Designer bag X was first released in 1950. The price, $V$, of the bag is related to $t$, the number of years since 1950, by the formula $V = ae^{kt},$ where $a$ and $k$ are constants. The table below gives the value of bag X in 1984, 2002, 2008, and 2012.

Year	1984	2002	2008	2012
t (years)	34	52	58	62
V ($)	1150	2850	4000	4900

1. Calculate the values of $\ln V$ for each given $t$

For $t = 34$, $\ln V = \ln(1150) \approx 7.05$
For $t = 52$, $\ln V = \ln(2850) \approx 7.96$
For $t = 58$, $\ln V = \ln(4000) \approx 8.29$
For $t = 62$, $\ln V = \ln(4900) \approx 8.50$

2. Plot the points and draw the line of best fit

Plot the points $(34, 7.05)$, $(52, 7.96)$, $(58, 8.29)$, and $(62, 8.50)$ on the graph paper using the specified scales and draw a straight line passing through or close to these points.

1. Express the given equation in linear form

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

2. Calculate the gradient to find $k$

Using the points $(34, 7.05)$ and $(62, 8.50)$ from the graph:
$k = \text{gradient} = \frac{8.50 - 7.05}{62 - 34}$
$k = \frac{1.45}{28} \approx 0.0518$

3. Find the $Y$-intercept to calculate $a$

From the graph, the $Y$-intercept (where $t = 0$) is approximately $5.29$.
$\ln a = 5.29$
$a = e^{5.29} \approx 198$

1. Substitute $V = 7000$ and the estimated values of $a$ and $k$ from part (a)(ii) into the equation

Using the estimated values from the graph (e.g., $a \approx 196$ and $k \approx 0.052$):
$V = ae^{kt}$
$7000 = 196e^{0.052t}$

2. Solve for $t$ and determine the corresponding year

$e^{0.052t} = \frac{7000}{196}$
$0.052t = \ln\left(\frac{7000}{196}\right)$
$0.052t \approx 3.5755$
$t \approx 68.76$

The year is $1950 + 68 = 2018$.

1. Rearrange the equation into a linear form $Y = mX + c$

$y = \frac{a}{x-b}$
$y(x-b) = a$
$xy - by = a$
$xy = by + a$

2. Identify the variables to plot and explain how to find the constants

Comparing $xy = by + a$ with $Y = mX + c$:

• Plot $Y = xy$ against $X = y$.
• The gradient $m$ corresponds to $b$.
• The vertical intercept $c$ corresponds to $a$.