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Math Help - What interest rate needed to double money in two years?

  1. #1
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    What interest rate needed to double money in two years?

    e^{5r}=2
    5r ln e=ln(2)
    r=ln(2)/5=.1386294361 (store in variable X)

    Using a calculator where Ans refers to the previous answer.

    1. 100+100X=113.8629436=Ans
    2. Ans+AnsX=129.6476993=new Ans
    3. Ans+AnsX=147.6206867=new Ans
    4. Ans+AnsX=168.0852593=new Ans
    5. Ans+AnsX=191.386824


    Shouldn't line 5 equal 200?
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    Re: What interest rate needed to double money in two years?

    please post the original problem as stated for you to solve ... what you have posted is complete jibberish.
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    Re: What interest rate needed to double money in two years?

    Sorry.

    The original problem is What constantly compounded interest rate is needed to double your money in two years?

    I setup the problem:
    2=e^{5r}

    I solved the problem:
    5r ln e=ln(2)
    r=ln(2)/5=.1386294361

    And got an interest rate of approximately 13.86%.

    Then I did a check, by starting with $100 and adding 13.86% times 100 to it.
    100+100(.1386294361)=113.8629436

    I took that answer and added 13.86% to it.
    (Ans)+(Ans)(.1386294361)=129.6476993

    Then I did that three more times.
    (Ans)+(Ans)(.1386294361)=147.6206867
    (Ans)+(Ans)(.1386294361)=168.0852593
    (Ans)+(Ans)(.1386294361)=191.386824

    Shouldn't I have $200 by the fifth time? Or I did the first part of the problem wrong.

    Thanks.
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    Re: What interest rate needed to double money in two years?

    Why use 5 r? What is the purpose of the 5?
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    Re: What interest rate needed to double money in two years?

    The original problem is What constantly compounded interest rate is needed to double your money in two years?
    let y_0 = initial investment

    2y_0 = y_0 e^{rt}

    2 = e^{rt}

    \ln{2} = rt

    \frac{\ln{2}}{t} = r

    substitute 2 years for t and find the required interest rate, r
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    Re: What interest rate needed to double money in two years?

    Quote Originally Posted by SammyS View Post
    Why use 5 r? What is the purpose of the 5?
    Sloppy title. Five years.

    Quote Originally Posted by skeeter View Post
    let y_0 = initial investment

    2y_0 = y_0 e^{rt}

    2 = e^{rt}

    \ln{2} = rt

    \frac{\ln{2}}{t} = r

    substitute 2 years for t and find the required interest rate, r
    That's what I did. But I wrote the title wrong. The correct time frame is 5 years.

    But then, when I apply that for five years, I get $191.39. Shouldn't I get $200?
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    Re: What interest rate needed to double money in two years?

    Quote Originally Posted by mathDad View Post
    Sloppy title. Five years.


    That's what I did. But I wrote the title wrong. The correct time frame is 5 years.

    But then, when I apply that for five years, I get $191.39. Shouldn't I get $200?
    I'm guessing that you are using $100 as the initial investment?

    r = \frac{\ln{2}}{5} = 0.1386...

    using this interest rate (and not rounding it)

    100e^{r \cdot 5} = 200

    if you do round r to 13.86% , you'll still get very close (199.97)
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    Re: What interest rate needed to double money in two years?

    Quote Originally Posted by mathDad View Post
    e^{5r}=2
    5r ln e=ln(2)
    r=ln(2)/5=.1386294361 (store in variable X)

    Using a calculator where Ans refers to the previous answer.

    1. 100+100X=113.8629436=Ans
    2. Ans+AnsX=129.6476993=new Ans
    3. Ans+AnsX=147.6206867=new Ans
    4. Ans+AnsX=168.0852593=new Ans
    5. Ans+AnsX=191.386824



    Shouldn't line 5 equal 200?
    Why should it? e^{rt} (and why are you using "5" for t when the problem asks about 2 years) refers to "continuous compounding". In the latter part of your conputation you are compounding annually (for 5 years, not 2).

    Compounding annually, with interest rate r, will give, in n years, A= P(1+ r)^n. The interest rate necessary to double the amount in 5 years (I assume that is what you meant), compounding annually, is given by solving (1+ r)^5= 2. Taking the logarithm of both sides, log(1+r)^5= 5log(1+ r)= log(2) so that log(1+r)= log(2)/5=0.1386. Then [tex]1+ r= e^{0.1386}= 1.1486983550[tex] and so r= 0.148693550.
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