The table shows the experimental values of two variables x and y.

1. Take logarithms on both sides of the equation

$y = ab^x$
$\lg y = \lg(ab^x)$

2. Apply laws of logarithms to separate the terms

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

1. Calculate the values of $\lg y$ for each given $x$

Using the linear form $\lg y = x \lg b + \lg a$, we compute $\lg y$ for each data point:

2. Choose suitable scales and plot the points

Set up the horizontal axis for $x$ (from 0 to 3.0 or 3.5) and the vertical axis for $\lg y$ (from 0 to 0.9 or 1.0). Plot the points $(0.5, 0.79)$, $(1.0, 0.67)$, $(1.5, 0.57)$, $(2.0, 0.46)$, $(2.5, 0.34)$, and $(3.0, 0.23)$ on the grid.

3. Draw the line of best fit

Draw a single straight line that best fits the plotted points, ensuring it passes as close as possible to all points.

1. Relate the y-intercept of the linear graph to the constant $a$

From the linearised equation $\lg y = (\lg b)x + \lg a$, the vertical intercept is $\lg a$.
Reading from the graph (or calculating from the points), the vertical intercept is approximately $0.904$.
$\lg a = 0.904$

2. Calculate the value of $a$

$a = 10^{0.904} \approx 8.02 \approx 8.0$ (to 2 s.f.)

3. Relate the gradient of the linear graph to the constant $b$ and calculate its value

The gradient of the line is $\lg b$.
Using points $(0.5, 0.792)$ and $(3.0, 0.230)$:
$\text{Gradient} = \frac{0.230 - 0.792}{3.0 - 0.5} = \frac{-0.562}{2.5} = -0.2248$
$\lg b = -0.2248$
$b = 10^{-0.2248} \approx 0.596 \approx 0.60$ (to 2 s.f.)

1. Read the value of $\lg y$ from the graph when $x = 0.8$

From the straight line graph of $\lg y$ against $x$, when $x = 0.8$, the corresponding value on the vertical axis is $\lg y \approx 0.71$.

2. Calculate the value of $y$

$\lg y = 0.71$
$y = 10^{0.71}$
$y \approx 5.13 \approx 5.1$ (to 1 d.p.)