Answer:
In 2009
Step-by-step explanation:
Since, the formula of population after t years,
![N=N_0 e^{rt}](https://tex.z-dn.net/?f=N%3DN_0%20e%5E%7Brt%7D)
Where,
r = rate of growing per year,
Here, r = 12% = 0.12,
So, the population formula would be,
![N=N_0 e^{0.12t}](https://tex.z-dn.net/?f=N%3DN_0%20e%5E%7B0.12t%7D)
If the population is estimated since 2003,
i.e. for 2003, t = 0,
We have N = 103, 800 for 2003,
![\implies 103800 = N_0 e^{0.12\times 0}=N_0 e^0 = N_0](https://tex.z-dn.net/?f=%5Cimplies%20103800%20%3D%20N_0%20e%5E%7B0.12%5Ctimes%200%7D%3DN_0%20e%5E0%20%3D%20N_0)
Thus, the function that shows the population after t years,
![N = 103800 e^{0.12t}](https://tex.z-dn.net/?f=N%20%3D%20103800%20e%5E%7B0.12t%7D)
If N = 209,000,
![209000 = 103800 e^{0.12t}](https://tex.z-dn.net/?f=209000%20%3D%20103800%20e%5E%7B0.12t%7D)
![\frac{209000}{103800} = e^{0.12t}](https://tex.z-dn.net/?f=%5Cfrac%7B209000%7D%7B103800%7D%20%3D%20e%5E%7B0.12t%7D)
![2.01349 = e^{0.12t}](https://tex.z-dn.net/?f=2.01349%20%3D%20e%5E%7B0.12t%7D)
Taking ln on both sides,
![\ln ( 2.01348 ) = 0.12t](https://tex.z-dn.net/?f=%5Cln%20%28%202.01348%20%29%20%3D%200.12t)
![\implies t = \frac{\ln(2.01348)}{0.12}=5.8\approx 6](https://tex.z-dn.net/?f=%5Cimplies%20t%20%3D%20%5Cfrac%7B%5Cln%282.01348%29%7D%7B0.12%7D%3D5.8%5Capprox%206)
∵ 2003 + 6 = 2009
Hence, in 2009, the population should reach 209,000.