![Approximately\:6\:years](https://img.qammunity.org/2023/formulas/mathematics/college/cu65d1zdx4ohrfeykn25feqbf0uzp1tx7s.png)
1) Since the initial population of rabbits consists of 200 animals and there is a constant growth of 30% so we can start out writing the exponential model for this problem, like this.
![P=200(1+0.3)^t](https://img.qammunity.org/2023/formulas/mathematics/college/exz4qtr6oeo47wplozvel02eoyesq8kqr0.png)
2) Let's plug P=1000 for the final population and solve for t:
![1000=200(1+0.3)^t](https://img.qammunity.org/2023/formulas/mathematics/college/p0u35v3mu8ftwuiepvgpq88625dem983rg.png)
So, let's count 2015 as our first year since we don't know when there will be 1,000 rabbits.
![\begin{gathered} 1000=200(1.3)^t \\ (1000)/(200)=(200(1.3)^t)/(200) \\ 5=(1.3)^t \\ \ln(5)=\ln(1.3)^t \\ \ln \left(5\right)=t\ln \left(1.3\right) \\ t=(\ln \left(5\right))/(\ln \left(1.3\right)) \\ t=6.13\approx6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b49i4a5j0jdlwx7tjmo8qz6kuzanw1u82d.png)
Note the property of the exponents of logarithms