Measured values of $x$ and $y$ are given in the following table.

$x$	1.2	1.6	2.0	2.6	3.5	4.5
$y$	11.1	6.29	3.98	2.36	1.55	0.79

It is known that $x$ and $y$ are to obey the formula $x^a y = c$, where $a$ and $c$ are constants to be found.

1. Take logarithms on both sides of the given equation and expand

$x^a y = c$
$\lg(x^a y) = \lg c$
$a \lg x + \lg y = \lg c$

2. Rearrange into the linear form $Y = mX + C$ and state the variables to plot

$\lg y = -a \lg x + \lg c$
This is in the form $Y = mX + C$, where $Y = \lg y$, $X = \lg x$, gradient $m = -a$, and $Y$-intercept $C = \lg c$.
Therefore, a straight line graph can be drawn by plotting $\lg y$ against $\lg x$.

1. Calculate the values of $\lg x$ and $\lg y$ (or $\ln x$ and $\ln y$) to 2 decimal places for plotting.

Taking logarithms to base 10:

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

Plot the points $(\lg x, \lg y)$ on the provided grid. Draw a straight line of best fit through the points.

3. Show that 5 of the readings are a close fit.

Observe that 5 of the plotted points lie on or very close to the drawn straight line, while one point (at $x = 3.5$) deviates significantly from the line.

1. Identify the anomalous point from the plotted data

From the graph plotted in part (ii), the point corresponding to $x = 3.5$ and $y = 1.55$ (which gives $\lg x \approx 0.544$ and $\lg y \approx 0.190$) lies significantly off the line of best fit.

2. Estimate the correct value of y from the line of best fit

At $x = 3.5$, $\lg x = \lg 3.5 \approx 0.544$.
Reading from the line of best fit at $\lg x = 0.544$, we find $\lg y \approx 0.05$.
Therefore, $y = 10^{0.05} \approx 1.12$.

1. Recall the linear form and identify gradient/intercept meaning

From part (i), taking $\lg$ of $x^a y = c$ gives
$\lg y = -a\lg x + \lg c.$
Comparing with $Y = mX + k$ (where $Y=\lg y$, $X=\lg x$): the gradient is $m = -a$ and the vertical intercept is $k = \lg c$.

2. Find $a$ from the gradient of the line

Read the gradient from the drawn straight line, e.g. using the points $(\lg 1.2,\ \lg 11.1)=(0.079,\ 1.045)$ and $(\lg 4.5,\ \lg 0.79)=(0.653,\ -0.102)$:
$\text{gradient} = \frac{-0.102 - 1.045}{0.653 - 0.079} \approx \frac{-1.147}{0.574} \approx -2.$
Since gradient $= -a$, we get $a = 2$.

3. Find $c$ from the intercept

Using $\lg y = -2\lg x + \lg c$ with a good point, e.g. $x=2.0,\ y=3.98$:
$\lg c = \lg y + 2\lg x = \lg 3.98 + 2\lg 2.0 \approx 0.600 + 0.602 = 1.202.$
$c = 10^{1.202} \approx 16.$
(Equivalently $x^2 y = c$ gives $2.0^2 \times 3.98 \approx 16$.)