0580 · 1.17Extended content only.

Multipliers and exponential change

0:00Two populations both start at one thousand, and both gain one hundred during their first year. After that, their totals begin to differ. One gains a fixed number each year, while the other grows by a fixed percentage. By the end of this video, you will understand why that difference produces exponential growth, and how the same idea explains depreciation. We need two starting ideas: percentage multipliers and powers. A percentage multiplier is the number we multiply the current amount by to get the new amount.

0:28A power records repeated multiplication by the same factor. We will use these ideas together, rather than treating exponential growth and decay as a completely new calculation. An increase of ten per cent means keeping the original one hundred per cent and adding another ten per cent. That gives one hundred and ten per cent of the current amount. Dividing by one hundred gives the multiplier one point one. Multiplying by this factor includes both the original amount and its increase.

0:55A decrease of ten per cent leaves ninety per cent of the current amount. The corresponding multiplier is zero point nine. Notice that this multiplier represents what remains, not what disappears. Multiplying by zero point one would calculate the amount lost, rather than the amount left. Exponential growth means multiplying by the same factor greater than one over each equal time interval. Exponential decay means multiplying by the same positive factor less than one.

1:21In both cases, the percentage change stays constant. The actual amount added or removed usually changes, because the percentage applies to a changing total. Return to the population of one thousand growing by ten per cent each year. Its first increase is one hundred, giving one thousand one hundred. During the second year, ten per cent of the new population is one hundred and ten. The population therefore becomes one thousand two hundred and ten, rather than one thousand two hundred.

1:49Adding the same number each year gives a constant difference between successive totals. Exponential change keeps the ratio between them constant instead. This is the essential distinction. The word exponential describes repeated multiplication, not simply something becoming large or changing quickly. After one period, multiply the initial amount by the multiplier once. After two periods, multiply that result by the same factor again. Repeating this n times gives the initial amount times the multiplier to the power n.

2:19Here, n counts how many times the change happens, which explains why it appears as the exponent. The period must match the rate given in the question. An annual rate uses the number of years, while a monthly rate uses the number of months. An annual percentage must not be applied once for every month. Also, this model assumes the stated percentage remains constant throughout the calculation.

Population growth example

2:40Our first exam-style example starts with two thousand people and an annual increase of ten per cent. Calculate the population after three years. The growth multiplier is one plus ten divided by one hundred, giving one point one. After the first year, 2000 × 1.1 gives two thousand two hundred. For the second year, multiply two thousand two hundred by one point one to obtain two thousand four hundred and twenty.

3:07Applying the multiplier again gives two thousand six hundred and sixty-two after the third year. Each calculation starts with the latest population. The final answer is therefore two thousand six hundred and sixty-two people. This example is suitable for Paper Two, which is the non-calculator paper for Extended students. You can find ten per cent by dividing the current population by ten. Add that increase, then repeat with the updated total.

3:32The multiplier method must work without a calculator, even though more awkward numbers need one on Paper Four. The compact calculation is 2000 × 1.1³. Squaring one point one gives one point two one. Multiplying that by one point one gives one point three three one. Finally, multiplying two thousand by this result gives the same population of two thousand six hundred and sixty-two. The final multiplier also reveals the overall percentage change.

4:00The population is now one hundred and thirty-three point one per cent of its starting size. That means the overall increase is thirty-three point one per cent. It is not thirty per cent, because later increases also build on earlier increases.

Depreciation and decay example

4:14Decay follows exactly the same repeated multiplication idea. A value falling by twenty per cent retains eighty per cent each year, so its multiplier is zero point eight. Starting from one thousand pounds, it becomes eight hundred pounds, then six hundred and forty pounds. The second loss is smaller because twenty per cent is taken from a smaller value. Depreciation means a reduction in an asset’s value over time. A car losing a fixed percentage of its current value each year follows exponential decay.

4:42Under this model, the value stays positive after any finite number of periods. Repeatedly multiplying a positive amount by a positive factor does not suddenly turn it into zero. Our second example concerns a car costing eighteen thousand pounds. It depreciates by fifteen per cent each year for three years. Calculate its final value to the nearest pound. Subtracting fifteen per cent from one hundred per cent leaves eighty-five per cent, giving a multiplier of zero point eight five.

5:09The calculation is 18000 × 0.85³. Squaring the multiplier gives zero point seven two two five. Multiplying this by zero point eight five gives zero point six one four one two five. Multiplying that result by eighteen thousand gives eleven thousand and fifty-four point two five pounds. The question requests the nearest pound, so the final answer is eleven thousand and fifty-four pounds. On Paper Four, the calculator paper, you can enter the complete expression directly.

5:40Keep the full calculator value until the final rounding step. Cambridge generally expects three significant figures for answers that are not exact, unless the question specifies another accuracy. A common error is using zero point one five as the multiplier for this car. That would leave only fifteen per cent of its value after one year. Another error is subtracting fifteen per cent of the original price every year. Exponential depreciation uses fifteen per cent of the current value, so the loss gets smaller.

Working backwards with growth

6:08A third exam-style question reverses the process. A population increases by five per cent each year and reaches nine thousand two hundred and sixty-one after three years. Calculate its initial population. The annual multiplier is one point zero five, so the starting population multiplied by this factor cubed equals the final population. Squaring one point zero five gives one point one zero two five. Multiplying once more gives one point one five seven six two five.

6:36Divide nine thousand two hundred and sixty-one by this three-year multiplier. The initial population is eight thousand, because division undoes the multiplication used for growth. Do not reverse a five per cent increase by applying a five per cent decrease. Those percentages would act on different starting amounts, so they do not undo each other. Divide by the growth multiplier instead. In an examination, show the multiplier, the correct power and the calculation, rather than writing only a final number.

7:05The central idea is a constant percentage applied to the changing amount over equal time intervals. Growth uses a multiplier greater than one, while decay uses a positive multiplier less than one. Powers count repeated applications of that multiplier. Match the exponent to the number of periods, keep intermediate values unrounded, and divide by the total multiplier when working backwards.