Exponential growth and decay

In exponential growth a quantity is multiplied by the same number every unit of time. In exponential decay the same is true, but the number is less than 1.

For a percentage change of R% per year, the multiplier is 1 + R/100 for an increase and 1 - R/100 for a decrease. A 4% increase has multiplier 1.04. A 10% decrease has multiplier 0.9.

Growth fits population change. Decay fits depreciation, where a value falls each year.

After n units of time the quantity is its starting value multiplied by the multiplier n times. That is a power of the multiplier.

On a calculator, enter the starting value, then ×, the multiplier, then the power key and n. The change each year is not constant: the amount added grows every year.

Growth gives a curve that gets steeper. Decay gives a curve that gets less steep.

With simple interest the same amount is added every year. With exponential growth the amount added changes each year, because the percentage is taken of a bigger value.

Read what the question asks for: the value of the investment includes the money invested at the start.

If the question says compound, or says the value changes by a percentage of its current value, use the multiplier and a power.

For depreciation the multiplier is less than 1. A value that falls by 10% each year is multiplied by 0.9.

To find an earlier value, go back one step at a time: divide by the multiplier once for each year. The reverse of a 10% fall is a division by 0.9, not a 10% rise.

After two years the value is multiplied by 0.9². So two years ago it was the current value divided by 0.9².

To find when a value first passes a target, write an inequality such as 22 680 × 0.9ⁿ < 10 000. Then find the whole number n that first satisfies it.

On a graphic display calculator, enter the value as Y1 = 22680 × 0.9^(X) and the target as Y2 = 10000. Use a table of values and read off the first X that works. Or try values of n and check each side of the target.

The answer is a number of complete years. If solving gives a value such as n = 7.8, round up to 8: after 7 years the target has not yet been passed.

A change over two different years can use two multipliers. Multiply them together to get the combined effect.

When the question asks for a year, find n first. Then add it to the starting year. The year is the final answer, not n.

Check the last step: the value at year n - 1 must be below the target and at year n above it.