I was working on this MLE problem which I derived to be \begin{equation*} f(x;\theta) = \frac{1}{x\cdot \ln \theta} \end{equation*} where 1 < x < $\theta$ \begin{equation*} f(x_1,x_2,x_3,...,x_n;\theta) = \prod_{i=1}^n\Big(\frac{1}{x_i\cdot \ln \theta} \Big) \end{equation*} \begin{equation*} f(x_1,x_2,x_3,...,x_n;\theta) = \Big(\frac{1^n}{\prod_{i=1}^n (x_i)\cdot \ln \theta^n} \Big) \end{equation*} \begin{equation*} L(\theta) = \ln(f(x_1,x_2,x_3,...,x_n;\theta)) = \ln\Big(\frac{1^n}{\prod_{i=1}^n (x_i)\cdot \ln \theta^n} \Big) \end{equation*} \begin{equation*} L(\theta) = \ln(1)- \ln\Big(\prod_{i=1}^n (x_i)\cdot \ln \theta^n\Big) = - \ln\Big(\prod_{i=1}^n (x_i)\Big) + \ln(n\ln \theta) \end{equation*} \begin{equation*} L(\theta) = -\Big(\sum_{i=1}^n (x_i)\Big) + \ln(n\ln \theta) \end{equation*} \begin{equation*} \frac{d}{d\theta}(L(\theta)) = -0 + \frac{n}{\theta\ln \left(\theta\right)} = \frac{n}{\theta\ln \left(\theta\right)} \end{equation*}
Obviously if I set the last equation equal to 0, I will get n to equal zero with $\theta$ being undefined. Is this even possible or did I make a mistake in my math?
You made some small mistakes: \begin{align} L(\theta) = \ln\Big(\frac{1^n}{\prod_{i=1}^n (x_i)\cdot \ln \theta^n} \Big) &= -\sum_{i=1}^n\ln(x_i) -n\ln(\ln(\theta)) \end{align} So the derivative for $\theta$ is: \begin{align} L'(\theta) = -n\frac{1}{\ln(\theta)}\cdot\frac{1}{\theta} \end{align} As you noted setting $L'$ to $0$ doesn't get you far. $L(\theta)$ is decreasing with $\theta$. You also noted that $1<x_i<\theta$ for all $i$. So $\hat{\theta}=\max_i x_i$ is the MLE.