An auction house claimed that it is worthwhile to invest in their art pieces as the value of one of their art pieces has been increasing exponentially since it was produced. The value, $\char36{}V$, of this art piece is related to $t$, the number of years since it was produced at the start of the year 1995. The variables $V$ and $t$ can be modelled by the equation $V = 10\;000 + ae^{kt}$, where $a$ and $k$ are constants. The table below gives values of $V$ and $t$ at the start of some of the years 2000 to 2015.

1. Transform the given non-linear equation into a linear form $Y = mX + c$.

$V = 10\;000 + ae^{kt}$
$V - 10\;000 = ae^{kt}$
Taking the natural logarithm on both sides:
$\ln(V - 10\;000) = \ln(ae^{kt})$
$\ln(V - 10\;000) = kt + \ln a$
This is in the linear form $Y = mX + c$, where $Y = \ln(V - 10\;000)$, $X = t$, gradient $m = k$, and $Y$-intercept $c = \ln a$.

2. Calculate the values of $Y = \ln(V - 10\;000)$ for each given $t$.

For $t = 5$, $V = 16\;000$:
$\ln(16\;000 - 10\;000) = \ln(6\;000) \approx 8.70$

For $t = 10$, $V = 20\;260$:
$\ln(20\;260 - 10\;000) = \ln(10\;260) \approx 9.24$

For $t = 15$, $V = 27\;545$:
$\ln(27\;545 - 10\;000) = \ln(17\;545) \approx 9.77$

For $t = 20$, $V = 40\;000$:
$\ln(40\;000 - 10\;000) = \ln(30\;000) \approx 10.31$

3. Plot the points and draw the line of best fit to verify the model.

Plot the points $(5, 8.70)$, $(10, 9.24)$, $(15, 9.77)$, and $(20, 10.31)$ on a Cartesian plane with $t$ on the horizontal axis and $\ln(V - 10\;000)$ on the vertical axis.
Draw a straight line passing through these points.
Since the plotted points lie on a straight line, the linear relationship is verified, which shows that the original model $V = 10\;000 + ae^{kt}$ is valid for the years 2000 to 2015.

1. Express the given equation in linear form

$V = 10000 + ae^{kt}$
$V - 10000 = ae^{kt}$
$\ln(V - 10000) = \ln(ae^{kt})$
$\ln(V - 10000) = kt + \ln a$

2. Calculate the gradient to estimate $k$

From the table, the points $(t, \ln(V - 10000))$ are approximately:
$(5, 8.700)$, $(10, 9.236)$, $(15, 9.773)$, $(20, 10.309)$
Gradient $k = \frac{10.309 - 8.700}{20 - 5} = \frac{1.609}{15} \approx 0.107$
(Note: Using a line of best fit from part (i), a typical value for $k$ is around $0.107$. Let's re-evaluate using the first and last points exactly: $\frac{\ln(30000) - \ln(6000)}{15} = \frac{\ln 5}{15} \approx 0.1073$)
$k \approx 0.107$

3. Calculate the vertical intercept to estimate $a$

The vertical intercept is $\ln a$.
Using point $(20, 10.309)$ and $k = 0.1073$:
$\ln a = 10.309 - 0.1073(20) = 8.163$
$a = e^{8.163} \approx 3509$
(Alternatively, using exact values: $\ln a = \ln(6000) - 5(\frac{\ln 5}{15}) = \ln(6000) - \ln(5^{1/3}) \approx 8.163$)
$a \approx 3510$ (to 3 s.f.)

1. Determine the values of a and k from the previous part

From part (ii), the linearised equation is $\ln(V - 10000) = kt + \ln a$. Using the points $(5, \ln 6000)$ and $(20, \ln 30000)$:
$k = \frac{\ln 30000 - \ln 6000}{20 - 5} = \frac{\ln 5}{15} \approx 0.1073$
$\ln a = \ln 30000 - 20(\frac{\ln 5}{15}) = \ln 30000 - \frac{4}{3}\ln 5 \approx 8.155$
$a = e^{8.155} \approx 3482$
So $V = 10000 + 3482e^{0.1073t}$.

2. Calculate the initial value of the art piece (at t = 0)

At $t = 0$ (year 1995),
$V_{1995} = 10000 + 3482e^{0} = 13482$

3. Calculate the value of the art piece in the year 2065

In the year 2065, $t = 2065 - 1995 = 70$.
$V_{2065} = 10000 + 3482e^{0.1073(70)} = 10000 + 3482e^{7.511} \approx 10000 + 3482(1828) \approx 6375000$

4. Compare the increase to the claim and conclude

Ratio $= \frac{V_{2065}}{V_{1995}} = \frac{6375000}{13482} \approx 473$
Since $473 \neq 500$, the claim is incorrect. (Note: Using exact values $a = 6000 / 5^{1/3} \approx 3508.8$ and $k = \frac{1}{15}\ln 5$, $V_{1995} = 13508.8$ and $V_{2065} = 10000 + 3508.8 e^{\frac{70}{15}\ln 5} = 10000 + 3508.8(5^{14/3}) \approx 6380000$. Ratio $\approx 472$).