Jane bought an electric car in January 2025. After purchase, the value of the car, $V, diminishes over time and can be modelled by $V = 38000e^{-kt}$, where $k$ is a positive constant and $t$ is time measured in years. The value of the car is expected to be $29000 after 3 years of driving. Jane intends to sell her car when its value drops to half of its original value. Showing clear mathematical calculations, find the year that Jane is likely to sell her electric car.

1. Substitute the given values to find the constant $k$.

When $t = 3$, $V = 29000$.
$29000 = 38000e^{-3k}$
$e^{-3k} = \frac{29000}{38000} = \frac{29}{38}$
$-3k = \ln\left(\frac{29}{38}\right)$
$k = -\frac{1}{3}\ln\left(\frac{29}{38}\right) \approx 0.089981$

2. Set up the equation for when the car's value drops to half.

Half of the original value is $\frac{1}{2} \times 38000 = 19000$.
$19000 = 38000e^{-kt}$
$e^{-kt} = 0.5$

3. Solve for $t$.

$-kt = \ln(0.5)$
$t = \frac{\ln(0.5)}{-k}$
$t = \frac{\ln(0.5)}{\frac{1}{3}\ln\left(\frac{29}{38}\right)} \approx 7.703 \text{ years}$

4. Determine the calendar year.

The car was bought in January 2025 ($t=0$).
$7.703$ years after January 2025 falls in the year $2025 + 7 = 2032$.