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Math Help - natural logarithim of a single quantitiy

  1. #1
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    natural logarithim of a single quantitiy

    stuck part way through on simplifying this problem....

    2/3[5ln(x+1) - 2ln(x-1) + 4ln(3x-1)] - 2ln(x+5) + 1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by RosieLaird View Post
    stuck part way through on simplifying this problem....

    2/3[5ln(x+1) - 2ln(x-1) + 4ln(3x-1)] - 2ln(x+5) + 1
    1. Parentheses do not match

    2. You know that A \ln(B)=\ln(B^A) that \ln(U)+\ln(V)=\ln(UV) and \ln(U)-\ln(V)=\ln(U/V), so tell us what problem you are having applying these laws to this simplification.

    CB
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  3. #3
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    Also keep in mind that  ln|e| = 1 .
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  4. #4
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    Hello, RosieLaird!

    This cannot be simplified to a single log.


    \frac{2}{3}\bigg[\sin(x+1) - 2\ln(x-1) + 4\ln(3x-1)\bigg] - 2\ln(x+5) + 1

    \frac{2}{3}\sin(x+1) - \frac{4}{3}\ln(x-1) + \frac{8}{3}\ln(3x-1) - 2\ln(x+3) + 1

    . . =\;\frac{8}{3}\ln(3x-1) - \frac{4}{3}\ln(x-1) - 2\ln(x+3) + \frac{2}{3}\sin(x+1) + 1

    . . =\;\frac{2}{3}\bigg[4\ln(3x-1) - 2\ln(x-1) - 3\lnm(x+3)\bigg] + \frac{2}{3}\sin(x+1) + 1

    . . =\;\frac{2}{3}\bigg[\ln(3x-1)^4 - \ln(x-1)^2 - \ln(x+3)^3\bigg] + \frac{2}{3}\sin(x+1) + 1

    . . =\;\frac{2}{3}\ln\left[\dfrac{(3x-1)^4}{(x-1)^2(x+3)^3}\right] + \frac{2}{3}\sin(x+1) + 1

    . . =\;\ln\left[\dfrac{(3x-1)^4}{(x-1)^2(x+3)^3}\right]^{\frac{2}{3}} + \frac{2}{3}\sin(x+1) + 1

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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    Hello, RosieLaird!

    This cannot be simplified to a single log.



    \frac{2}{3}\sin(x+1) - \frac{4}{3}\ln(x-1) + \frac{8}{3}\ln(3x-1) - 2\ln(x+3) + 1

    . . =\;\frac{8}{3}\ln(3x-1) - \frac{4}{3}\ln(x-1) - 2\ln(x+3) + \frac{2}{3}\sin(x+1) + 1

    . . =\;\frac{2}{3}\bigg[4\ln(3x-1) - 2\ln(x-1) - 3\lnm(x+3)\bigg] + \frac{2}{3}\sin(x+1) + 1

    . . =\;\frac{2}{3}\bigg[\ln(3x-1)^4 - \ln(x-1)^2 - \ln(x+3)^3\bigg] + \frac{2}{3}\sin(x+1) + 1

    . . =\;\frac{2}{3}\ln\left[\dfrac{(3x-1)^4}{(x-1)^2(x+3)^3}\right] + \frac{2}{3}\sin(x+1) + 1

    . . =\;\ln\left[\dfrac{(3x-1)^4}{(x-1)^2(x+3)^3}\right]^{\frac{2}{3}} + \frac{2}{3}\sin(x+1) + 1

    Just as I misread the parentheses you have misread the first term in the square brackets; it id 5\ln(x+1) not \sin(x+1)

    CB
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  6. #6
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    Ha!

    That's a new blunder for me . . .

    . . Reading 5ln(x+1) to be sin(x+1)



    Sheesh, that's embarrassing . . . Hope no one saw it.
    .
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  7. #7
    Senior Member yeKciM's Avatar
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    lol I see there this :

     \displaystyle \frac {2}{3[5\ln(x+1)-2\ln(x-1)+4\ln(3x-1)]} -2\ln(x+5)+1

    heheheh confusing

    P.S. at first glance that 5ln is sin hehehehe
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