## 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

1. theEric Group Title

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 Group Title

Oh, answers: (a) $$\ln(2)\div r$$ year (b) $$9.90$$ years (c) $$8.66\%$$

3. wio Group Title

Is it something like this?$2S_0=S_0(1+r)^T$

4. wio Group Title

Then solve for $$T$$ first the first problem.

5. theEric Group Title

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 Group Title

Oh it's compounded continuously, I see.

7. wio Group Title

So you used a differential equation to solve this?

8. theEric Group Title

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 Group Title

What is your original differential equation?

10. theEric Group Title

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 Group Title

12. AccessDenied Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

Oh wait, maybe I was on the right track! One second!

17. theEric Group Title

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 Group Title

Yes! You are welcome! The other two should follow up pretty nicely from your new equation now. :)

19. theEric Group Title

Is part (b) a follow-up to part (a)? Should I leave $$S=S_0$$, do you think?

20. theEric Group Title

$$S=2S_0$$, I mean.

21. AccessDenied Group Title

yeah, I think they want you to simply apply your new equation. I was able to get the correct answers doing so.

22. theEric Group Title

(b) $$T=\dfrac{\ln(2)}{.07}\approx 9.90$$ Thank you very much!

23. theEric Group Title

Thank you for checking the answers before I got to them, I mean! :D

24. AccessDenied Group Title

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 $$r=\dfrac{\ln(2)}{8}=.08664...\approx.0866=8.66\%$$ $$\huge\color{#44BB33}{\large\ \ o\ o\ \\\smile}$$