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Math Help - Solving for x....when x is both an exponent and denominator

  1. #1
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    Solving for x....when x is both an exponent and denominator

    How do you solve this equation?

    [1-(1/1.05)^30]0.0635 = [(1/1.05)^30] * [(1-(1/1.05)^(x-65)] * [30/x]

    See how x is both a denominator, and an exponent?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Is that it?

    <br />
\left[1-\left(\dfrac{1}{1.05}\right)^{30}\right]0.0635 = \dfrac{30}{x}\left[\left(\dfrac{1}{1.05}\right)^{30}\right]\left[1-\left(\dfrac{1}{1.05}\right)^{x-65}\right]

    Can you simplify this more? Work out for example all the parts containing the values without x?
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  3. #3
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    All of the numbers are irrational; they get kind of messy :/ is there a way to do it, though?
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  4. #4
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    Quote Originally Posted by RU091 View Post
    How do you solve this equation?

    [1-(1/1.05)^30]0.0635 = [(1/1.05)^30] * [(1-(1/1.05)^(x-65)] * [30/x]

    See how x is both a denominator, and an exponent?
    Something tells me you cannot solve this algebraically...

    Moving the x's to one side and the values to the other, we get:


    \displaystyle \dfrac{[1-(\frac{1}{1.05})^{30}]0.0635}{30[(\frac{1}{1.05})^{30}]} = \dfrac{[1-(\frac{1}{1.05})^{x-65}]}{x}

    Taking logarithms of both sides won't help us. The best idea may be to use an approximation method to solve for x.
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  5. #5
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    What7s an approximation method? How do you do that?

    Thanks for all your help
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  6. #6
    Senior Member Educated's Avatar
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    An approximation method approximates x, as you cannot get the precise answer. It gets a decimal approximation down to however many decimal places of accuracy you want.

    The easiest one I can think of is the Newton Raphson method, but finding the derivative of the function isn't too easy.
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  7. #7
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    Quote Originally Posted by RU091 View Post
    How do you solve this equation?

    [1-(1/1.05)^30]0.0635 = [(1/1.05)^30] * [(1-(1/1.05)^(x-65)] * [30/x]

    See how x is both a denominator, and an exponent?
    1. Simplify the coefficients.

    2. Do you want real roots? It does not have any! If you are interested in where it changes sign then that would be at x=0, but it is undefined there.

    CB
    Last edited by CaptainBlack; November 13th 2010 at 12:39 AM.
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Educated View Post
    Something tells me you cannot solve this algebraically...

    Moving the x's to one side and the values to the other, we get:


    \displaystyle \dfrac{[1-(\frac{1}{1.05})^{30}]0.0635}{30[(\frac{1}{1.05})^{30}]} = \dfrac{[1-(\frac{1}{1.05})^{x-65}]}{x}

    Taking logarithms of both sides won't help us. The best idea may be to use an approximation method to solve for x.
    Can be solved using Lambert's W function (which won't give a real root in this case) but that would not really help the OP so I won't do that here.

    CB
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  9. #9
    Senior Member Educated's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    2. Do you want real roots? It does not have any!
    CB
    Really? I got 2 real roots from the equation: \displaystyle \dfrac{[1-(\frac{1}{1.05})^{30}]0.0635}{30[(\frac{1}{1.05})^{30}]} - \dfrac{[1-(\frac{1}{1.05})^{x-65}]}{x} = 0
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Educated View Post
    Really? I got 2 real roots from the equation: \displaystyle \dfrac{[1-(\frac{1}{1.05})^{30}]0.0635}{30[(\frac{1}{1.05})^{30}]} - \dfrac{[1-(\frac{1}{1.05})^{x-65}]}{x} = 0

    Well the Maxima simplifies (evaluates coefficients numerically because of decimal coeficients):


    KK:(1-(20/21)^30)*(0.0635)/30/(20/21)^30-(1-(20/21)^(x-65))/x;

    0.0070314446940689-\frac{1-{21}^{65-x}\,{20}^{x-65}}{x}


    Which now gives a root at ~=-183.5, but no other (why I now have a root I don't know but appears to be something to do with using 20/21 for 1/1.05), but I think it is wrong - more checking needed me-thinks

    (Switching back to 1/1.05 takes us back to the previous situation with no roots and plots now plot what the function evaluates to

    I have changed the definition of KK to functional form which gives LaTeX to compare with your's:

    \mathrm{KK}\left( x\right) :=\dfrac{\frac{\left( 1-{\left( \frac{1}{1.05}\right) }^{30}\right) \,0.0635}{30}}{{\left( \frac{1}{1.05}\right) }^{30}}-\dfrac{1-{\left( \frac{1}{1.05}\right) }^{x-65}}{x}

    Which looks OK to me)

    CB
    Last edited by CaptainBlack; November 13th 2010 at 01:42 AM.
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