The table shows experimental values of two variables, x and y, which are connected by an equation of the form $y^{n} x^{k} = 10$.

1. Calculate the logarithms of the given $x$-values.

Using common logarithms,

2. Calculate the logarithms of the corresponding $y$-values.

Using common logarithms,

So the points to plot are
$(0.079,0.343),\ (0.204,0.255),\ (0.301,0.185),\ (0.380,0.130).$

3. Draw the graph with the required scale and line of best fit.

On graph paper:

• horizontal axis: lg x,
• vertical axis: lg y,
• scale on each axis: 4 cm represents 0.1 unit,
• plot the four points accurately,
• draw a straight line of best fit through the points.

1. Convert the given equation into a linear form involving $\lg y$ and $\lg x$.

Given
$y^n x^k=10,$
take $\lg$ on both sides:
$\lg(y^n x^k)=\lg 10$
$n\lg y+k\lg x=1.$

2. Write the equation in the form of the plotted straight line, $\lg y=m\lg x+c$.

Rearrange
$n\lg y+k\lg x=1$
so that $\lg y$ is the subject:
$n\lg y=1-k\lg x$
$\lg y=-\frac{k}{n}\lg x+\frac{1}{n}.$
Comparing with
$Y=mX+c,$
where $Y=\lg y$ and $X=\lg x$,
$m=-\frac{k}{n}, \qquad c=\frac{1}{n}.$

3. Estimate the gradient and vertical intercept from the graph.

From the straight line of best fit on the graph, the gradient is approximately
$m\approx -0.71$
and the vertical intercept is approximately
$c\approx 0.40.$

4. Use the gradient and intercept to calculate $n$ and $k$.

Since
$c=\frac{1}{n},$
$0.40\approx \frac{1}{n}$
$n\approx \frac{1}{0.40}=2.5.$
Also,
$m=-\frac{k}{n},$
so
$-0.71\approx -\frac{k}{2.5}$
$k\approx 0.71(2.5)=1.775\approx 1.8.$

1. Recall the gradient of the original graph from part (ii).

From part (ii), the graph of $\lg y$ against $\lg x$ has gradient
$m=-\frac{k}{n}.$
Using $k\approx1.8$ and $n\approx2.5$,
$m\approx -\frac{1.8}{2.5}=-0.72.$

2. Find the reciprocal gradient when the axes are interchanged.

When $\lg x$ is plotted against $\lg y$, the gradient is the reciprocal of the original gradient:
$\text{new gradient}=\frac{1}{-0.72}\approx -1.39.$