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theEric
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Hi! This is a financebased 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
theEric Group Title
Hi! This is a financebased 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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theEric Group TitleBest ResponseYou've already chosen the best response.1
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

theEric Group TitleBest ResponseYou've already chosen the best response.1
Oh, answers: (a) \(\ln(2)\div r\) year (b) \(9.90\) years (c) \(8.66\%\)
 10 months ago

wio Group TitleBest ResponseYou've already chosen the best response.0
Is it something like this?\[ 2S_0=S_0(1+r)^T \]
 10 months ago

wio Group TitleBest ResponseYou've already chosen the best response.0
Then solve for \(T\) first the first problem.
 10 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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

wio Group TitleBest ResponseYou've already chosen the best response.0
Oh it's compounded continuously, I see.
 10 months ago

wio Group TitleBest ResponseYou've already chosen the best response.0
So you used a differential equation to solve this?
 10 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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

wio Group TitleBest ResponseYou've already chosen the best response.0
What is your original differential equation?
 10 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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

wio Group TitleBest ResponseYou've already chosen the best response.0
Okay then your answer for (a) is correct.
 10 months ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
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

theEric Group TitleBest ResponseYou've already chosen the best response.1
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

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
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

theEric Group TitleBest ResponseYou've already chosen the best response.1
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

theEric Group TitleBest ResponseYou've already chosen the best response.1
Oh wait, maybe I was on the right track! One second!
 10 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
Yes! You are welcome! The other two should follow up pretty nicely from your new equation now. :)
 10 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Is part (b) a followup to part (a)? Should I leave \(S=S_0\), do you think?
 10 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\(S=2S_0\), I mean.
 10 months ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
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

theEric Group TitleBest ResponseYou've already chosen the best response.1
(b) \(T=\dfrac{\ln(2)}{.07}\approx 9.90\) Thank you very much!
 10 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Thank you for checking the answers before I got to them, I mean! :D
 10 months ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
Yep! I am glad to have been helpful! :)
 10 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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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