The table shows experimental values of $x$ and $y$.

$x$	2	3	4	5	6
$y$	11.8	16.2	23.5	35.5	55.2

It is known that $x$ and $y$ are related by the equation $y = Ae^{kx} + 5$, where $A$ and $k$ are constants.

1. Rearrange the equation and take natural logarithms to convert it to a linear form $Y = mX + c$.

$y = Ae^{kx} + 5$
$y - 5 = Ae^{kx}$
$\ln(y - 5) = \ln(Ae^{kx})$
$\ln(y - 5) = kx + \ln A$

2. Identify the variables to plot and relate the constants to the features of the straight line.

Comparing $\ln(y - 5) = kx + \ln A$ with $Y = mX + c$:

• Plot $Y = \ln(y - 5)$ against $X = x$.
• The gradient $m$ corresponds to $k$.
• The vertical intercept $c$ corresponds to $\ln A$.

1. Calculate the values of $\ln(y - 5)$ for the given data points

From (i), the linear form is $\ln(y - 5) = kx + \ln A$.
We compute $Y = \ln(y - 5)$ for each $x$:

• $x = 2$: $\ln(11.8 - 5) = \ln 6.8 \approx 1.92$
• $x = 3$: $\ln(16.2 - 5) = \ln 11.2 \approx 2.42$
• $x = 4$: $\ln(23.5 - 5) = \ln 18.5 \approx 2.92$
• $x = 5$: $\ln(35.5 - 5) = \ln 30.5 \approx 3.42$
• $x = 6$: $\ln(55.2 - 5) = \ln 50.2 \approx 3.92$

2. Plot the graph of $\ln(y - 5)$ against $x$ and draw the line of best fit

Plot the points $(2, 1.92)$, $(3, 2.42)$, $(4, 2.92)$, $(5, 3.42)$, and $(6, 3.92)$ on a Cartesian plane and draw a straight line passing through them.

3. Calculate the gradient to find $k$

Using the points $(2, 1.917)$ and $(6, 3.916)$ from the line:
$k = \text{gradient} = \frac{3.916 - 1.917}{6 - 2} = \frac{1.999}{4} \approx 0.50$

4. Find the vertical intercept to determine $A$

The vertical intercept is $\ln A$.
Using $Y = mX + c$ with point $(2, 1.917)$:
$1.917 = 0.50(2) + \ln A$
$\ln A = 1.917 - 1.00 = 0.917$
$A = e^{0.917} \approx 2.50$

1. Calculate the value on the vertical axis corresponding to $y = 30$

The vertical axis represents $\ln(y - 5)$.
When $y = 30$,
$\ln(30 - 5) = \ln 25 \approx 3.22$

2. Read the corresponding $x$-value from the graph

From the graph of $\ln(y - 5)$ against $x$, reading the $x$-coordinate when the vertical coordinate is $3.22$ gives:
$x \approx 4.6$

1. Relate the given equation to the variables of the linear law graph

The graph plotted in (ii) has $Y = \ln(y - 5)$ against $X = x$.
Given $Ae^{kx} = 30(2^{-x})$, and knowing from the original equation that $y - 5 = Ae^{kx}$, we have:
$y - 5 = 30(2^{-x})$

2. Take natural logarithms to find the equation of the straight line to draw

$\ln(y - 5) = \ln(30(2^{-x}))$
$\ln(y - 5) = \ln 30 + \ln(2^{-x})$
$\ln(y - 5) = -x \ln 2 + \ln 30$
So the suitable straight line to draw is $Y = -(\ln 2)x + \ln 30$, which is approximately $Y = -0.693x + 3.40$.

3. Find the solution by determining the intersection point

By drawing the line $Y = -0.693x + 3.40$ on the same axes, the $x$-coordinate of the point of intersection with the original line gives the solution.
From the graph, the intersection occurs at $x \approx 2.08$.