Take the natural log of both sides to get $\displaystyle \ln d= \ln\left(ae^{-kt}\right)\implies \ln d= \ln a-kt$. Thus, if we take $\displaystyle y=\ln d$, $\displaystyle m=-k$ and $\displaystyle c=\ln a$, we have the linear equation $\displaystyle y=mt+c$
Can you continue from here?
thankyou for your input. If I wanted to make a graph with the data from the table, I presume this is possible, I don't want you to create a graph for me or anything but could you please give me a example for t = 2 about where i would hit the y axis. thanks