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Natural log of product

  1. Jul 18, 2013 #1
    1. The problem statement, all variables and given/known data

    L(θ) = ∏(θ/(2√xi)*e^(-θ√xi)),i=1, n

    2. Relevant equations



    3. The attempt at a solution

    -> θ2∏(1/(2√xi)*e^(-θ√xi))

    taking natural log of both sides

    lnL(θ) = nlnθ + ln∏(1/(2√xi)*e^(-θ√xi))

    = nlnθ + Ʃln(1/(2√xi)*e^(-θ√xi))

    Ok so from what I understand the ln of a product is the sum. But i'm not sure how exactly to simplify from here
     
  2. jcsd
  3. Jul 18, 2013 #2

    haruspex

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    You can apply that again to the product [1/(2√xi)]*[e^(-θ√xi)]. You can then also use ln(xa) = a ln(x) for cases like a = -1, a = 1/2. What can you do with ln(ex)?
     
  4. Jul 18, 2013 #3

    Zondrina

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    ##ln(\frac{a}{b}) = ln(a) - ln(b)##
     
    Last edited: Jul 18, 2013
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