In 2005, Andrew bought a painting. In 2008, the value of the painting is estimated to be $3878.58. The value $V(t)$, in dollars, of the painting $t$ years after being purchased is given by the function $V(t) = 2400 e^{kt}$, where k is a constant.

1. Set up the equation to find the constant $k$ using the given value in 2008.

In 2005, $t = 0$. In 2008, $t = 3$.
$V(3) = 2400e^{3k} = 3878.58$

2. Solve for $k$.

$e^{3k} = \frac{3878.58}{2400}$
$e^{3k} = 1.616075$
$3k = \ln(1.616075)$
$3k \approx 0.4800$
$k = 0.16$

3. Calculate the value of the painting after 10 years.

Substitute $t = 10$ and $k = 0.16$ into the model:
$V(10) = 2400e^{0.16(10)}$
$V(10) = 2400e^{1.6}$
$V(10) \approx 11887.27$

1. Find the value of the constant $k$ using the given information

From the context, $t=0$ in 2005, so $V(0) = 2400$.
In 2008, $t=3$ and $V(3) = 3878.58$.
$2400e^{3k} = 3878.58$
$e^{3k} = \frac{3878.58}{2400}$
$3k = \ln\left(\frac{3878.58}{2400}\right)$
$k \approx 0.16$

2. Set up the inequality for the value to at least double

The initial value is $\char36{}2400$. For the value to double, $V(t) \ge 4800$.
$2400e^{0.16t} \ge 4800$
$e^{0.16t} \ge 2$

3. Solve for $t$ and determine the calendar year

$0.16t \ge \ln 2$
$t \ge \frac{\ln 2}{0.16}$
$t \ge 4.332$
Since $t=0$ corresponds to the year 2005, $t=4.332$ corresponds to the year $2005 + 4.332 = 2009.332$.
The earliest year he can sell it is 2009.