Ah. That doesn't surprise me in the least. And yes, that is a semilog plot. Probably the easiest thing to do is to fit a straight line to the semilog plot of your data. The idea is this: assume your function (on the original plot) is of the form
$\displaystyle y=A\,e^{-k t}.$
Here it is essential that both $\displaystyle A>0$ and $\displaystyle k>0.$
Now, plotting in a semilog plot is the same as taking the logarithm of both sides of this equation. The result is the following:
$\displaystyle \ln(y)=\ln(A)+\ln(e^{-kt})=\ln(A)-kt.$
Your semilog plot has essentially plotted $\displaystyle \ln(y)$ versus $\displaystyle t.$ Therefore, you can find the slope and intercept of a straight-line-fit. The slope will be $\displaystyle -k$, and the intercept $\displaystyle \ln(A).$
Make sense?