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Math Help - Help Simplifying Logarithms

  1. #1
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    Help Simplifying Logarithms

    I have a problem here involving natural logs, please show your work so I can learn if I'm making a mistake.

    Simplify the following expression as completely as possible:
    5x + ln (3x) - ln (9x^2) + ln (3) - ln (x) + ln (x^2) -4x

    I know I can put the 2 exponential numbers in front of their natural logs ex. 2 ln (9x), 2 ln (x), but I'm having trouble deciding where to divide/multiply/etc.
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  2. #2
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    Quote Originally Posted by Farnsworth View Post
    I have a problem here involving natural logs, please show your work so I can learn if I'm making a mistake.

    Simplify the following expression as completely as possible:

    5x + ln (3x) - ln (9x^2) + ln (3) - ln (x) + ln (x^2) -4x

    I know I can put the 2 exponential numbers in front of their natural logs ex. 2 ln (9x), 2 ln (x), but I'm having trouble deciding where to divide/multiply/etc.
    You can evaluate ln(3) with your calculator, best not to though!.
    5x-4x=x.

    You need the division law of logarithms for the subtraction of logarithms
    and you have 2 choices to start with.

    ln(3x)-ln(9x^2)\ and\ ln(x^2)-ln(x)

    or

    ln(3x)-ln(x)\ and\ ln(x^2)-ln(9x^2)

    Use the logarithm law

    log_e(a)-log_e(b)=log_e\frac{a}{b}
    Last edited by Archie Meade; August 4th 2010 at 05:30 PM.
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  3. #3
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    First you have a mistake: \ln(9x^2)=2\ln(3x).
    Do you see why?
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  4. #4
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    so going off ln(3x) - ln(x) and ln(x^2) - ln (9x^2) the answer would come out as ln(3) - ln(9), then I could add the former ln(3) and x to get

    2 ln(3) - ln(9) + x which would cancel out to x?

    Man, if that's right I'm going to be so glad I asked you guys. *hopes*
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  5. #5
    Senior Member eumyang's Avatar
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    Quote Originally Posted by Farnsworth View Post
    so going off ln(3x) - ln(x) and ln(x^2) - ln (9x^2) the answer would come out as ln(3) - ln(9)
    I'm getting \ln\,3 + \ln \left( \frac{1}{9} \right) here.
    Which equals
    \ln\,3 -2 \ln\,3 = -\ln\,3
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  6. #6
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    Quote Originally Posted by eumyang View Post
    I'm getting \ln\,3 + \ln \left( \frac{1}{9} \right) here.
    Which equals
    \ln\,3 -2 \ln\,3 = -\ln\,3
    Yeah, but ln (3) + ln (1/9) and ln(3) - ln(9) both come out as the answer to -ln(3), so adding the first ln(3) would leave me with x, right?
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  7. #7
    Senior Member eumyang's Avatar
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    Yep. I misread your post.
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  8. #8
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    Quote Originally Posted by Farnsworth View Post
    so going off ln(3x) - ln(x) and ln(x^2) - ln (9x^2) the answer would come out as ln(3) - ln(9), then I could add the former ln(3) and x to get

    2 ln(3) - ln(9) + x which would cancel out to x?

    Man, if that's right I'm going to be so glad I asked you guys. *hopes*
    good work!
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