The table shows experimental values of two variables, $x$ and $y$.

$x$	0.5	1.0	1.5	2.0
$y$	15.9	19.1	23.4	30.2

It is known that $x$ and $y$ are related by the equation $y = 10 + Ab^x$.

1. Calculate the values of $\lg(y-10)$ for each given $x$.

For $x = 0.5$, $y = 15.9 \implies \lg(15.9 - 10) = \lg(5.9) \approx 0.771$
For $x = 1.0$, $y = 19.1 \implies \lg(19.1 - 10) = \lg(9.1) \approx 0.959$
For $x = 1.5$, $y = 23.4 \implies \lg(23.4 - 10) = \lg(13.4) \approx 1.127$
For $x = 2.0$, $y = 30.2 \implies \lg(30.2 - 10) = \lg(20.2) \approx 1.305$

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

Plot the points $(0.5, 0.77)$, $(1.0, 0.96)$, $(1.5, 1.13)$, and $(2.0, 1.31)$ on the grid and draw a straight line passing through or close to all points.

1. Express the given equation in linear form

$y = 10 + Ab^x$
$y - 10 = Ab^x$
$\lg(y - 10) = \lg(Ab^x)$
$\lg(y - 10) = \lg A + x \lg b$

2. Identify the gradient and Y-intercept from the linear form

Comparing $\lg(y - 10) = (\lg b)x + \lg A$ with $Y = mX + c$:
Gradient $m = \lg b$
$Y$-intercept $c = \lg A$

3. Calculate the gradient and Y-intercept from the data points

Using points $(0.5, \lg 5.9)$ and $(2.0, \lg 20.2)$:
$X_1 = 0.5, Y_1 \approx 0.771$
$X_2 = 2.0, Y_2 \approx 1.305$
Gradient $m = \frac{1.305 - 0.771}{2.0 - 0.5} = \frac{0.534}{1.5} = 0.356$
$Y$-intercept $c = Y_1 - mX_1 = 0.771 - 0.356(0.5) = 0.593$

4. Solve for A and b

$\lg b = 0.356 \implies b = 10^{0.356} \approx 2.27$
$\lg A = 0.593 \implies A = 10^{0.593} \approx 3.92$
(Note: Values depend on the line of best fit drawn in part (i). Using the first and last points directly gives $A \approx 3.9, b \approx 2.3$. A typical line of best fit might yield $A \approx 4.5, b \approx 2.1$.)

1. Relate the given equation to the variables plotted on the graph

The original equation is $y = 10 + Ab^x$, which gives $y - 10 = Ab^x$.
The equation to solve is $Ab^x = 10^{2x}$.
Substituting $y - 10$ for $Ab^x$, we get $y - 10 = 10^{2x}$.

2. Transform the equation to match the vertical axis of the graph

Take the common logarithm ($\lg$) of both sides:
$\lg(y - 10) = \lg(10^{2x})$
$\lg(y - 10) = 2x$

3. Identify the line to draw and state the solution

Draw the straight line $Y = 2x$ (where $Y = \lg(y - 10)$ and $X = x$) on the same graph.
The $x$-coordinate of the point of intersection between this line and the original graph gives the solution.
From the graph, $x \approx 0.36$.