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Math Help - Simplifying log expressions

  1. #1
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    Simplifying log expressions

    Hi,
    I am having trouble with this one problem. If someone could help I would be extremely grateful...

    Use the properties of logs to simplify the following log expression :

    log9 (1/8)

    My math book says the answer is -1 -log9 2, but I just can not figure it out.


    Thanks
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  2. #2
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    Quote Originally Posted by scutt View Post
    Hi,
    I am having trouble with this one problem. If someone could help I would be extremely grateful...

    Use the properties of logs to simplify the following log expression :

    log9 (1/8)

    My math book says the answer is -1 -log9 2, but I just can not figure it out.


    Thanks
    recheck that "book" answer ... it's not even close.

    \log_9\left(\frac{1}{8}\right) = -\log_9(8)

    you sure that the base of the log is 9 ?
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  3. #3
    Member integral's Avatar
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    \log_9\left(\frac{1}{8}\right)

    Your book says:

    -1-log_92



    log_9\frac{1}{8}=-\textrm{log}_98\,\,\,\because\,\,\,log_ab^n=nlog_a  b
    -log_98=-log_9(2^3)\,\,\because\,\,\,2^3=8
    -log_9(2^3)=-3log_92\,\,\,\because\,\,\,log_ab^n=nlog_ab
    -3log_92=\frac{-3log_32}{2}\,\,\,\because\,\,\, log_{b^n}a=\frac{log_ba}{n}

    -1-log_92=-\frac{log_32}{2}-1
    -\frac{log_32}{2}-1\neq\frac{-3log_32}{2}

    I would say your book is wrong.

    You can most likely use anything past the first step and get credit for simplification. :P

    My idea of simplification is the lowest possible base and coefficient numbers.
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