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theEric Group Title

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.

  • 10 months ago
  • 10 months ago

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  1. theEric Group Title
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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?

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

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

    • 10 months ago
  4. wio Group Title
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    Then solve for \(T\) first the first problem.

    • 10 months ago
  5. theEric Group Title
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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 \(S(0)=S_0\)

    • 10 months ago
  6. wio Group Title
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    Oh it's compounded continuously, I see.

    • 10 months ago
  7. wio Group Title
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    So you used a differential equation to solve this?

    • 10 months ago
  8. theEric Group Title
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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...

    • 10 months ago
  9. wio Group Title
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    What is your original differential equation?

    • 10 months ago
  10. theEric Group Title
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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!

    • 10 months ago
  11. wio Group Title
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    Okay then your answer for (a) is correct.

    • 10 months ago
  12. AccessDenied Group Title
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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. :)

    • 10 months ago
  13. theEric Group Title
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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!

    • 10 months ago
  14. AccessDenied Group Title
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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.

    • 10 months ago
  15. theEric Group Title
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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!

    • 10 months ago
  16. theEric Group Title
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    Oh wait, maybe I was on the right track! One second!

    • 10 months ago
  17. theEric Group Title
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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! :)

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

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

    • 10 months ago
  20. theEric Group Title
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    \(S=2S_0\), I mean.

    • 10 months ago
  21. AccessDenied Group Title
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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.

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

    • 10 months ago
  23. theEric Group Title
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    Thank you for checking the answers before I got to them, I mean! :D

    • 10 months ago
  24. AccessDenied Group Title
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    Yep! I am glad to have been helpful! :)

    • 10 months ago
  25. theEric Group Title
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    :)

    • 10 months ago
  26. theEric Group Title
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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 \(r=\dfrac{\ln(2)}{8}=.08664...\approx.0866=8.66\%\) \(\huge\color{#44BB33}{\large\ \ o\ o\ \\\smile}\)

    • 10 months ago
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