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Math Help - Value growing

  1. #1
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    Value growing

    A value grows exponentially with a fixed percent per years. The value after 14 and 20 years have grown to 1819 and 6839. Determine the percentage it is growing per year.
    Is there an equation for this or something? or do i have to guess and see if it end up at 1819 after 14 years? ..
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  2. #2
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    Re: Value growing

    Quote Originally Posted by Greenid View Post
    A value grows exponentially with a fixed percent per years. The value after 14 and 20 years have grown to 1819 and 6839. Determine the percentage it is growing per year.
    Is there an equation for this or something? or do i have to guess and see if it end up at 1819 after 14 years? ..
    The word exponentially means you're going have e in there.

    The general equation for exponential growth is A(t) = A_0e^{kt} where:

    • A(t) = Value at time t
    • A_0 = Initial value (t=0)
    • k = growth constant
    • t = time


    In your case sub (14,1819) and (20,6839) for (t,A(t)) respectively into A(t) = A_0e^{kt}

    You will then have a simultaneous equation with A_0 and k as your unknowns. The question wants you to find k
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    Re: Value growing

    Not sure if i understood right.
    But, A(t) = 6839-1819 =5020
    A0 = 1819
    t = 6 ?
    And then do ln 5020 = 1819 * k * ln e^6
    k= ln 5020 / 1819 * ln e^6 ?
    Kinda know this is wrong since my answer become so weird..
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  4. #4
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    Re: Value growing

    6839 = A_0 e^{20k}

    1819 = A_0 e^{14k}

    divide the first equation by the second ...

    \frac{6839}{1819} = e^{6k}

    solve for k
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    Re: Value growing

    don't kill me, but no ln?
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  6. #6
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    Re: Value growing

    Quote Originally Posted by Greenid View Post
    don't kill me, but no ln?
    what do you mean "no ln" ?
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  7. #7
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    Re: Value growing

    Quote Originally Posted by Greenid View Post
    don't kill me, but no ln?
    I don't understand? Are you trying to solve this question without using the natural logarithm? If so that cannot be done AFAIK

    Your next step is to take skeeter's equation from post 4 and take the natural logarithm of both sides noting that \ln(e^{f(x)}) = f(x)
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  8. #8
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    Re: Value growing

    I was just abit confused
    ln e^6k = ln (6839/1819)
    6k/6 = 1.3242/6
    k= 0.2207 *100 = 22.07% ?
    Correct now?
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  9. #9
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    Re: Value growing

    Quote Originally Posted by Greenid View Post
    I was just abit confused
    ln e^6k = ln (6839/1819)
    6k/6 = 1.3242/6
    k= 0.2207 *100 = 22.07% ?
    Correct now?
    Yep. Looks good to me although you may want to put 22.07% per year
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  10. #10
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    Re: Value growing

    Not sure what's going on here...but the "annual equivalent rate" seems to be
    what's asked for; so it matters not HOW (weekly,daily, hourly, continuously...)
    the interest is calculated; we have an opening and ending balance; so:
    r = (6839/1819)^(1/6) - 1 = ~.24698 or ~24.7 %
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