The value of a car is modelled by the formula $$ V = 16000e^{-kt} + A, \quad t > 0, \quad t \in \mathbb{R} $$ where $V$ is the value of the car in pounds, $t$ is the age of the car in years, and $k$ and $A$ are positive constants - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 5

To derive the value of $k$, we use the information that the car's value is £13500 when $t = 2$:

$13500 = 16000e^{-2k} + 1500$

Rearranging the equation yields:

$13500 - 1500 = 16000e^{-2k}$ $12000 = 16000e^{-2k}$

Dividing both sides by 16000 gives:

$e^{-2k} = \frac{12000}{16000} = \frac{3}{4}$

Taking the natural logarithm of both sides:

$-2k = \ln(\frac{3}{4})$

Thus:

$k = -\frac{1}{2}\ln(\frac{3}{4}) = \ln(\frac{2}{\sqrt{3}})$

To find the age when the value is £6000, we set:

$6000 = 16000e^{-kt} + 1500$

Rearranging gives:

$6000 - 1500 = 16000e^{-kt}$ $4500 = 16000e^{-kt}$

Thus:

$e^{-kt} = \frac{4500}{16000} = \frac{9}{32}$

Taking the natural logarithm of both sides:

$-kt = \ln(\frac{9}{32})$

Substituting $k = \ln(\frac{2}{\sqrt{3}})$ gives:

$-t \ln(\frac{2}{\sqrt{3}}) = \ln(\frac{9}{32})$

Thus:

$t = -\frac{\ln(\frac{9}{32})}{\ln(\frac{2}{\sqrt{3}})}$

Calculating this numerically, we find:

$t \approx 8.82$

So the age of the car when its value is £6000 is approximately 8.82 years.