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Math Help - Need Help with Basic Log

  1. #1
    Member
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    Need Help with Basic Log

    Evaluate:
    1) \log_2 128 - \log_2 32
    2) \log_8 \sqrt8


    Can someone please help me with these log equations! Please show me what to do IN DETAIL (STEP BY STEP)! I really don't get what to do. I know you're suppose to do error and trial for the first equation, but how do I do this? And I don't know how to use the calculator "log" function when the base is not 10!

    Answers:

    1) 2
    2) 0


    P.S.
    Are the following the correct way of writing an inverse of y = 4^x?
    Exponential:
     f^-1(x) = 4^x
    Logarithmic:
     y = \log_4 x
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  2. #2
    Senior Member DivideBy0's Avatar
    Joined
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    From
    Melbourne, Australia
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    Quote Originally Posted by Macleef View Post
    Evaluate:
    1) \log_2 128 - \log_2 32
    2) \log_8 \sqrt8


    Can someone please help me with these log equations! Please show me what to do IN DETAIL (STEP BY STEP)! I really don't get what to do. I know you're suppose to do error and trial for the first equation, but how do I do this? And I don't know how to use the calculator "log" function when the base is not 10!

    Answers:

    1) 2
    2) 0


    P.S.
    Are the following the correct way of writing an inverse of y = 4^x?
    Exponential:
     f^-1(x) = 4^x
    Logarithmic:
     y = \log_4 x
    When trying to simplify basic logs of the form \log_a b, try to turn it into the form \log_a a^x, as this equals a.

    1) \log_2 128 -\log_2 32

    =\log_2 2^7 - \log_2 2^5

    Now, think of what this means.

    x=\log_2 2^7 \iff 2^x = 2^7

    x must then be 7.

    Also, x=\log_2 2^5 \iff 2^x = 2^5

    \Rightarrow x = 5

    So we then have 7-5=2

    2) \log_8 \sqrt{8}

    Recall that \sqrt{8} = 8^\frac{1}{2}

    So, \log_8 \sqrt{8} = \log_8 8^{\frac{1}{2}} = \frac{1}{2} ... i think the answers must be wrong.
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  3. #3
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    Thanks, it makes sense to me now
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