The table below shows the annual global production of plastics, P, measured in millions of tonnes per year, for six selected years - AQA - A-Level Maths Mechanics - Question 9 - 2021 - Paper 1

The completed table is as follows:

t	0	5	10	15	20	25

log$_{10} P$ 1.88 1.97 2.08 2.19 2.31 2.41

When plotting log$_{10} P$ against t, ensure that at least four points are plotted accurately, with a ruled line drawn from t=0 to t=25 to represent the line of best fit.

From the line of best fit, take two points from the graph, for example (0, 1.88) and (25, 2.41). The formula for the gradient (k) is:

$k = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2.41 - 1.88}{25 - 0} = \frac{0.53}{25} \approx 0.0212 \approx 0.02$

Using the log$_{10} P$ for t=0:

$log_{10} P = log_{10} A + kt$

At t = 0, log$_{10} P = 1.88$:

$1.88 = log_{10} A + 0 \implies log_{10} A = 1.88 \implies A = 10^{1.88} \approx 75$

For 2030, t = 50 (since 2030 is 50 years after 1980):

$P = A \times 10^{kt} = 75 \times 10^{0.02 \times 50} \approx 75 \times 10^{1} = 750 \text{ million tonnes.}$

We set:

$8000 = A \times 10^{kt} \implies 8000 = 75 \times 10^{0.02t}$

Dividing both sides by 75:

$106.67 = 10^{0.02t}$

Taking the logarithm:

$log_{10} (106.67) = 0.02t \implies t = \frac{log_{10}(106.67)}{0.02} \approx 2082$

Thus, the year is 2082 (2082 + 1980).

The model assumes constant growth which may not hold true in future due to varying factors such as environmental regulations, changes in plastic demand, or technological advancements. Thus, using this model for long-term predictions may lead to inaccuracies.