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Math Help - logs?

  1. #1
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    logs?

    HEllo,
    Im just trying to remember the algebraic method of sloving for r in an equ. such as this.

    a = (1 + (r/m) )^m - 1
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jason03 View Post
    HEllo,
    Im just trying to remember the algebraic method of sloving for r in an equ. such as this.

    a = (1 + (r/m) )^m - 1
    a = \left ( 1 + \frac{r}{m} \right )^m - 1

    a + 1 = \left ( 1 + \frac{r}{m} \right )^m

    1 + \frac{r}{m} = (a + 1)^{1/m}

    \frac{r}{m} = (a + 1)^{1/m} - 1

    r = \frac{1}{m} \left ( (a + 1)^{1/m} - 1 \right )

    -Dan
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  3. #3
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    Hello,
    I tried your equ. but im not coming up with the correct answer.

    heres the equ with the numbers

    .0931 = (1 + r/4)^4 - 1

    the answer is .09 and it works when plugged into the above equ.

    But I tried using your equ. and I get a different answer
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  4. #4
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    Hello,

    Quote Originally Posted by topsquark View Post
    r = \frac{1}{m} \left ( (a + 1)^{1/m} - 1 \right )

    -Dan
    There is just a small mistake here.

    It's r={\color{red}m} \left ( (a + 1)^{1/m} - 1 \right )
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  5. #5
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    ohh Thank you!


    If you dont mind me being a pest, how did you arive at that. I follow you up until the exponents change.
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  6. #6
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    Quote Originally Posted by topsquark View Post
    a = \left ( 1 + \frac{r}{m} \right )^m - 1

    a + 1 = \left ( 1 + \frac{r}{m} \right )^m

    power everything to 1/m. As (a^b)^c=a^{bc}, ((\dots)^m)^{1/m}=(\dots)

    1 + \frac{r}{m} = (a + 1)^{1/m}

    \frac{r}{m} = (a + 1)^{1/m} - 1

    multiply by m in each side

    r = {\color{red} m} \left ( (a + 1)^{1/m} - 1 \right )

    -Dan
    Is it ok ?
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