let $\displaystyle {X_1,...,X_n}$ be a random sample from an exponential distribution with the density function $\displaystyle f(x|\theta)=\theta exp(-\theta x)$. Derive a likelihood ratio test of $\displaystyle H_0:\theta=\theta_0$versus $\displaystyle H_A:\theta \neq \theta_0$, and show that the rejection region is of the form $\displaystyle (\overline{X}exp[-\theta_0\overline{X}]\leq c)$

What I did:
$\displaystyle \frac{f_0(x)}{f_1(x)}=\frac{\prod_{i=1}^{n}\theta_ 0exp[-\theta_0 X_i]}{\prod_{i=1}^{n}\hat{\theta}exp[-\hat{\theta} X_i]}$

=$\displaystyle \prod_{i=1}^{n}\frac{\theta_0}{\hat{\theta}}exp(\h at{\theta}X_i-\theta_0X_i)$

=$\displaystyle log\frac{f_0(x)}{f_1(x)}=nlog(\frac{\theta_0}{\hat {\theta}})+\hat{\theta}\sum_{i=1}^{n}X_i-\theta_0\sum_{i=1}^{n}X_i$
=$\displaystyle nlog\theta_0-nlog\hat{\theta}+n\hat{\theta}\overline{X}-n\theta_0\overline{X}$

But it is not even close to what I want to show

Yeah I also don't know which theta should I use for alternative equation, so I use theta hat