Mr Lu invested a sum of money in a bank in 1980. The value of the investment, $V$ in thousands dollars can be modelled by an equation of the form $V = V_0 k^t$, where $V_0$ and $k$ are constants and $t$ is the number of years since Mr Lu invested the sum of money.
The table below gives some values of $V$ and $t$.

$t$ (years)	5	10	15	20	25
$V$ (in thousands $)	192	370	714	1374	2642

1. Calculate the values of $\lg V$ for each corresponding value of $t$ to 2 decimal places.

When $t = 5$, $\lg 192 \approx 2.28$
When $t = 10$, $\lg 370 \approx 2.57$
When $t = 15$, $\lg 714 \approx 2.85$
When $t = 20$, $\lg 1374 \approx 3.14$
When $t = 25$, $\lg 2642 \approx 3.42$

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

Plot the points $(5, 2.28)$, $(10, 2.57)$, $(15, 2.85)$, $(20, 3.14)$, and $(25, 3.42)$ on a grid with the horizontal axis representing $t$ and the vertical axis representing $\lg V$ (starting at 1.8).
Draw a straight line passing through or as close as possible to all the plotted points.

1. Relate the initial value to the $y$-intercept of the linear graph

Taking logarithms on both sides of $V = V_0 k^t$ gives:
$\lg V = (\lg k)t + \lg V_0$
The initial value occurs at $t = 0$, which corresponds to the vertical intercept of the $\lg V$ against $t$ graph.

2. Read the intercept from the graph and calculate the initial value

From the plotted graph, the $\lg V$-intercept is approximately $2.00$.
$\lg V_0 = 2.00$
$V_0 = 10^{2.00} = 100$
Since $V$ is in thousands of dollars, the initial value is $100 \times 1000 = 100\,000$.

1. Express the given equation in linear form to identify the gradient

Given $V = V_0 k^t$, taking logarithms on both sides gives:
$\lg V = \lg (V_0 k^t)$
$\lg V = \lg V_0 + t \lg k$
Comparing this to $Y = mX + c$, where $Y = \lg V$ and $X = t$, the gradient of the line is $\lg k$.

2. Calculate the gradient of the line using points from the data

Using the points $(10, \lg 370)$ and $(20, \lg 1374)$ from the table:
$\text{Gradient} = \frac{\lg 1374 - \lg 370}{20 - 10}$
$\text{Gradient} = \frac{3.138 - 2.568}{10} = 0.057$

3. Solve for $k$

$\lg k = 0.057$
$k = 10^{0.057} \approx 1.14$

1. Calculate the target value of $V$ after a 50% increase

From part (ii), the initial value $V_0 \approx 100$ (since $\lg V_0 \approx 2.00$).
A 50% increase means the new value is $V = 1.5 \times 100 = 150$.

2. Find the corresponding $\lg V$ and read $t$ from the graph

$\lg V = \lg 150 \approx 2.18$
Reading from the graph at $\lg V = 2.18$, we find $t \approx 3.1$.