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Math Help - Logarithms Properties/Identities/Etc.

  1. #1
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    Logarithms Properties/Identities/Etc.

    Need some help on these, just to make sure I understand what I'm doing...

    1) If \log_ba = x then \log_ab =

    \frac{1}{\log_ab}

    2) If \log_ba = x then \log_\frac{1}{b}a =

    -\log_ba

    3) \log(2 - x) + \log(3 - x) = \log12 solve for x

     x = -1

    4)  y = \log_bx + \log_b(x - 6) solve for x

    b^y = x^2 - 6x

    then...

    would the quadratic formula be used?

     x^2 - 6x - b^y = 0

    ??
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  2. #2
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    Quote Originally Posted by JSB1917 View Post
    4)  y = \log_bx + \log_b(x - 6) solve for x

    b^y = x^2 - 6x

    then...

    would the quadratic formula be used?

     x^2 - 6x - b^y = 0

    ??
    To solve for x yes, use a=1, b=-6 and c = - b^y
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  3. #3
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    Quote Originally Posted by pickslides View Post
    To solve for x yes, use a=1, b=-6 and c = - b^y
    Ok, so it would come out to...

    x = \frac {6 \pm \sqrt{36 + 4b^y}}{2}

     x > 6
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  4. #4
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    Quote Originally Posted by JSB1917 View Post
    Need some help on these, just to make sure I understand what I'm doing...

    1) If \log_ba = x then \log_ab =

    \frac{1}{\log_ab}

    2) If \log_ba = x then \log_\frac{1}{b}a =

    -\log_ba
    Ok, I'm confused. My teacher marked both of these wrong. Instead he wanted \frac {1}{x} for the first and -x for the second, but isn't what I put essentially the logical equivalence of the answers?

    I got a zero for both of these. I had to turn back in the HW sheet because he completely missed a problem and gave a zero for it and I didn't have time to think about these two at the moment.

    I can't understand why I shouldn't get at least 1 point for both of the problems, even though he wanted the work shown.
    Last edited by JSB1917; September 22nd 2009 at 07:47 AM.
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  5. #5
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    Quote Originally Posted by JSB1917 View Post
    Need some help on these, just to make sure I understand what I'm doing...

    1) If \log_ba = x then \log_ab =


    2) If \log_ba = x then \log_\frac{1}{b}a =



    3) \log(2 - x) + \log(3 - x) = \log12 solve for x



    (1) \frac{1}{\log_ab}=x \Rightarrow \log_ab=\frac{1}{x}

    (2) \frac{1}{log_a\frac{1}{b}}=\frac{1}{-log_ab}=-\frac{1}{x}

    (3)log (2-x)(3-x)=log 12

    x^2-5x-6=0

    (x-6)(x+1)=0

    Continue solving here .
    Last edited by mathaddict; September 22nd 2009 at 08:06 AM.
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  6. #6
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    Quote Originally Posted by mathaddict View Post
    ..
    wait what?

    Also I noticed a typo earlier, it's supposed to be for number one.

    \frac {1}{\log_b a}
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  7. #7
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    Quote Originally Posted by JSB1917 View Post
    wait what?

    Also I noticed a typo earlier, it's supposed to be for number one.

    \frac {1}{\log_b a}
    edited , so clear now ?
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