let {X_1,...,X_n} be a random sample from an exponential distribution with the density function f(x|\theta)=\theta exp(-\theta x). Derive a likelihood ratio test of H_0:\theta=\theta_0versus H_A:\theta \neq \theta_0, and show that the rejection region is of the form (\overline{X}exp[-\theta_0\overline{X}]\leq c)

What I did:
\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]}

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

= 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
= 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