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Thread: 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 $\displaystyle \log_ba = x$ then $\displaystyle \log_ab =$

    $\displaystyle \frac{1}{\log_ab}$

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

    $\displaystyle -\log_ba$

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

    $\displaystyle x = -1$

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

    $\displaystyle b^y = x^2 - 6x$

    then...

    would the quadratic formula be used?

    $\displaystyle x^2 - 6x - b^y = 0$

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

    $\displaystyle b^y = x^2 - 6x$

    then...

    would the quadratic formula be used?

    $\displaystyle x^2 - 6x - b^y = 0$

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

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

    $\displaystyle x > 6$
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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 $\displaystyle \log_ba = x$ then $\displaystyle \log_ab =$

    $\displaystyle \frac{1}{\log_ab}$

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

    $\displaystyle -\log_ba$
    Ok, I'm confused. My teacher marked both of these wrong. Instead he wanted $\displaystyle \frac {1}{x}$ 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; Sep 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 $\displaystyle \log_ba = x$ then $\displaystyle \log_ab =$


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



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



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

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

    (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; Sep 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.

    $\displaystyle \frac {1}{\log_b a}$
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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.

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