Answer:
The value of the account in the year 2009 will be $682.
Step-by-step explanation:
The acount's balance, in t years after 1999, can be modeled by the following equation.
![A(t) = Pe^{rt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20Pe%5E%7Brt%7D)
In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.
$330 in an account in the year 1999
This means that ![P = 330](https://tex.z-dn.net/?f=P%20%3D%20330)
$590 in the year 2007
2007 is 8 years after 1999, so P(8) = 590.
We use this to find r.
![A(t) = Pe^{rt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20Pe%5E%7Brt%7D)
![590 = 330e^{8r}](https://tex.z-dn.net/?f=590%20%3D%20330e%5E%7B8r%7D)
![e^{8r} = \frac{590}{330}](https://tex.z-dn.net/?f=e%5E%7B8r%7D%20%3D%20%5Cfrac%7B590%7D%7B330%7D)
![e^{8r} = 1.79](https://tex.z-dn.net/?f=e%5E%7B8r%7D%20%3D%201.79)
Applying ln to both sides:
![\ln{e^{8r}} = \ln{1.79}](https://tex.z-dn.net/?f=%5Cln%7Be%5E%7B8r%7D%7D%20%3D%20%5Cln%7B1.79%7D)
![8r = \ln{1.79}](https://tex.z-dn.net/?f=8r%20%3D%20%5Cln%7B1.79%7D)
![r = \frac{\ln{1.79}}{8}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5Cln%7B1.79%7D%7D%7B8%7D)
![r = 0.0726](https://tex.z-dn.net/?f=r%20%3D%200.0726)
Determine the value of the account, to the nearest dollar, in the year 2009.
2009 is 10 years after 1999, so this is A(10).
![A(t) = 330e^{0.0726t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20330e%5E%7B0.0726t%7D)
![A(10) = 330e^{0.0726*10} = 682](https://tex.z-dn.net/?f=A%2810%29%20%3D%20330e%5E%7B0.0726%2A10%7D%20%3D%20682)
The value of the account in the year 2009 will be $682.