Natural Log of Product Solution

[Phox]

Homework Statement

L(θ) = ∏(θ/(2√x_i)*e^(-θ√x_i)),i=1, n

Homework Equations

The Attempt at a Solution

-> θ²∏(1/(2√x_i)*e^(-θ√x_i))

taking natural log of both sides

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

= nlnθ + Ʃln(1/(2√x_i)*e^(-θ√x_i))

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

 

[haruspex]

= nlnθ + Ʃln(1/(2√x_i)*e^(-θ√x_i))

Ok so from what I understand the ln of a product is the sum.

You can apply that again to the product [1/(2√x_i)]*[e^(-θ√x_i)]. You can then also use ln(x^a) = a ln(x) for cases like a = -1, a = 1/2. What can you do with ln(e^x)?

 

[STEMucator]

Homework Statement

L(θ) = ∏(θ/(2√x_i)*e^(-θ√x_i)),i=1, n

Homework Equations

The Attempt at a Solution

-> θ²∏(1/(2√x_i)*e^(-θ√x_i))

taking natural log of both sides

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

= nlnθ + Ʃln(1/(2√x_i)*e^(-θ√x_i))

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

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