I include a bit of an introduction, even though my main question is more mathematical. I was tasked with finding the Maximum Likelihood Estimate for $\theta$ in $$\mathrm P(X>x) = \left(\frac ax \right)^\theta $$ where $X$ is a variable, and $x$ represents a value that variable can take on. The Probability Density Function is $\newcommand\diff[2]{\frac{\mathrm d#1}{\mathrm d#2}}\diff Fx=\frac{-\theta a^\theta}{x^{\theta + 1}}$, where $F = \mathrm P(X>x)$. I maximise the loglikelihood function $l = \ln(-\theta) + \theta \ln a - (\theta + 1)\ln x\ $ to get $\hat\theta(x_i) = \frac 1{\ln x_i - \ln a}$, where the $\hat.$ indicates that $\hat\theta$ is an estimate of $\theta$, based on the data sample. Now, the answer is supposed to be $$\hat\theta = \frac 1{\overline {\ln x} - \ln a}$$ where $\overline {\phantom{x}}$ indicates the average: $\overline{\ln x} = \frac 1n \sum_i \ln x_i$. I am stumped as to how to get this answer directly from $\hat\theta(x_i)$.
Does $$\frac 1n \sum_i \frac 1{\ln x_i - \ln a} = \frac 1{\overline {\ln x} - \ln a}\qquad ?$$
I think $\frac 1n \sum_i \widehat{\frac 1{\theta(x_i)}} = \frac 1n\sum_i (\ln x_i - \ln a) =\overline{\ln x} - \ln a = \widehat {\frac 1\theta} \implies \hat\theta = \frac 1{\overline{\ln x} - \ln a}$, but is this the only way to show the above?
Usually the maximum likelihood estimator is calculated based on the pdf of $X$.
If $P(X> x)=\left( \frac{\alpha}{x}\right)^{\theta}$ then $P(X\leq x)=1-\left( \frac{\alpha}{x}\right)^{\theta}$
Differentating w.r.t. $x$
$f_X(x)=\frac{\theta a^\theta}{x^{\theta + 1}}$
Then the likelihood function is
$L(\theta)=\Pi_{i=1}^n \frac{\theta a^\theta}{x_i^{\theta + 1}}$
$L(\theta)=\theta^n a^{\theta\cdot n}\cdot \Pi_{i=1}^n x_i^{-1-\theta}$
Taking ln
$$\ln L=n\cdot \ln (\theta)+n\cdot \theta\cdot \ln (\alpha)+\sum\limits_{i=1}^{n} (-1-\theta)\cdot \ln (x_i)$$
I think you can differentiate $\ln L$, set it equal to $0$ and solve the equation for $\theta$.
The result is indeed $\hat\theta = \frac 1{\overline {\ln x} - \ln a}$, where $\overline {\ln x}=\frac{ 1}{n}\sum\limits_{i=1}^{n} \ln (x_i)$