Lets say that I have N observations that are poisson and i.i.d. The prior is an exponential with parameter 2. I know that the exponential distribution is given by

$
\lambda e^{(-\lambda x)}
$

But how does it work with x when you multiply the prior times the likelihood to get the maximum a posteriori (MAP) estimate? In most places, they seem to just set $x=1$, but I don't understand why.

This would lead to the the MAP estimate to be

$
\operatorname{argmax} \quad \lambda^{\sum x_n}e^{-\lambda N}e^{-2\lambda}
$


So my question is, how do you handle $x$ in the exponential distribution when it is a prior?