Can you show a plot of the data? That looks like an exponential curve.
Anyway, there are a number of ways to fit $a e^{-bt}$ to data to get
values for $a$ and $b$. An easy way to start is to take the log of the
data and do a linear fit of whatever type you like
(least squares, $L_1$, ...). This can give you an initial estimate
for $a$ and $b$ which can be put into a
non-linear fitting routine to fit $a e^{-bt}$.
If the initial data points are "bad", you can remove them when you
do the fitting.
Something I discovered many years ago (in the context of Newton's law of cooling)
is that
if we have three equally spaced (in time) data points,
we can fit an exponential to them. Suppose the points
are $(t-d, p), (t, q), $ and $(t+d, r)$,
and we want to fit $a e^{bt}+c$ to them.
We have $a e^{bt}+c = q$,
$a e^{b(t-d)}+c = p$,
and $a e^{b(t+d)}+c = r$.
Subtracting the first two,
$q-p =a (e^{bt}-e^{b(t-d)}) =a e^{bt}(1-e^{-bd}) $.
Subtracting the last two,
$r-p =a (e^{b(t+d)}-e^{bt}) =a e^{bt}(e^{bd}-1) $.
Dividing these two,
$\frac{r-p}{q-p} = \frac{e^{bd}-1}{1-e^{-bd}}
$. Since we know $p, q, r,$ and $d$, we can solve for $b$.
We can then get $a$ and $c$.
(This is modulo any errors on my part.)