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.
On
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.
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.