Exponential growth and decay

Exponential growth is what you have when something grows by the same percentage every step: money in a bank earning interest, or bacteria that double every hour. We describe it with the equation $f(x) = b \times a^x$, where $a = 1 + r$. The usual hurdle is turning the percentage into the growth factor $a$ correctly.

When do I use this?

When you meet tasks with a percentage rise or fall of one kind or another, it's a good idea to think exponential growth. That could be the example with interest in the bank, or maybe bacteria that double, so bacteria becoming twice as many per hour or per day. GCSE calls these growth and decay problems, and compound interest is the classic one. The equation itself, and what $a$ and $b$ mean, is explained in exponential functions.

The procedure

We build the equation $f(x) = b \times a^x$ from the information in the question, then put in the number of steps.

1. Find the start value $b$. That's the amount you start with, before any growth or fall. It's where the graph crosses the y-axis.
2. Turn the percentage into the growth factor $a$. The growth factor is $a = 1 + r$, where $r$ is the percentage as a decimal. We add $1$ because we're putting the percentage rise or fall on top of the $100\%$ we already have. A rise of $2\%$ gives $a = 1 + 0.02 = 1.02$. A fall is a negative $r$: a fall of $20\%$ gives $a = 1 + (-0.20) = 0.80$.
3. Write the equation. $f(x) = b \times a^x$, where $x$ counts the steps: years, hours, days, whatever the percentage is per.
4. Put in the number of steps. Work out $a^x$ first, then multiply by $b$.

Worked examples

Money in the bank (growth)

Let's say you start with £$1000$ in the bank, and the bank gives you $2\%$ interest a year, so every year you get $2\%$ added on top. The start value is £$1000$ and $r = 0.02$, so

$$f(x) = 1000 \times (1 + 0.02)^x = 1000 \times 1.02^x$$

Here $x$ is the number of years, because you get $2\%$ per year. To find out how much is in the account after $2$ years, we put $2$ in for $x$:

$$f(2) = 1000 \times 1.02^2 = 1000 \times 1.0404 = 1040.40$$

So after $2$ years at $2\%$ interest there's £$1040.40$ in the account.

Bacteria that double (growth)

Say a dish starts with $500$ bacteria and the number doubles every hour. Doubling is a rise of $100\%$, so $r = 1$ and the growth factor is $a = 1 + 1 = 2$. With a start value of $500$:

$$f(x) = 500 \times 2^x$$

After $5$ hours:

$$f(5) = 500 \times 2^5 = 500 \times 32 = 16000$$

A value that falls (decay)

Now something that falls by a percentage. Say a value of £$2000$ falls by $20\%$ a year. A fall of $20\%$ means $r = -0.20$, so the growth factor is $a = 1 + (-0.20) = 0.80$, and

$$f(x) = 2000 \times 0.80^x$$

After $1$ year that's $2000 \times 0.80 = 1600$, which is just the usual "take $20\%$ off" from percentage increase and decrease. After $3$ years:

$$f(3) = 2000 \times 0.80^3 = 2000 \times 0.512 = 1024$$

So after $3$ years the value is £$1024$. Notice that $a$ is still a positive number. It's below $1$, and that's what makes the function fall instead of grow. The graph is the red, falling curve in the exponential functions glossary.

Common mistakes

• Using the percentage itself as the growth factor. $2\%$ growth is $a = 1.02$, not $a = 0.02$. The $1$ is the $100\%$ you already had, and the $0.02$ is what gets added on top.
• Making $a$ negative for a fall. A fall of $20\%$ gives $a = 0.80$, which is positive but below $1$. The growth factor is always greater than $0$.
• Forgetting that the percentage applies every step. Each year's $2\%$ is added on top of the new, bigger amount. That's why $x$ sits in the index: $1.02^x$ means $1.02$ multiplied by itself $x$ times.

Related

• Exponential functions: growth and decay
• Compound interest: the interest formula
• Percentage increase and decrease with a multiplier