Log of products and densities By taking the log of the function:  $$\prod_{t=1}^T n_t! \prod_{i=1}^{n_t} \frac{f_v[b_{i:n_t}]}{[1-Fv(r)]},$$ it is possible to end up with the following expression: 
$$
\log(n_t!)+\sum_{i=1}^{n_t} \log\{f_v[b_{i:n_t}]\}-n_t \log[1-F_v(r)],
$$
where $f_v$ is a pdf and $F_v$ is the CDF.
How is that transformation possible? and how is it done? 
EDIT 1: Here is a picture from the book. 

 A: Guide:
Use the following properties


*

*Logarithm of the product is equal to the sum of the logarithm, that is $$\log(\prod_{i=1}^na_i)=\sum_{i=1}^n\log a_i $$

*$$\log a^b = b\log a$$

*Also notice that the denominator term of the original expression is independent from $i$.
Also, you left out a summation over $t$ in the final answer.
A: Let's ignore the first product (could it be a typo?), and call the rest $x_t$. Then use the fact that $\ln (a \cdot b) = \ln a + \ln b$ and $\ln\left(a^b\right) = b \ln a$ to get 
$$
\begin{split}
\ln x_t
 &= \ln \left( n_t!
               \prod_{i=1}^{n_t} \frac{f_v[b_{i:n_t}]}{[1-F_v(r)]}
        \right) \\
 &= \ln \left( \frac{n_t!}{[1-F_v(r)]^{n_t}} \prod_{i=1}^{n_t} f_v[b_{i:n_t}] \right) \\
 &= \ln (n_t!) -\ln\left( [1-F_v(r)]^{n_t} \right)
    + \ln \left(\prod_{i=1}^{n_t} f_v[b_{i:n_t}] \right) \\
 &= \ln (n_t!) -n_t \ln( 1-F_v(r))
    + \sum_{i=1}^{n_t} \ln ( f_v[b_{i:n_t}] )
\end{split}
$$
UPDATE
On your question about how $(1-F_v(r))$ ended up with an exponent of $n_t$, note that $(1-F_v(r))$ is a function of $r$ but not a function of $i$, and therefore
$$
\prod_{i=1}^{n_t} [1-F_v(r)]
 = \underbrace{[1-F_v(r)] \times \ldots \times [1-F_v(r)]}_{n_t \text{ times}}
 = [1-F_v(r)]^{n_t}.
$$
Moreover, consider now the double product as your likelihood function
$$
L(r) = \prod_{t=1}^T n_t!
       \prod_{i=1}^{n_t} \frac{f_v[b_{i:n_t}]}{[1-F_v(r)]}
     = \prod_{t=1}^T x_t
$$
then your log-likelihood is
$$
\ln L(r)
 = \ln\left(\prod_{t=1}^T x_t \right)
  = \sum_{t=1}^T \ln(x_t),
$$
where $\ln x_t$ for each $t$ is given in the original answer.
