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Math Help - Equation

  1. #1
    Newbie joshirabi's Avatar
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    Equation

    solve it pls
    8^2x+1 = 9^2X-1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by joshirabi
    solve it pls
    8^2x+1 = 9^2X-1
    Please use brackets to resolve the ambiguity in what the exponents are,
    also resolve the doubt about the intended identity of x and X.

    (I can guess what the question is supposed to be but don't see why the
    helpers here should have to reconstruct the problem).


    RonL
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  3. #3
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    Quote Originally Posted by joshirabi
    solve it pls
    8^2x+1 = 9^2X-1
    8^2=64
    9^2=81
    Thus,
    64x+1=81x-1
    Subtract 64x from both sides,
    1=17x-1
    Add 1 to both sides,
    2=17x
    Thus,
    x=\frac{2}{17}
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  4. #4
    Newbie joshirabi's Avatar
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    Question re equation

    solve this equation pls
    8^(2x+1) = 9^(2X-1)
    where (2x+1) is power of 8 and (2x-1) is power of 9.
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  5. #5
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    Quote Originally Posted by joshirabi
    solve this equation pls
    8^(2x+1) = 9^(2X-1)
    where (2x+1) is power of 8 and (2x-1) is power of 9.
    Okay,
    \log 8^{2x+1}=\log 9^{2x-1}
    Thus,
    (2x+1)\log 8=(2x-1)\log 9
    2x\log 8+\log 8=2x\log 9-\log 9
    2x\log 8-2x\log 9=-\log 8-\log 9
    2x\log 9-2x\log 8=\log 9+\log 8
    2x(\log 9 - \log 8)=\log 9 +\log 8

    2x=\frac{\log 9+\log 8}{\log 9-\log 8}
    But,
    \log 9+\log 8 =\log (72)
    \log 9 -\log 8=\log (9/8)
    Thus,
    x=\frac{\log 72\cdot \log 9/8}{2}
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  6. #6
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    Of course, The Perfect Hacker meant: . x \:=\:\frac{\log72}{2\log\frac{9}{8}}

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  7. #7
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    Quote Originally Posted by Soroban
    Of course, The Perfect Hacker meant: . x \:=\:\frac{\log72}{2\log\frac{9}{8}}

    That is correct.
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  8. #8
    Member Glaysher's Avatar
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    Alternatively written

    8^{2x + 1} = 9^{2x - 1}

    e^{(\ln 8)^{2x + 1}} = e^{(\ln 9)^{2x - 1}}

    e^{\ln 8(2x + 1)} = e^{\ln 9(2x - 1)}

    \ln 8(2x + 1)} = \ln 9(2x - 1)

    2x \ln 8 + \ln 8 = 2x \ln 9 - \ln 9

    2x \ln 9 - 2x \ln 8 = \ln 8 + \ln 9

    2x(\ln 9 - \ln 8) = \ln 72

    2x = \frac{\ln 72}{\ln \frac{9}{8}}

    x = \frac{\ln 72}{2\ln\frac{9}{8}}
    Last edited by Glaysher; August 25th 2006 at 05:14 AM.
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  9. #9
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    Quote Originally Posted by Glaysher
    Alternatively written...
    I know I have a problem of being sometimes complicated. But I would not post something like that. It will confuse the student.
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  10. #10
    Member Glaysher's Avatar
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    Quote Originally Posted by ThePerfectHacker
    I know I have a problem of being sometimes complicated. But I would not post something like that. It will confuse the student.
    Don't know why I didn't just take logs of both sides would have saved a few steps but I don't understand why you put

    (3x + 7) + (4x - 9) + (5x - 6) = 12x - 8

    at the start of your solution.

    If it wasn't for that I wouldn't have bothered posting and I wanted something different. Nothing wrong with a different approach. Just go with the method you understand best.
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  11. #11
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    Quote Originally Posted by Glaysher
    but I don't understand why you put

    (3x + 7) + (4x - 9) + (5x - 6) = 12x - 8

    at the start of your solution.
    LaTex error.
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  12. #12
    Member Glaysher's Avatar
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    That makes a lot more sense.
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