# Natural log of product

1. Jul 18, 2013

### Phox

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. Jul 18, 2013

### haruspex

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)?

3. Jul 18, 2013

### Zondrina

$ln(\frac{a}{b}) = ln(a) - ln(b)$

Last edited: Jul 18, 2013