Sufficient Statistic for $f(x,\theta)=\dfrac{2}{\theta^{2}} (\theta-x) \cdot 1_{(0,\theta)}(x), \;\forall \theta \in (0,\theta) $ Let $X_{1},\ldots, X_{n}$ be random variables independent and identically distributed; show that the following density function is in the exponential family and find the sufficient statistic for $\theta$.
$$ f(x, \theta)=f(x,\theta)=\dfrac{2}{\theta^{2}} (\theta-x) \cdot 1_{(0,\theta)}(x), \;\forall x \in (0,\theta) $$
My attempt is
$f(x, \theta)=e^{\ln(\theta-x)+\ln(2/\theta^{2})}\cdot 1_{(0,\theta)}(x)$.
My question is: can I consider $T(x_{1},\ldots,x_{n})=\sum_{i=1}^{n}\ln(\theta-x_{1})$ a sufficient statistic for $\theta?$ If the sufficient statistic cannot depend on the parameter, does this density function not have a sufficient statistic?
How to reduce this expression so that the statistic found does not depend on $\theta$?

Hint: Look at the likelihood ratio
$$
\text{LR}=\prod_i \frac{(2/\theta^2)(\theta-x_i)\cdot \mathbb{1}_{(0,\theta)}(x_i)}{(2/\psi^2)(\psi-x_i)\cdot \mathbb{1}_{(0,\psi)}(x_i)}
$$
defined for $\max_i x_i < \min(\theta, \psi)$. Use that the likelihood ratio is a function of any sufficient statistic. The above can be simplified to
$$
\text{LR}=\frac{\psi^2}{\theta^2}\prod_i \frac{\theta-x_i}{\psi-x_i}.
$$
