# MLE for the mean of a distribution?

Let $X_1, \ldots, X_n$ be a random sample of size $n$ from a distribution having the pdf $$f(x)=\frac{2x}{\theta^n}.$$ I need to find the MLE for the mean of the distribution but am not sure how.

The idea behind MLE is to pick the value of the parameter to maximize the chances that the currently observed sample would be generated. Assuming $\{X_k\}$ are iid and actually result in the values $\{x_k\}$, we have for $\epsilon \to 0$ $$\begin{split} p(\theta) &= \mathbb{P}\left[ X_1 \in [x_1 - \epsilon, x_1 + \epsilon], \ldots, X_n \in [x_n - \epsilon, x_n + \epsilon] \right] \\ &= \prod_{k=1}^n \mathbb{P}\left[ X_k \in [x_k - \epsilon, x_k + \epsilon] \right] \\ &= \prod_{k=1}^n f(x_k) = \prod_{k=1}^n \frac{2 x_k}{\theta^n} = \left(\frac{2}{\theta^n}\right)^n \prod_{k=1}^n x_k. \end{split}$$ It helps to maximize $$L(x) = \ln f(x) = n \ln \left(\frac{2}{\theta^n}\right) + \sum_{k=1}^n \ln(x_k)$$ with respect to $\theta$ getting the best value for $\theta$ in terms of the $x_k$'s...