Answer
Using technology, the best exponential model is $V(t)=1.4638(0.970244)^{t}$,
which provides a very good fit, but not better than the linear model of part (a).
Work Step by Step
Using the desmos project created in part (a), we change the model to $y\sim Ab^{x_{1}}$ and obtain
$A=1.4638$
$b=0.970244$
Changing the third column to calculate $Ab^{x_{1}}$, we find that the model is a good fit, but not better than the linear model in part (a).
