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Math Help - Just want to make sure about this logarithm problem

  1. #1
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    Just want to make sure about this logarithm problem

    I had a problem similar to this one.

    Write expression as a logarithm of a single quantity and then simplify if possible.

    (1/4)[log (x-1)-log (x+1)]+3log x

    Please show the steps.
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by dc52789 View Post
    I had a problem similar to this one.

    Write expression as a logarithm of a single quantity and then simplify if possible.

    (1/4)[log (x-1)-log (x+1)]+3log x

    Please show the steps.

    Use log properties

     \log{x}^y=y\log{x}

     \log{xy}=\log{x}+\log{y}

     \log{\frac{x}{y}}=\log{x}-\log{y}
    Last edited by 11rdc11; October 25th 2008 at 11:48 AM. Reason: my answer is wrong to hard input data on a ps3 lol
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  3. #3
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    Quote Originally Posted by 11rdc11 View Post
    it simplfies to  \log{(x^4-x^3)}
    Um...I still need to know the steps to how you get that answer.
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  4. #4
    Moo
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    Hello,
    Quote Originally Posted by dc52789 View Post
    I had a problem similar to this one.

    Write expression as a logarithm of a single quantity and then simplify if possible.

    (1/4)[log (x-1)-log (x+1)]+3log x

    Please show the steps.
    Note that x>0 and |x|>1, so x>1. Otherwise it is not defined.

    notice that x-1=(x-1)(x+1)

    So \log(x^2-1)-\log(x+1)=\log \left(\frac{x^2-1}{x+1}\right)=\log(x-1)

    \frac 14 \cdot \log(x-1)+3 \log(x)=\log \left((x-1)^{1/4}\right)+\log(x^3)

    =\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)

    now it depends on how you want it to be simplified.
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,

    Note that x>0 and |x|>1, so x>1. Otherwise it is not defined.

    notice that x-1=(x-1)(x+1)

    So \log(x^2-1)-\log(x+1)=\log \left(\frac{x^2-1}{x+1}\right)=\log(x-1)

    \frac 14 \cdot \log(x-1)+3 \log(x)=\log \left((x-1)^{1/4}\right)+\log(x^3)

    =\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)

    now it depends on how you want it to be simplified.
    o k...which would be an easier way to simplify it?
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  6. #6
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    Quote Originally Posted by 11rdc11 View Post
    =\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)

    Moo how are these equal?
    I don't know. I get lost whenever I have these questions.
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Note that x>0 and |x|>1, so x>1. Otherwise it is not defined.

    notice that x-1=(x-1)(x+1)

    So \log(x^2-1)-\log(x+1)=\log \left(\frac{x^2-1}{x+1}\right)=\log(x-1)

    \frac 14 \cdot \log(x-1)+3 \log(x)=\log \left((x-1)^{1/4}\right)+\log(x^3)

    =\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)

    now it depends on how you want it to be simplified.

    =\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)

    Here is how moo did it

    \log \left((x^{13}-x^{12})^{1/4}\right)

    ~^4\sqrt{x^{12}(x-1)}

    x^3(^4\sqrt{x-1})
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  8. #8
    Moo
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    Quote Originally Posted by 11rdc11 View Post
    =\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)

    Here is how moo did it

    \log \left((x^{13}-x^{12})^{1/4}\right)

    ~^4\sqrt{x^{12}(x-1)}

    x^3(\sqrt^4{x-1})
    Not exactly the way I did^^

    x^3=(x^{12})^{1/4} and then you can distribute it in the (x-1)^{1/4} part


    By the way, dc52789, it would be nice if you thought a little instead of just asking all the steps =)
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  9. #9
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    Quote Originally Posted by Moo View Post
    Not exactly the way I did^^

    x^3=(x^{12})^{1/4} and then you can distribute it in the (x-1)^{1/4} part


    By the way, dc52789, it would be nice if you thought a little instead of just asking all the steps =)
    I know but it really got me lost and the professor didn't go over it before the test.
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