For example, I'm trying to find the MLE for $\theta$ by taking the log-likelihood of pdf:
$ f(x|\theta)=\frac{(\theta-1)log(x)}{x^\theta} $
$ \log \prod_{i=1}^n\frac{(\theta-1) \log(x)}{x^\theta} $
Is there a way to take the $\log(x)$ out? I don't want to take the logarithm of a logarithm:
$ \log(\theta-1)+ \log \log(x)-\theta\log(x) $
I'm not very good at differentiating, so I'm not sure when I differentiate by $\theta$ if the $\log \log(x)$ would just go away since there's no theta attached?
First, the log-likelihood function is $$g(\theta) = \log\left(\prod_{i=1}^{n}\frac{(\theta-1)}{x_i^\theta}\log(x_i)\right)=n\log(\theta-1)-\theta\sum_{i=1}^{n}\log(x_i)+\sum_{i=1}^{n}\log(\log(x_i)).$$ Observe that you need to take the derivative w.r.t. $\theta$. Consequently, $$g'(\theta) = \frac{\partial}{\partial\theta}\left(n\log(\theta-1)-\theta\sum_{i=1}^{n}\log(x_i)+\sum_{i=1}^{n}\log(\log(x_i))\right)\\=\frac{\partial}{\partial\theta}\left(n\log(\theta-1)-\theta\sum_{i=1}^{n}\log(x_i)\right) + 0,$$ as $\frac{\partial}{\partial \theta}\log(\log(x_i))=0$.