Since you are given the rate of growth, the population must be computed from the integral ...
[tex]\displaystyle p(t)=2000+\int_{0}^{t}{\left(1000\cdot 5^{t}\right)}\,dt=2000+\frac{1000}{\ln{(5)}}\cdot \left(5^{t}-1\right)\\\\=2000+\frac{1000}{\ln{(5)}}\cdot 4\approx 4485\qquad\mbox{after 1 hour}[/tex]
