The population $P$, in millions, of a country was recorded on January of the various years and the results are shown in the table below.

Year	2005	2010	2015	2020
$P$	12.95	14.67	17.52	22.11

Given that $P = 10 + ab^t$, where $t$ is the time measured in years from 2000 and $a$ and $b$ are constants.
$\lg(P - 10) = \lg a + t \lg b$

1. Calculate the values of $t$ and $Y = \lg(P - 10)$ for each data point.

The year 2000 corresponds to $t = 0$.
For 2005, $t = 5$. $P = 12.95 \implies \lg(12.95 - 10) = \lg(2.95) \approx 0.470$
For 2010, $t = 10$. $P = 14.67 \implies \lg(14.67 - 10) = \lg(4.67) \approx 0.669$
For 2015, $t = 15$. $P = 17.52 \implies \lg(17.52 - 10) = \lg(7.52) \approx 0.876$
For 2020, $t = 20$. $P = 22.11 \implies \lg(22.11 - 10) = \lg(12.11) \approx 1.083$

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

Plot the points $(5, 0.470)$, $(10, 0.669)$, $(15, 0.876)$, and $(20, 1.083)$ on a Cartesian plane with $t$ on the horizontal axis and $\lg(P - 10)$ on the vertical axis. Draw a straight line passing through these points.

1. Identify the gradient and vertical intercept from the linear equation

Comparing $\lg(P - 10) = (\lg b)t + \lg a$ with $Y = mX + c$, the gradient is $\lg b$ and the vertical intercept is $\lg a$.

2. Calculate the gradient to find $b$

Using points from the line of best fit (e.g., $(5, 0.47)$ and $(20, 1.08)$):
Gradient $m = \frac{1.08 - 0.47}{20 - 5} = \frac{0.61}{15} \approx 0.0407$
$\lg b = 0.0407 \implies b = 10^{0.0407} \approx 1.10$

3. Find the vertical intercept to determine $a$

From the graph, the vertical intercept $c \approx 0.265$.
$\lg a = 0.265 \implies a = 10^{0.265} \approx 1.84$