ML estimation for Weibull What are the maximum likelihood estimators of $\eta$ and $\beta$ ($\eta>0$ and $\beta>0$) for an i.i.d. sample of size $n$ from the following density: $f(x_i)=\frac{\beta x_i^{\beta-1} }{\eta ^ {\beta} } \cdot \exp\left(-(x_i/{\eta})^{\beta}\right)$ ?
Ps. According to Forbes et al 2011, p.196, they are the solution to the following equation system, but I don't manage to get the same result... 
$$\eta=\left[\frac{\sum_{i=1}^n{x_i^{\beta}}}{n}\right]^{1/\beta}, \beta=\frac{n}{\frac{1}{\eta}\sum_{i=1}^{n}{x_i^{\beta} \cdot \log(x_i)}-\sum_{i=1}^{n}{\log(x_i)}}.$$ 
Is there an error in the book, or am I making a mistake? 
 A: The log-likelihood is 
$$\log L=n\log\beta-n\beta\log\eta+(\beta-1)\sum_i\log x_i-\eta^{-\beta}\sum_ix_i^\beta$$
You have to maximize this, so you can first make partial derivatives vanish, which leads to two equations. But note the equations can only be solved numerically.
$$\frac{\partial\log L}{\partial \eta}=-\frac{n\beta}{\eta}+\frac{\beta}{\eta^{\beta+1}}\sum_ix_i^\beta=0$$
Hence
$$\eta=\left[\frac1n\sum_ix_i^\beta\right]^{1/\beta}$$
The other partial derivative is:
$$\frac{\partial\log L}{\partial \beta}=\frac n\beta-n\log\eta+\sum_i\log x_i+\log(\eta)\;\eta^{-\beta}\sum_ix_i^\beta-\eta^{-\beta}\sum_i\log (x_i)x_i^\beta=0$$
By reusing the first equation it's possible to simplify this a bit:
$$\log(\eta)\;\eta^{-\beta}\sum_ix_i^\beta=\log\eta\frac{1}{\frac1n\sum_ix_i^\beta}\sum_ix_i^\beta=n\log\eta$$
Hence
$$\frac n\beta+\sum_i\log x_i-\eta^{-\beta}\sum_i\log (x_i)x_i^\beta=0$$
$$\beta=\frac{n}{\frac1{\eta^\beta}\sum_ix_i^\beta\log x_i-\sum_i\log x_i}$$
It's almost the expression in your question, but I get $\eta^\beta$ instead of $\eta$ in the second equation.
It's also possible to remove entirely $\eta$ from the second equation:
$$\frac1\beta=\frac1n\left[\frac{n\sum x_i^\beta\log x_i}{\sum_i x_i^\beta}-\sum_i\log x_i\right]=\frac{\sum x_i^\beta\log x_i}{\sum_i x_i^\beta}-\frac1n\sum_i\log x_i$$
