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Math Help - Stuck at the end of an interest problem involving ln

  1. #1
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    Stuck at the end of an interest problem involving ln

    in short..the problem is 7500=5000(1+r/12)^36

    i'm down to this part e^.01126 = (1+r/12)

    where do i go from here?
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  2. #2
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    7500=5000(1+r/12)^36
    I guess you're solving for "r" ?

    1.5 = \left(1 + \frac{r}{12}\right)^{36}

    1.5^{\frac{1}{36}} = 1 + \frac{r}{12}

    r = 12\left(1.5^{\frac{1}{36}} - 1\right) \approx 13.6%
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  3. #3
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    thats the right answer but my book does it differently...but at the very end it loses me...

    the book does (i'm going to use the book example)

    18000=10000(1+r/12)^12(5)
    = 1.8 = (1+r/12)60
    then use logarithm

    ln 1.8=ln(1+r/12)60
    =ln 1.8 = 60ln(1+r/12)
    =ln1.8/60 = (1+r/12)
    =.009796 = (1+r/12)

    =(1+r/12) = e^.009796
    = 1.009844 <--------

    here's where i get stuck...where did they get 1.009844

    and then it continues to r/12=1.009844-1 =.1181 also 11.81%

    u're answer is right, i just want to understand where they got the 1.00944 from e^.009796?
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  4. #4
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    e^{.009796} = 1.009844...
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  5. #5
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    ahh, i got it..i wasn't even thinking about using my calculator thanks
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