Let [tex]P(t)[/tex] denote the population at time [tex]t[/tex] in millions.
The population's growth rate is modeled by the differential equation
[tex]P'(t)=kP(t)[/tex]
This equation is separable, and we can solve easily:
[tex]\dfrac{\mathrm dP(t)}{\mathrm dt}=kP(t)\iff\dfrac{\mathrm dP(t)}{P(t)}=k\,\mathrm dt[/tex]
[tex]\implies\displaystyle\int\frac{\mathrm dP}P=k\int\mathrm dt[/tex]
[tex]\implies \ln|P(t)|=kt+C[/tex]
[tex]\implies P(t)=e^{kt+C}=e^{kt}e^C=Ce^{kt}[/tex]
Let's assume 1950 corresponds to year [tex]t=0[/tex], so that 2000 would correspond to [tex]t=50[/tex]. Then [tex]P(0)=2560[/tex], which means we have
[tex]2560=Ce^{0k}=C[/tex]
and [tex]P(50)=6080[/tex], which means
[tex]6080=2560e^{50k}\implies \dfrac{19}8=e^{50k}[/tex]
[tex]\implies 50k=\ln\dfrac{19}8[/tex]
[tex]\implies k=\dfrac1{50}\ln\dfrac{19}8\approx0.0173[/tex]
So the population is modeled by
[tex]P(t)=2560e^{0.0173t}[/tex]
(assuming [tex]t=0[/tex] refers to the year 1950)
