The equation will be of the form:
\(A(t)= A_{0} e^{rt}\)
where A is the amount after t hours, and r is the decay constant.
To find the value of r, we plug the given values into the equation, giving:
\(40=320e^{3r}\)
Rearranging and taking natural logs of both sides, we get:
\(ln\ 0.125=3r\)
\(r=\frac{ln\ 0.125}{3}=-0.693\)
The required model is:
\(A=A _{0} e^{-0.693t}\)
