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Math Help - logarithm

  1. #1
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    logarithm

    write as a single logarithm

    -2log (x-2) + 1/4 log((x^2) -4) - 3log (2) + 6 log (x) =

    so far I have gotten this far:

    = log [ ((x^2)-4)^1/4 the whole thing over 8(x^6)(x-2)^2 ]

    I am not sure what to do with the nominator I assume it will be (x-2)(x+2)^1/4 but even if that is correct I still don't know what to do next.

    I am sorry about the messy writing but i don't think I have a text editor option here that would make my life easier....

    Thank you
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zdenka View Post
    write as a single logarithm

    -2log (x-2) + 1/4 log((x^2) -4) - 3log (2) + 6 log (x) =

    so far I have gotten this far:

    = log [ ((x^2)-4)^1/4 the whole thing over 8(x^6)(x-2)^2 ]

    I am not sure what to do with the nominator I assume it will be (x-2)(x+2)^1/4 but even if that is correct I still don't know what to do next.

    I am sorry about the messy writing but i don't think I have a text editor option here that would make my life easier....

    Thank you
    Only one change: the exponent on the 6 log(x) term is positive so it goes in the numerator:
    log \left ( \frac{x^6(x^2 - 4)^{1/4}}{8(x - 2)^2} \right )

    This is all you need to do with it.

    -Dan
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  3. #3
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    so there is nothing I can do with the Difference of squares in the nominator??


    thanks
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by zdenka View Post
    so there is nothing I can do with the Difference of squares in the nominator??


    thanks
    Do you mean numerator or denominator? I assume the latter.
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  5. #5
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    Hello, zdenka!

    Well, we can take a step further, but I don't think it's expected of us.


    \log\left[\frac{x^6(x^2-4)^{\frac{1}{4}}}{8(x-2)^2}\right]

    We have: . \log\left[\frac{x^6(x-2)^{\frac{1}{4}}(x+2)^{\frac{1}{4}}}{8(x-2)^2}\right] \;=\;\log\left[\frac{x^6(x+2)^{\frac{1}{4}}}{8(x-2)^{\frac{7}{4}}}\right]


    But it's great that you noticed it and felt obligated to do something!

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  6. #6
    Member Jonboy's Avatar
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    True that Soroban, and good call zdenka !
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  7. #7
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    It all makes sense now..


    thanks !!!!
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