Given $(X_n)_n$ be i.i.d. random variables with pdf $$\ f(x\mid\theta) = \frac{\theta\ x^{\theta-1}}{3^{\theta}}I_{(0,3)}(x)$$ where $\ \theta > 0$. Using the WLLN, show that the maximum likelihood estimate (MLE) converges to $\theta$ in probability.
My progress: By definition, $$L(\theta\mid x) = \prod_{i=1}^{n} f(x_i\mid\theta) = \frac{\theta^n(x_1\ldots x_n)^{\theta-1}}{3^{n\theta}}$$ From there, taking derivative of $L(\theta\mid x)$ with respect to $\theta$ and setting it equal to $0$ to solve for $\theta$, I found that the MLE of $\theta$ is $$\hat\theta_n= \frac{n}{\ln(\frac{3^n}{x_1\ldots x_n})}$$ But I could not see how the MLE converges to $\theta$ using WLLN. I tried dividing both numerator and denominator by $n$ to get: $$\hat\theta_n=\frac{1}{\ln(3) - \frac{\ln(x_1\ldots x_n)}{n}}$$ On the other hand, $E(x_k) = \frac{3\theta}{\theta+1}$ for every $k$, but it's not true that $\frac{\ln(x_1)+\ln(x_2) + \ldots + \ln(x_n)}{n}\rightarrow E(\ln(x_1))$ in probability.
My question: Could anyone please give me some hints on this problem? Did I get the wrong MLE? I reviewed my computation several times and could not find error though.