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Math Help - Can someone tell what I'm doing wrong with this logerithm?

  1. #1
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    Can someone tell what I'm doing wrong with this logerithm?

    The problem is
    Log6^18- Log 6^3.

    When you have subtraction in a log the properties say divide. So I would assume I would divide 18/3 which is 6. But the answer to the problem is 1.

    They both have a base of 6 so I know i don't have to change that. On my help cd that came with the text book says divide M/N which is what I have done.

    How is the answer 1?


    Thanks

    Jason
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  2. #2
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    Is this the reason? When you divide 18/3 =6 and the base is also 6 means if the base is 6 and the power is 6 then it= 1?
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  3. #3
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    This is not an answer. It is a complaint.
    Why do you not learn to use notation?
    Frankly, I cannot read your posting.
    I do not know what you are asking.
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  4. #4
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     \log_{6}18 - \log_{6}3 = \log_{6}6 = 1 .
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  5. #5
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    Quote Originally Posted by shilz222 View Post
     \log_{6}18 - \log_{6}3 = \log_{6}6 = 1 .
    Why do you think that the question meant that?
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  6. #6
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    Quote Originally Posted by Darkhrse99 View Post
    Log6^18- Log 6^3.
    I'm with Plato. I would have thought that this problem was log_{10}(6^{18}) - log_{10}(6^3)
    which is decidedly not equal to 1.

    @ Darkhrse99:
    To type out log_b(a) type in:
    [ math ] <-- Without the spaces
    log_{b}(a)
    [ /math ] <-- Again, without the spaces.

    I realize that learning LaTeX should not really be necessary to ask questions here, but I can't think of a good way to use the ASCII codes to do logarithms.

    -Dan
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  7. #7
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    Quote Originally Posted by shilz222 View Post
     \log_{6}18 - \log_{6}3 = \log_{6}6 = 1 .
    This is the correct question. I couldn't gt the little 6 in the equation right so [tex] formula didn't look right. Sorry I couldn't type it more clearly. On the other hand, I did get it right.
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