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

You can rewrite the differential equation as
[tex]50\cdot\dfrac{dA}{A}=dt\\\\50\ln{(A)}+C=t\qquad\mbox{integrating}\\\\t=50\ln{\left(\dfrac{A}{4000}\right)}\qquad\mbox{applying the boundary condition}[/tex]
This last expression looks like the one you describe. If you solve for A instead of t, you get
[tex]A=4000\cdot e^{0.02t}[/tex]
This is the same as your other answer.
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)