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Math Help - compounded interest

  1. #1
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    compounded interest

    what sum of money must be invested at 7.3% interest compounded continuously to yield $20,000 in 12 years

    how long will it take for the investment to triple in value

    A=Pe^{rt}
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  2. #2
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     20000 = P\times e^{0.073\times 12}

    Find P.

    After this find t for 3\times P
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  3. #3
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    i found P and after that i multiplied it by 3 and solved for t and ended up with a negative number am i suppose to get a negative number because t is years
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     20000 = Pe^{0.073\times 12} \implies P = \frac{20000}{e^{0.073\times 12}} \implies P = 8329

    Now solve for t where 3\times 8329 = 8329 e^{0.073t}

    Is this what you found?

    Spoiler:
    t\approx 15
    Last edited by pickslides; December 21st 2009 at 03:33 PM. Reason: bas latex
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  5. #5
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    Quote Originally Posted by math321 View Post
    i found P and after that i multiplied it by 3 and solved for t and ended up with a negative number am i suppose to get a negative number because t is years
    would a negative value for t make sense?


    note you don't need to know the value of P to find how long it takes for the investment, P , to triple

    3P = Pe^{rt}

    divide both sides by P ...

    3 = e^{rt}

    use logs and solve for t
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  6. #6
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    this is wat i did
    this is how i managed to get a negative number

    20000=8323.93*e^{.073*12}
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  7. #7
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    Quote Originally Posted by math321 View Post
    this is wat i did
    this is how i managed to get a negative number

    20000=8323.93*e^{.073*12}
    first, the original investment is P = 8328.91.

    now, read the original question again ...

    what sum of money must be invested at 7.3% interest compounded continuously to yield $20,000 in 12 years

    how long will it take for the investment to triple in value
    the equation should be ...

    3(\textcolor{red}{8328.91})=\textcolor{red}{8328.9  1} \cdot e^{.073t}

    once again, note that the value of P will cancel, leaving you with the equation 3 = e^{0.073t}
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  8. #8
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    Thumbs up

    or i now see
    wat i did wrong
    when im finish ill get back to you to see if it is correct
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  9. #9
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    for my final answer i got 15.05 years
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  10. #10
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    Quote Originally Posted by math321 View Post
    this is wat i did
    this is how i managed to get a negative number

    20000=8323.93*e^{.073*12}
    I have no idea what you mean by "this is how i managed to get a negative number". What you show is an equation (which is only very approximately true by the way- the right hand side is about 19988). What did you do to get a negative number?
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  11. #11
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    Quote Originally Posted by HallsofIvy View Post
    I have no idea what you mean by "this is how i managed to get a negative number". What you show is an equation (which is only very approximately true by the way- the right hand side is about 19988). What did you do to get a negative number?
    who knows?

    the OP has been bouncing back and forth between sites to find a solution ...

    S.O.S. Mathematics CyberBoard :: View topic - compounded interest
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  12. #12
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    logs problem

    log2 (x+1) + log2 (x-1) = 3 ..... log to base 2
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  13. #13
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    This should be posted in a separate thread

     \log_2 (x+1) + \log_2 (x-1) = 3

    When adding logs you can mulitply

     \log_2 ((x+1)(x-1))  = 3

     (x+1)(x-1)  = 2^3

     (x+1)(x-1)  = 8

    expand and solve...
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  14. #14
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    Quote Originally Posted by skeeter View Post
    who knows?

    the OP has been bouncing back and forth between sites to find a solution ...

    S.O.S. Mathematics CyberBoard :: View topic - compounded interest
    Way to be on the lookout skeeter. This sort of thing is common I bet but in cases like these can be very aggravating as the OP has no intention of showing any effort.
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