Suppose that an observation $X$ is drawn from the following distribution
$f(x|\theta) = \begin{cases} \frac{1}{\theta} & \text{ if } 0 < x < \theta \\ 0 & \text{ if } otherwise \end{cases}$
The prior pdf is
$\xi(\theta) = \begin{cases} \theta e^{-\theta} & \text{ if } \theta > 0 \\ 0 & \text{ if } otherwise \end{cases}$
So I just multiply the likelihood function and the prior to get the posterior
$\xi(\theta|x) \propto \theta e^{-\theta} \frac{1}{\theta}$
$\propto e^{-\theta}$.
Up to this point everything makes sense. Then the solution proceeds to state that
$\xi(\theta|x) = \begin{cases} e^{x - \theta} & \text{ if } \theta > x \\ 0 & \text{ if } otherwise \end{cases}$
Where did the extra $e^x$ in the front come from?