A pot of melted chocolate is cooled from its initial temperature to a temperature of $T^\circ\text{C}$ in $x$ minutes and follows the equation of the form $T = A(B^x)$, where $A$ and $B$ are constants. The freezing point of the chocolate is $17^\circ\text{C}$. The table below shows the corresponding values of $T$ and $x$ recorded.

$x$	5	10	15	20	25
$T$	35.429	20.921	12.353	7.295	4.307

1. Calculate the values of $\lg T$ for each given $x$.

For $x = 5$, $\lg T = \lg 35.429 \approx 1.55$
For $x = 10$, $\lg T = \lg 20.921 \approx 1.32$
For $x = 15$, $\lg T = \lg 12.353 \approx 1.09$
For $x = 20$, $\lg T = \lg 7.295 \approx 0.86$
For $x = 25$, $\lg T = \lg 4.307 \approx 0.63$

2. Set up the axes according to the given scales.

$x$-axis: 2 cm for 5 units, starting from 0 to 25.
$\lg T$-axis: 1 cm for 0.1 units. Since values range from 0.63 to 1.55, the axis can start from 0.6 or 0 to accommodate the points.

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

Plot the points $(5, 1.55)$, $(10, 1.32)$, $(15, 1.09)$, $(20, 0.86)$, and $(25, 0.63)$ on the grid and draw a straight line passing through them.

1. Read the value of $\lg T$ from the graph at $x = 13$

From the graph of $\lg T$ against $x$, at $x = 13$, the value of $\lg T$ is approximately $1.17$.

2. Calculate the temperature $T$ and compare it with the freezing point

$\lg T = 1.17$
$T = 10^{1.17} \approx 14.8^\circ\text{C}$
Since $14.8^\circ\text{C} < 17^\circ\text{C}$, the chocolate is already frozen.

1. Linearise the given equation to relate it to the graph of $\lg T$ against $x$

$T = A(B^x)$
Taking $\lg$ on both sides:
$\lg T = \lg A + \lg(B^x)$
$\lg T = (\lg B)x + \lg A$
This is in the form $Y = mX + c$, where the gradient $m = \lg B$ and the vertical intercept $c = \lg A$.

2. Calculate the gradient of the line to find $B$

Using the points $(5, 1.549)$ and $(25, 0.634)$ from the table:
$\text{Gradient} = \frac{0.634 - 1.549}{25 - 5} = \frac{-0.915}{20} = -0.04575$
$\lg B = -0.04575$
$B = 10^{-0.04575} \approx 0.900$

3. Find the vertical intercept to calculate $A$

Substitute $(5, 1.549)$ into the linear equation:
$1.549 = -0.04575(5) + \lg A$
$1.549 = -0.22875 + \lg A$
$\lg A = 1.77775$
$A = 10^{1.77775} \approx 60.0$

1. Identify the original temperature and set up the equation for half the original temperature

The original temperature is at $x = 0$, which gives $T = A(B^0) = A$.
Half the original temperature is $\frac{A}{2}$.
$A(B^x) = \frac{A}{2}$
$B^x = 0.5$

2. Solve for $x$ using logarithms and the value of $B$ from part (c)

Taking logarithms on both sides:
$x \lg B = \lg 0.5$
Using $B \approx 0.900$ (from part c):
$x \lg 0.900 = \lg 0.5$
$x = \frac{\lg 0.5}{\lg 0.900} \approx 6.58$