8.) Find the amount A in an account after t years given the following conditions.

This is a separable differential equation so can be written as
[tex]\dfrac{dA}{A}=0.02\,dt[/tex]
Integrating, then taking antilogs, you have
[tex]\displaystyle\int{\frac{dA}{A}}=0.02\int{dt}\\\\\ln(A)+C=0.02t\\\\A=C\cdot e^{0.02t}[/tex]
The exponential factor is 1 at t=0, so the multiplying constant is 4000.
[tex]A=4000e^{0.02t}[/tex]
dA/dt = 0.02A
dA/A = 0.02dt
integrating left n right
lnA = 0.02t + C where C is a constant
at t=0, A=4000
C=ln4000
lnA = 0.02t + ln4000
0.02t = lnA-ln4000
t = ln(A/4000) / 0.02
t = 40ln(A/4000)