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Math Help - Approximate values of these logs?

  1. #1
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    Approximate values of these logs?

    given that log_b 2=(approximately)0.756 and that log_b 10=(approximately)2.513, find the approximate values of each of ther following

    a) log_b 40

    for this one I got log_b 2^2x10. (x is times) then I got (.756)^2x10 =5.715. I have no idea if this is right? I am also unsure of..

    b)log_b 1/2

    I know I have to flip the fraction, so would that make it -log_b 2?

    c) log_b 81/log_b 27

    no idea what to do with this one?

    d)log_b 1/b^3
    same here. i just don't these.

    there are a few more to this problem, but I got those ok. these have me stumped??
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  2. #2
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    Hello, Comical!

    Given: . \log_b(2)\:\approx\:0.756,\;\;\log_b(10)\:\approx\  : 2.513
    find the approximate value of each of the following:

    a)\;\;\log_b(40)
    \log_b(40) \:=\:\log_b(2^2\cdot10) \:=\:\log_b(2^2) + \log_b(10) \:=\:2\log_b(2) + \log_b(10)

    Therefore: . \log_b(40) \;=\;2(0.756) + 2.513 \;=\;4.025




    b)\;\;\log_b\!\left(\frac{1}{2}\right)

    \log_b\!\left(\frac{1}{2}\right) \;=\;\log_b(2^{-1}) \;=\;-1\cdot \log_b(2) \;=\;-1(0.756) \;=\;-0.756




    c)\;\;\frac{\log_b(81)}{\log_b(27)}
    This one has nothing to do with the given decimals . . .

    \frac{\log_b(81)}{\log_b(27)} \;=\;\frac{\log_b(3^4)}{\log_b(3^3)} \;=\;\frac{4\cdot\log_b(3)}{3\cdot\log_b(3)} \;=\; \frac{4\cdot{\color{red}\rlap{//////}}\log_b(3)}{3\cdot{\color{red}\rlap{//////}}\log_b(3)} \;=\;\frac{4}{3}




    d)\;\;\log_b\!\left(\frac{1}{b^3}\right)
    \log_b\!\left(\frac{1}{b^3}\right) \;=\;\log_b\left(b^{-3}\right) \;=\;-3\cdot\underbrace{\log_b(b)}_{\text{This is 1}} \;=\;-3\cdot1 \;=\;-3

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  3. #3
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    Woops, wrong account
    Last edited by Chinnie15; September 17th 2009 at 07:31 PM.
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