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theEric

  • one year ago

Hi! This is a finance-based problem, which might be my difficulty. I have the answers from the book. Here is the prompt: "Suppose that a sum \(S_0\) is invested at an annual rate of return \(r\) and compounded continuously. (a) Find the time \(T\) required for the original sum to double in value as a function of \(r\). (b) Determine \(T\) if \(r=7\%\). (c) Find the return rate that must be achieved if the initial investment is to double in \(8\) years.

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  1. theEric
    • one year ago
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    I think I should start out by finding \(S'(t)\), but that's where I think I'm wrong. It just comes from understanding the situation, I know.. I was thinking \(S'=S\ r\). Is that correct?

  2. theEric
    • one year ago
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    Oh, answers: (a) \(\ln(2)\div r\) year (b) \(9.90\) years (c) \(8.66\%\)

  3. wio
    • one year ago
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    Is it something like this?\[ 2S_0=S_0(1+r)^T \]

  4. wio
    • one year ago
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    Then solve for \(T\) first the first problem.

  5. theEric
    • one year ago
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    That doesn't look familiar... I used separation of variables to get \(S=e^{r\ t}\ C\), and then I was going to treat it like an initial value problem where \(S(0)=S_0\)

  6. wio
    • one year ago
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    Oh it's compounded continuously, I see.

  7. wio
    • one year ago
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    So you used a differential equation to solve this?

  8. theEric
    • one year ago
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    Yeah... I think that's what the class needs me to do! So, I need to think about what the rate is. Is it just the the \(S\) times the \(r\)? I was less sure when I asked... I think I'll see if I can get to a correct answer...

  9. wio
    • one year ago
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    What is your original differential equation?

  10. theEric
    • one year ago
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    I was thinking to use \(S'=S\ r\), and then from there I got to \(S=e^{rt}C\). I'm sorry, I'm doing a couple things at once, now. Don't worry about getting back to me - I'm only half here!

  11. wio
    • one year ago
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    Okay then your answer for (a) is correct.

  12. AccessDenied
    • one year ago
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    Have you had luck with this problem so far? I think what you've had so far seems like it's on the right track, just curious. :)

  13. theEric
    • one year ago
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    Hi! I'm actually confused! The answers are from the book. I typed up my work here, but I hit a button and it all went away.. So I'm typing it up in notepad and will paste it here in a moment. Thanks for checking!

  14. AccessDenied
    • one year ago
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    Alright, and yes that's a good idea. I do that from time to time as well in case of emergencies or when it feels laggy typing LaTeX stuff.

  15. theEric
    • one year ago
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    Then I did \(S(0)=S_0=e^{r\ (0)}C=C\implies S_0=C\) Then \(S=e^{r\ t}S_0\). I just realized I did some work but it doesn't pertain to the question!

  16. theEric
    • one year ago
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    Oh wait, maybe I was on the right track! One second!

  17. theEric
    • one year ago
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    Now I try to solve for \(T\). \(S=e^{r\ T}S_0\\\implies\dfrac{S}{S_0}=e^{r\ T}\\\implies\ln\left(\dfrac{S}{S_0}\right)=\ln(e^{r\ T})=r\ T\\\implies T=\ln\left(\dfrac{S}{S_0}\right)\ r^{-1}\) \(S=2S_0\\\implies T=\ln\left(\dfrac{2S_0}{S_0}\right)\ r^{-1}=\ln(2)\ r^{-1}\) Mission one, success! Thanks for sticking around to see my accomplishment! :)

  18. AccessDenied
    • one year ago
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    Yes! You are welcome! The other two should follow up pretty nicely from your new equation now. :)

  19. theEric
    • one year ago
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    Is part (b) a follow-up to part (a)? Should I leave \(S=S_0\), do you think?

  20. theEric
    • one year ago
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    \(S=2S_0\), I mean.

  21. AccessDenied
    • one year ago
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    yeah, I think they want you to simply apply your new equation. I was able to get the correct answers doing so.

  22. theEric
    • one year ago
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    (b) \(T=\dfrac{\ln(2)}{.07}\approx 9.90\) Thank you very much!

  23. theEric
    • one year ago
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    Thank you for checking the answers before I got to them, I mean! :D

  24. AccessDenied
    • one year ago
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    Yep! I am glad to have been helpful! :)

  25. theEric
    • one year ago
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    :)

  26. theEric
    • one year ago
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    At this point, I found the answers! Thank you everyone! Just for completion, I'll show my work to get (c). Maybe this can help someone else, if they search for it or something. Or maybe if someone wants to use it as a reference! \(r=\ln(2)\ t^{-1}\) and \(t=8\) so \(r=\dfrac{\ln(2)}{8}=.08664...\approx.0866=8.66\%\) \(\huge\color{#44BB33}{\large\ \ o\ o\ \\\smile}\)

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