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Mathematics 13 Online
OpenStudy (anonymous):

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.

OpenStudy (bakonloverk):

multiply 1,500 * 0.03 * 5

OpenStudy (bakonloverk):

then add tht 2 1,500

OpenStudy (bakonloverk):

so it would b 1,500 + 225

OpenStudy (bakonloverk):

so the first answer is $1,725

OpenStudy (anonymous):

Thank you so much but i got one question how did you get the 225 and also what does it mean natural logarithm?

OpenStudy (bakonloverk):

the 225 is 1,500 * 0.03 * 5

OpenStudy (bakonloverk):

and no i dont know wht the natural logarithm is

OpenStudy (anonymous):

okay thank you so much seriously you have no idea!:)

OpenStudy (bakonloverk):

lol ur welcome

OpenStudy (anonymous):

\[1500(1.03)^5\] for the first one second one solve \[2=(1.03)^t\] via \[t=\frac{\ln(2)}{\ln(1.03)}\]

OpenStudy (tkhunny):

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 \(t\) \(t = \dfrac{ln(2)}{0.03} = 23.105\)

OpenStudy (tkhunny):

@greenandblue @bakonloverk @satellite73

OpenStudy (bakonloverk):

u called 4 me?

OpenStudy (tkhunny):

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. :-)

OpenStudy (bakonloverk):

ohhhh kk

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