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greenandblue
Group Title
PLEASE HELP ME ON THIS!!
Joe deposits $1,500 in an account that pays 3% annual interest compounded continuously.
a. How much will Joe have in his account after 5 years?
b. How long will it take Joe to double his money? Use natural logarithms and explain your answer.
 one year ago
 one year ago
greenandblue Group Title
PLEASE HELP ME ON THIS!! Joe deposits $1,500 in an account that pays 3% annual interest compounded continuously. a. How much will Joe have in his account after 5 years? b. How long will it take Joe to double his money? Use natural logarithms and explain your answer.
 one year ago
 one year ago

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bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
multiply 1,500 * 0.03 * 5
 one year ago

bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
then add tht 2 1,500
 one year ago

bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
so it would b 1,500 + 225
 one year ago

bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
so the first answer is $1,725
 one year ago

greenandblue Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much but i got one question how did you get the 225 and also what does it mean natural logarithm?
 one year ago

bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
the 225 is 1,500 * 0.03 * 5
 one year ago

bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
and no i dont know wht the natural logarithm is
 one year ago

greenandblue Group TitleBest ResponseYou've already chosen the best response.0
okay thank you so much seriously you have no idea!:)
 one year ago

bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
lol ur welcome
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[1500(1.03)^5\] for the first one second one solve \[2=(1.03)^t\] via \[t=\frac{\ln(2)}{\ln(1.03)}\]
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.2
We seem to be missing an important word in the original problem statement. "CONTINUOUSLY." a) \(1500\cdot e^{0.03\cdot5} = 1742.75\) b) \(1500\cdot e^{0.03\cdot t} = 2\cdot 1500 = 3000\) or \(e^{0.03\cdot t} = 2\) Solving for \(t\) \(t = \dfrac{ln(2)}{0.03} = 23.105\)
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.2
@greenandblue @bakonloverk @satellite73
 one year ago

bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
u called 4 me?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.2
Yup. You did Simple Interest, Satellite73 did Annual Compound Interest. The problem statement requires Continuous Compounding. That's all. Lot's of ways to tackle these things. That's why banks and insurance companies are so confusing. :)
 one year ago

bakonloverk Group TitleBest ResponseYou've already chosen the best response.1
ohhhh kk
 one year ago
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