Exponential Growth and Decay Modeling

Repeated percentage change

Exponential growth and decay describe situations in which the same multiplier is applied in every equal time period. This means the same percentage change is repeated, not the same numerical change.

For example, if a population grows by 20% each year, its multiplier is

$1+\frac{20}{100}=1.2.$

Starting from 100, the successive amounts are 100, 120, 144 and 172.8. The increases are 20, 24 and 28.8, so they are not equal. Each increase is nevertheless 20% of the amount at the start of that year.

For a decrease of 20%, the multiplier is

$1-\frac{20}{100}=0.8.$

Repeatedly multiplying by 0.8 leaves 80% of the current amount each time. A multiplier greater than 1 produces growth; a multiplier between 0 and 1 produces decay.

Building an exponential formula

Let $P$ be the initial amount, $k$ the multiplier per time period and $n$ the number of periods. The amount $A$ after $n$ periods is

$A=P\times k^n.$

The power $n$ records how many times the multiplier is applied. At the start, $n=0$, and $k^0=1$, so $A=P$. After one period, ; after two periods, .

For growth at $r\%$ per period,

$A=P\left(1+\frac{r}{100}\right)^n.$

For decay at $r\%$ per period,

$A=P\left(1-\frac{r}{100}\right)^n.$

The letters used may change. For instance, $y=ka^x$ has the same structure: $k$ is the initial value, $a$ is the multiplier and counts the time periods.

An amount of £100 growing by 10% per year can therefore be modelled by

$A=100\times1.1^n,$

where $n$ is the number of years. The formula is meaningful only when the unit of $n$ matches the period attached to the multiplier.

Recognising the graph

An exponential graph curves because its numerical change becomes larger or smaller as the current amount changes. Growth becomes increasingly steep. Decay falls quickly at first and then more slowly, while remaining positive in this model.

Data for Repeated growth and decay multipliers

A straight-line model instead adds or subtracts a fixed amount each period. Therefore, a statement such as “increases by 12 each year” is linear, whereas “increases by 12% each year” indicates exponential growth.

Using a formula to predict an amount

First identify the initial amount, percentage change, length of one period and number of periods. Convert the percentage to a multiplier, substitute into $A=P\times k^n$, and evaluate the power before multiplying by $P$.

Suppose a colony initially contains 750 organisms and grows by 6% each day. Its model is

$A=750\times1.06^n,$

where $n$ is measured in days. After four days, the model gives $750\times1.06^4\approx946.86$. If organisms must be counted as whole individuals, the context requires a suitable whole-number answer; the unrounded decimal should be retained until the final step.

A model is an assumption about what will happen if the same rate continues. A long-term prediction is not a guarantee: real populations, investments and depreciating objects may stop following a constant percentage rate.

Working backwards to the initial amount

If the final amount, multiplier and number of periods are known, rearrange

$A=P\times k^n$

to obtain

$P=\frac{A}{k^n}.$

For example, if a quantity is 864 after three periods of growth with multiplier 1.2, then its initial value is

$P=\frac{864}{1.2^3}=500.$

This reverses all three applications of the multiplier at once.

Finding the multiplier and percentage rate

If an amount changes from $P$ to $A$ over $n$ equal periods, then

$k^n=\frac{A}{P}$

and hence

$k=\sqrt[n]{\frac{A}{P}}.$

Once $k$ is known, convert it into a percentage change:

• for growth, $r=(k-1)\times100\%$;
• for decay, $r=(1-k)\times100\%$.

Suppose a value falls from 500 to 320 over two years at a constant percentage rate. Then

$k=\sqrt{\frac{320}{500}}=0.8.$

The yearly rate of decrease is therefore $(1-0.8)\times100\%=20\%$. Dividing 320 by 500 gives 0.64, but that is the multiplier over the whole two-year interval, not the annual multiplier.

Finding a number of periods

When $P$, $A$ and $k$ are known but $n$ is not, form the equation

$A=P\times k^n.$

At GCSE, whole-number time periods can be found by testing successive values of $n$ until the amount reaches or passes the required value. Interpret the wording carefully. “After how many complete years is the value below £2,000?” asks for the first whole year for which the model gives a value below £2,000.

Interpreting the result

The calculation is only part of the answer. Include the correct unit and relate the result to the situation. Money is normally given to the nearest penny unless the question specifies otherwise. Counts such as people or vehicles may require a whole number, but whether to round up, down or to the nearest integer depends on the wording.

Also distinguish an amount from a change. If a model gives a car value of £8,400 after five years, that is its remaining value. Its total loss in value is the original price minus £8,400.