The value of a car is modelled by the formula $$ V = 16000e^{-kt} + A $$ 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 4 - 2018 - Paper 5

At $t = 2$ years, the value is £13500:

$V = 16000e^{-2k} + A = 13500.$
Substituting $A = 1500$ gives:

$13500 = 16000e^{-2k} + 1500$
This simplifies to:

$13500 - 1500 = 16000e^{-2k}$
$12000 = 16000e^{-2k}$
Dividing both sides by 16000:

$e^{-2k} = \frac{12000}{16000} = \frac{3}{4}.$
Taking the natural logarithm on both sides:

$-2k = \ln \left( \frac{3}{4} \right)$
Thus:

$k = -\frac{1}{2} \ln \left( \frac{3}{4} \right)$
Using the properties of logarithms:

$k = \ln \left( \frac{2}{\sqrt{3}} \right).$

We start with the equation for value:

$6000 = 16000e^{-kt} + A$
Substituting $A = 1500$ gives:

$6000 = 16000e^{-kt} + 1500$
This simplifies to:

$6000 - 1500 = 16000e^{-kt}$
$4500 = 16000e^{-kt}$
Dividing both sides by 16000:

$e^{-kt} = \frac{4500}{16000} = \frac{45}{160} = \frac{9}{32}.$
Taking the natural logarithm on both sides:

$-kt = \ln \left( \frac{9}{32} \right).$
Substituting $k = \ln \left( \frac{2}{\sqrt{3}} \right)$, we have:

$-t \ln \left( \frac{2}{\sqrt{3}} \right) = \ln \left( \frac{9}{32} \right).$
Solving for $t$:

$t = -\frac{\ln \left( \frac{9}{32} \right)}{\ln \left( \frac{2}{\sqrt{3}} \right)}.$
Calculating gives:

$t \approx 8.82$
Thus, the age of the car is approximately 8.82 years.