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Math Help - working with logarithms?

  1. #1
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    working with logarithms?

    so my advanced functions teacher didn't really explain this lesson well and i can't seem to figure it out on my own.



    if you could be kind enough to explain the process i would greatly appreciate it!
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  2. #2
    MHF Contributor harish21's Avatar
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    The logarithm of a number y with respect to a base b is the exponent to which we have to raise b to obtain y.

    the definition would be:

     x = log_{b} y \Longleftrightarrow b^x = y

    and we say that x is the logarithm of y with base b if and only if b to the power x equals y.

    For example:

    10^2 = 100  \Longleftrightarrow log_{10} 100 = 2

    and, 10^{-2} = 0.01 \Longleftrightarrow log_{10} 0.01 = -2

    Based on this, can you now complete your question?!
    Last edited by harish21; October 7th 2010 at 06:16 PM.
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  3. #3
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    Quote Originally Posted by yolanda View Post
    so my advanced functions teacher didn't really explain this lesson well and i can't seem to figure it out on my own.



    if you could be kind enough to explain the process i would greatly appreciate it!
    \displaystyle\ \log_9\left(\frac{1}{3}\right)=x

    \displaystyle\ \log_3\left(\frac{1}{9}\right)=y

    Imagine a "fog" surrounding the "log" and the base slides over to the other side,
    so we get

    \displaystyle\frac{1}{3}=9^x

    \displaystyle\frac{1}{9}=3^y

    Now x and y will be easier to derive.
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  4. #4
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    thank you
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