一家拍卖行声称，投资他们的艺术品是值得的，因为其中一件艺术品自创作以来，其价值一直呈指数级增长。 这件艺术品的价值 $\char36{}V$ 与 $t$ 相关，其中 $t$ 是自1995年初创作以来的年数。 变量 $V$ 和 $t$ 可以用方程 $V = 10\;000 + ae^{kt}$ 来建模，其中 $a$ 和 $k$ 是常数。下表给出了2000年至2015年期间某些年份年初的 $V$ 和 $t$ 的值。

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 $\approx 472$).