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Math Help - solve 3

  1. #1
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    solve 3

    Last edited by CaptainBlack; August 5th 2007 at 02:10 PM.
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  2. #2
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    I'm not completely sure how to do this as I han't covered it :P
    but here's some aid:

    from that we get: (I'm not entirely sure)

    1^x + 1^x + 6^x + 12^x = 32 (by adding/subtracting in the fractions)
    and 1^anything is always 1
    so

    1 + 1 + 6^x + 12^x = 32
    2 + 6^x + 12^x = 32
    6^x + 12^x = 30

    and that's all I've gotten up to since I haven't done this before

    hope someoone else can help and please correct me if I've done anything wrong
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  3. #3
    Senior Member DivideBy0's Avatar
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    I think you might be mistaken in that

    6^{\frac{1}{2}}=\sqrt{6} \not= 6 + \frac{1}{2}

    The whole \frac{1}{2} is an exponent, and they can't be added or subtracted easily.
    By the way, looks like this is a really hard problem :/
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  4. #4
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    Quote Originally Posted by DivideBy0 View Post
    I think you might be mistaken in that

    6^{\frac{1}{2}}=\sqrt{6} \not= 6 + \frac{1}{2}

    The whole \frac{1}{2} is an exponent, and they can't be added or subtracted easily.
    By the way, looks like this is a really hard problem :/
    oh sorry, I thought it meant 1/2
    thanks for telling me
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  5. #5
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    Hello, perash!

    I have an "eyeball" solution . . .


    \left(6^{\frac{1}{2}} - 5^{\frac{1}{2}}\right)^x + \left(3^{\frac{1}{2}}<br />
- 2^{\frac{1}{2}}\right)^x + \left(3^{\frac{1}{2}} + 2^{\frac{1}{2}}\right)^x + \left(6^{\frac{1}{2}} + 5^{\frac{1}{2}}\right)^x \;=\;32
    We have: . \left(\sqrt{6}-\sqrt{5}\right)^x + \left(\sqrt{6} + \sqrt{5}\right)^x + \left(\sqrt{3} - \sqrt{2}\right)^x + \left(\sqrt{3} + \sqrt{2}\right)^x \;=\;32


    Since the left side equals an integer, x must be even: . x = 2n

    We have: . \left(\sqrt{6}-\sqrt{5}\right)^{2n} + \left(\sqrt{6} + \sqrt{5}\right)^{2n} + \left(\sqrt{3} - \sqrt{2}\right)^{2n} + \left(\sqrt{3} + \sqrt{2}\right)^{2n} \;=\;32

    . . \left[\left(\sqrt{6}-\sqrt{5}\right)^2\right]^n + \left[\left(\sqrt{6} + \sqrt{5}\right)^2\right]^n + \left[\left(\sqrt{3} - \sqrt{2}\right)^2\right]^n + \left[\left(\sqrt{3} + \sqrt{2}\right)^2\right]^n \;=\;32

    . . \left(11 - 2\sqrt{30}\right)^n + \left(11 + 2\sqrt{30}\right)^n + \left(5 - 2\sqrt{6}\right)^n + \left(5 + 2\sqrt{6}\right)^n \;=\;32

    And we see that n = 1 is a solution.


    Therefore: . x \:=\:2n \:=\:\boxed{2}

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  6. #6
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    Graphing it shows that -2 is also a value
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by perash View Post
    Graphing it shows that -2 is also a value
    In retrospect we should have expected that because:
    \sqrt{6} - \sqrt{5} = \frac{\sqrt{6} - \sqrt{5}}{1} \cdot \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}}

    = \frac{6 - 5}{\sqrt{6} + \sqrt{5}} = \frac{1}{\sqrt{6} + \sqrt{5}}

    The same holds for the \sqrt{3} - \sqrt{2} term so we have:
    (\sqrt{6} - \sqrt{5})^x + (\sqrt{6} + \sqrt{5})^x + (\sqrt{3} - \sqrt{2})^x + (\sqrt{3} + \sqrt{2})^x

    = \frac{1}{(\sqrt{6} + \sqrt{5})^x} + (\sqrt{6} + \sqrt{5})^x + \frac{1}{(\sqrt{3} + \sqrt{2})^x} + (\sqrt{3} + \sqrt{2})^x

    So if we have a + x solution, we should also have the - x solution also.

    -Dan
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