 100 = P e ^{r}\] \[\Large 200 = P e ^{2r}\]

OpenStudy (anonymous):

yeah i had up to that part i get stuck on how to simplify it here do i divide by P

OpenStudy (agent0smith):

Yeah you can solve either way. Maybe divide the first one by P, to get \[\Large \frac{ 100 }{P }= e ^{r}\] And remember you can write the second equation like this \[\Large 200 = P (e ^{r})^2\]

OpenStudy (anonymous):

so for the first one i dont know if i can do this but i wrote it as ln 100/p= r

OpenStudy (agent0smith):

You can do that, but you don't need to. Might be just easier to use substitution (see my reply above).

OpenStudy (anonymous):

oh so then i substitute the 100/p into the second ok let me try it

OpenStudy (anonymous):

so after i plugged it in i got 200=p(200/p^2) i can simplify one p so i get that p=1 not sure if thats correct seems wrong for some reason

OpenStudy (agent0smith):

P won't be 1... i think you used the wrong substitution, it was 100/P not 200/P

OpenStudy (agent0smith):

\[\Large 200 = P \left( \frac{ 100 }{ P }\right) ^2\]

OpenStudy (anonymous):

would i cancel out the p's

OpenStudy (agent0smith):

Do the squaring of everything inside the brackets first.

OpenStudy (anonymous):

so is 200/p^2

OpenStudy (anonymous):

i tried to solve it i got that P is 50 is that correct

OpenStudy (agent0smith):

100^2 isn't 200 :P

OpenStudy (anonymous):

sorry too late to be doing math lol 1000/p then i multiple p to both sides aftr i divide by 200

OpenStudy (agent0smith):

haha i know :P but it's not 1000 either Bell :P\[\Large 200 = P \times \frac{ 10000 }{ P^2 }\] i had to check 100^2 on a calculator... i wrote 100,000 first... oops :P

OpenStudy (anonymous):

i messed up again so from there i can simplify by canceling the p's

OpenStudy (agent0smith):

yep\[\Large 200 = \frac{ 10000 P }{ P^2 }\]

OpenStudy (anonymous):

so p is 50

OpenStudy (anonymous):

so now for the rate i plug in the principle and simplify

OpenStudy (agent0smith):

Yep! \[\Large \frac{ 100 }{P }= e ^{r}\] you'll prob need to take logs

OpenStudy (anonymous):

so i write is as ln 2 =r is it 69.31 or 6.93

OpenStudy (agent0smith):

r=0.69, so multiply by 100 to get a percentage

OpenStudy (anonymous):

my answers are multiple choice and i have to choose between those two or none so would it be none since it would have been 69%

OpenStudy (agent0smith):

0.6931*100=69.31%

OpenStudy (agent0smith):

r was 0.6931, i just rounded to 0.69

OpenStudy (anonymous):

oh ok wow this problem took me forever to understand well thank you so much i need to get some rest lol bye and thanks again

OpenStudy (agent0smith):

You're welcome. Good night!

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