A drug is administered to a patient through an injection. The concentration of drug, $x$ mg/$l$ in a patient's bloodstream after time, $t$ hours, is believed to be related by the equation
$x = x_o e^{-kt},$
where $x_o$ and $k$ are constants. Some values of $x$ and $t$ are recorded in the following table:

$t$ (hours)	2	4	6	8	10
$x$ (mg/$l$)	48.3	41.4	35.6	30.8	26.4

1. Convert the given equation into linear form to confirm the variables to be plotted

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

2. Calculate the values of $\ln x$ for the given values of $t$

Construct a table for $t$ and $\ln x$ (values rounded to 2 decimal places for plotting):

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

Using a suitable scale (e.g., $1\text{ cm}$ to $1\text{ unit}$ on the $t$-axis and $1\text{ cm}$ to $0.1\text{ units}$ on the $\ln x$-axis starting from $3.0$):

• Plot the points $(2, 3.88)$, $(4, 3.72)$, $(6, 3.57)$, $(8, 3.43)$, and $(10, 3.27)$.
• Draw a straight line of best fit passing through or as close as possible to these points.

1. Express the given equation in linear form $Y = mX + c$.

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

2. Calculate the gradient of the line to estimate $k$.

Using the points $(2, \ln 48.3)$ and $(10, \ln 26.4)$ from the table:
Gradient $m = \frac{\ln 26.4 - \ln 48.3}{10 - 2}$
$m = \frac{3.2734 - 3.8774}{8} \approx -0.0755$
Since $m = -k$, we have $k \approx 0.0755$.

3. Calculate the $Y$-intercept to estimate $x_o$.

Using the point $(2, 3.8774)$ and $m = -0.0755$:
$c = Y - mX = 3.8774 - (-0.0755)(2) = 4.0284$
Since $c = \ln x_o$, we have:
$\ln x_o = 4.0284$
$x_o = e^{4.0284} \approx 56.1$

1. Set up the equation for the concentration being halved

Let the initial concentration be $x_o$. When the concentration is halved, $x = \frac{1}{2}x_o$.
$\frac{1}{2}x_o = x_o e^{-kt}$
$e^{-kt} = 0.5$

2. Solve for $t$ using the value of $k$ from part (ii)

Taking natural logarithms on both sides:
$-kt = \ln 0.5$
$t = \frac{-\ln 0.5}{k}$
Using $k \approx 0.0755$ (from part ii):
$t = \frac{-\ln 0.5}{0.0755} \approx 9.18 \text{ hours}$

1. Take natural logarithms of both sides of the given equation to relate it to the plotted graph of $\ln x$ against $t$.

Given $x_o e^{-kt} = e^{0.04t + 3.2}$.
Since $x = x_o e^{-kt}$, the equation becomes $x = e^{0.04t + 3.2}$.
Taking natural logarithms on both sides:
$\ln x = \ln(e^{0.04t + 3.2})$
$\ln x = 0.04t + 3.2$

2. Identify the line to draw and state the method to find the solution.

Draw the straight line $Y = 0.04t + 3.2$ on the same graph (where $Y = \ln x$).
The solution is the $t$-coordinate of the point of intersection between the original line and this new line.

3. Calculate the intersection point algebraically to simulate reading from the graph.

From part (ii), the original line is $\ln x = -0.076t + 4.03$ (using $k \approx 0.076$ and $\ln x_o \approx 4.03$).
Equating the two lines:
$-0.076t + 4.03 = 0.04t + 3.2$
$0.116t = 0.83$
$t \approx 7.16$ (Accept values read from the graph, typically around $7.1$ to $7.2$)