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Math Help - Natural Log Seperation

  1. #1
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    Natural Log Seperation

    I worked these math problems out, however the system says they are incorrect. I would appreciate it if you could please check over my work, and maybe even tell me where I went wrong:

    1. ln(a)=2 , ln(b)=3 , ln(c)=10

    \frac{ln(a^{-2}b^1)}{ln(bc)^1}

    -2ln(a)+1ln(b)-1ln(b)-1ln(c)

    -2(2)+1(2)-1(3)-1(5)

    9


    2. ln(a)=2 , ln(b)=3 , ln(c)=10

    ln(c^{-4})*ln(\frac{a}{b^{-4}})^4

    -4ln(c)+4ln(a)+16ln(b)

    -4(5)+4(2)+16(3)

    36
    Last edited by qbkr21; January 29th 2007 at 01:32 PM.
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  2. #2
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    Quote Originally Posted by qbkr21 View Post
    1. ln(a)=2 , ln(b)=3 , ln(c)=10
    #1 should be:
    \frac{ln(a^{-2}b^1)}{ln(bc)^1}= \frac{-2ln(a) + ln(b)} {ln(b) + ln(c)}
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  3. #3
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    Hello, qbkr21!

    You are misusing some log rules . . .


    1.\;\ln(a)=2,\;\ln(b)=3,\;\ln(c)=10

    \frac{\ln(a^{-2}b^1)}{\ln(bc)^1} \;= \;-2\ln(a)+1\ln(b)-1\ln(b)-1\ln(c) . . . no

    \frac{\ln(a^{-2}b)}{\ln(bc)} \:=\:\frac{-2\ln(a) + \ln(b)}{\ln(b) + \ln(c)} \:=\:\frac{-2(2) + 3}{3 + 10} \:=\:-\frac{1}{13}



    2.\;\ln(a)=2,\;\ln(b)=3,\;\ln(c)=10

    \ln(c^{-4})\cdot\ln(\frac{a}{b^{-4}})^4 \;= \;-4\ln(c) + 4\ln(a) + 16\ln(b) . . . no

    \ln(c^{-4})\cdot\ln\left(\frac{a}{b^{-4}}\right)^4\:=\:-4\ln(c)\cdot 4\ln\left(\frac{a}{b^{-4}}\right)

    . . =\:-16\ln(c)\left[\ln(a) - \ln(b^{-4})\right] \:= \:-16\ln(c)\left[\ln(a) + 4\ln(b)\right]

    . . =\:-16(10)\left[2 + 4\cdot3\right] \;=\;-160(14)\:=\:-2240

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  4. #4
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    These answers are both off a bit... The answer to #1 should have been \frac{-1}{8}
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  5. #5
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    Quote Originally Posted by qbkr21 View Post
    These answers are both off a bit... The answer to #1 should have been \frac{-1}{8}
    Their answers are correct. I not only derived their expressions, but I plugged the original problem into my calculator and verified their results. Could there be a typo in your problem? That "1" in the denominator is in a strange place.

    -Dan
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  6. #6
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    Quote Originally Posted by qbkr21 View Post
    These answers are both off a bit... The answer to #1 should have been \frac{-1}{8}
    If that is true, then you have given the problem incorrectly.
    Please review what you have written.
    It is also possible that the text is wrong!
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  7. #7
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    ok thanks give me a moment... The latex software is still real new to me
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  8. #8
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    Ok don't worry about the 1st one I figured it out and I know from now on thanks to Soroban that you treat ln on top and bottom as two separate pieces. Problem #2 is a follows:

    I think that the values might be off, use these values instead:

     \ ln(a)=2 \ ln(b)=3 \ ln(c)=5

    (\ln\ c^{-4})(\ln \frac{a}{b^{-4}})^{4}
    Last edited by qbkr21; January 29th 2007 at 06:21 PM.
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