The table shows experimental values of two variables, $x$ and $y$, which are connected by the equation $y = ae^{bx - 1}$, where $a$ and $b$ are constants.

$x$	1	2	3	4	5
$y$	1.89	2.3	2.82	3.44	4.20

1. Calculate the values of $\ln y$ for each corresponding $x$ value.

Using the given $y$ values, we compute $\ln y$ to 2 decimal places:

• $x = 1$: $\ln(1.89) \approx 0.64$
• $x = 2$: $\ln(2.3) \approx 0.83$
• $x = 3$: $\ln(2.82) \approx 1.04$
• $x = 4$: $\ln(3.44) \approx 1.24$
• $x = 5$: $\ln(4.20) \approx 1.44$

2. Set up the axes and plot the points.

Draw the horizontal axis for $x$ with a scale of 2 cm to 1 unit (from $x=0$ to $x=5$).
Draw the vertical axis for $\ln y$ with a scale of 2 cm to 0.2 units (from $\ln y = 0.6$ to $\ln y = 1.6$).
Plot the points $(1, 0.64)$, $(2, 0.83)$, $(3, 1.04)$, $(4, 1.24)$, and $(5, 1.44)$.

3. Draw the line of best fit.

Draw a straight line passing through or as close as possible to all the plotted points.

1. Transform the non-linear equation into linear form

$y = ae^{bx - 1}$
$\ln y = \ln(ae^{bx - 1})$
$\ln y = \ln a + \ln(e^{bx - 1})$
$\ln y = \ln a + (bx - 1)$
$\ln y = bx + (\ln a - 1)$

2. Calculate the gradient to find $b$

From the table, the points $(x, \ln y)$ are approximately:
$(1, 0.637), (2, 0.833), (3, 1.037), (4, 1.235), (5, 1.435)$
Gradient $m = b = \frac{1.435 - 0.637}{5 - 1} = \frac{0.798}{4} = 0.1995 \approx 0.20$

3. Calculate the vertical intercept to find $a$

Vertical intercept $c = \ln a - 1$
Using point $(1, 0.637)$ and $m = 0.20$:
$0.637 = 0.20(1) + c \implies c = 0.437$
$\ln a - 1 = 0.437$
$\ln a = 1.437$
$a = e^{1.437} \approx 4.21$
(Note: Reading the intercept directly from the graph at $x=0$ gives $c \approx 0.44$, so $\ln a - 1 = 0.44 \implies \ln a = 1.44 \implies a = e^{1.44} \approx 4.22$. Acceptable range for $a$ is typically $4.1$ to $4.3$, and $b$ is $0.19$ to $0.21$.)

1. Identify the equation of the line to be drawn on the graph

The original equation is $y = ae^{bx - 1}$.
We are asked to solve $2.46 = ae^{bx - 1}$, which corresponds to $y = 2.46$.
Since the vertical axis of the graph is $\ln y$, we take the natural logarithm:
$\ln y = \ln 2.46$
$\ln y \approx 0.900$

2. Determine the solution from the graph

Draw the horizontal line $\ln y = 0.90$ on the graph.
Find the $x$-coordinate of the point of intersection between this horizontal line and the plotted straight line.
Reading from the graph, $x \approx 2.32$.