Exponential Growth and Decay

Linear change and exponential change

A quantity changes linearly when you add the same amount every period: a machine that loses 300 dollars of value each year, a jar that gains 50 pounds a month. A quantity changes exponentially when you multiply by the same factor every period: a population rising by 3% a year, a car losing 20% of its value each year.

The arithmetic follows the description. Adding a fixed amount is addition, so after n periods the total change is n times the amount. Multiplying by a fixed factor is repeated multiplication, so after n periods the change is the factor to the power n. ${1.03}^{10}$ is not $1.3$; it is $1.34$, because each year's growth is calculated on the year before.

Any percentage change taken from the value at the start of each period is exponential. That is why the second year's growth is bigger than the first: there is more there to grow.

The multiplier: 1 + r for growth, 1 - r for decay

Every percentage change can be written as a single multiplier. A rise of r per cent has multiplier $1+\frac{r}{100}$; a fall of r per cent has multiplier $1-\frac{r}{100}$.

So a 15% rise has multiplier $1.15$, a 6% rise $1.06$, a 20% fall $0.8$ and a 3% fall $0.97$. A multiplier above $1$ grows the quantity, a multiplier below $1$ shrinks it, and a multiplier of exactly $1$ leaves it unchanged.

Reading a multiplier backwards is just as quick. A multiplier of $1.08$ is an 8% rise, and a multiplier of $0.72$ is a 28% fall, because $1-0.72=0.28$.

The formula, and what each letter means

For a quantity that changes by the same percentage each period, the value after n periods is $A=P(1+r{)}^n$. Here $P$ is the starting value, $r$ is the change per period written as a decimal (negative for decay), $n$ is the number of periods, and $A$ is the value at the end.

A machine bought for $20000$ dollars that loses 15% of its value each year has $P=20000$ and multiplier $0.85$. After 3 years $A=20000\times {0.85}^3=20000\times 0.614125=12282.50$ dollars. The line of working has to show the multiplier and the power, because those are the two decisions being marked.

Two checks every time. First, does the power match the rate? A rate of 1% a month applied for two years means $n=24$, not $2$. Second, does the answer make sense? A machine that depreciates must end up worth less than it cost, and a population that grows must end up larger.

• Machine bought for 20000 dollars, losing 15% a year, after 3 years: $20000\times {0.85}^3=12282.50$ dollars
• Car worth 12000 dollars, losing 20% a year, after 4 years: $12000\times {0.8}^4=4915.20$ dollars
• Town of 45000 people growing 2% a year, after 5 years: $45000\times {1.02}^5=49684$ people
• Village of 8000 people falling 3% a year, after 6 years: $8000\times {0.97}^6=6664$ people

Working backwards: the starting value and the number of periods

Some questions give the value after n periods and ask what the quantity started at. Reverse the calculation by dividing by the multiplier to the power n. A machine worth $9720$ dollars after two years of losing 10% a year started at $9720÷{0.9}^2=9720÷0.81=12000$ dollars.

Other questions ask you to compare two points in time or to check a claim. Working out the multiplier over the whole period does that in one step: ${1.03}^{15}=1.557967$ to six decimal places, so a town of $50000$ people growing 3% a year reaches $50000\times 1.557967=77898$ people after 15 years, which is more than half again as many.

The power must always match the rate. If the rate is given per year, n counts years; if it is given per month, n counts months. Reading the units of the rate before you press the power button prevents most errors in this topic.

• Value after n periods: multiply by the factor n times, which is the factor raised to the power n
• Starting value: divide the final value by the factor to the power n
• Number of periods: count the periods the rate applies to, not the number of years by habit

Getting the details right in the exam

Keep the full value on the calculator display and round once, at the end. Rounding the multiplier before you raise it to a power, or rounding each year's value as you go, changes the final answer - and a method error of that kind takes the accuracy mark with it.

Round to suit the context: money to 2 decimal places or the nearest cent, a population to a whole number of people, and anything the question specifies to the accuracy it asks for.

Write the multiplier and the power as a line of working. $20000\times {0.85}^3$ shows the examiner both decisions - the factor and the number of periods - so a slip in the arithmetic then costs one mark instead of all of them.