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Math Help - Simple log question please help if you can

  1. #1
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    Simple log question please help if you can

    Without a calculator evaluate:
    Log(base 16)1/8

    .... I just used the change of base forumla to get Log1/8 over Log16

    is this all that can be done? Thanks for help
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  2. #2
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    Quote Originally Posted by vagina69 View Post
    Without a calculator evaluate:
    Log(base 16)1/8

    .... I just used the change of base forumla to get Log1/8 over Log16

    is this all that can be done? Thanks for help
    One of the principles of logs is \log\frac{x}{y}=\log x-\log y.

    Since the \log_{x}1=0 for any x, you can simplify to 0-\log_{16}8.

    There are probably a few ways of going on, but the easiest to me is splitting 8 into \frac{16}{2}. The \log_{16}16=1 and so we are left with 0-1-\log_{16}2.

    As 2 is the fourth root of 16, 16^\frac{1}{4}=2, and so the \log_{16}2=\frac{1}{4}. We can now solve 0-1-\frac{1}{4}=-\frac{3}{4}
    Last edited by ChristopherDunn; May 2nd 2010 at 05:40 PM.
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  3. #3
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    Quote Originally Posted by ChristopherDunn View Post
    One of the principles of logs is \log\frac{x}{y}=\log x-\log y.

    Since the \log_{x}1=0 for any x, you can simplify to 0-\log_{16}8.

    There are probably a few ways of going on, but the easiest to me is splitting 8 into \frac{16}{2}. The \log_{16}16=1 and so we are left with 0-1-\log_{16}2.

    As 2 is the fourth root of 16, 16^\frac{1}{4}=2, and so the \log_{16}2=\frac{1}{4}. We can now solve 0-1-\frac{1}{4}=-\frac{3}{4}
    Hi Cris,

    Answer could be clearer and more direct.

    log base 16 (1/8 ) =logbase 16 ( 2/16)= log base 16 (2) -logbase16 (16)

    = 1/4 - 1



    bjh
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  4. #4
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    Quote Originally Posted by bjhopper View Post
    Hi Cris,

    Answer could be clearer and more direct.

    log base 16 (1/8 ) =logbase 16 ( 2/16)= log base 16 (2) -logbase16 (16)

    = 1/4 - 1



    bjh

    Ah. Right you are. I did add an extra step in there.
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