Exponential Growth and Decay

When something grows or shrinks by the same percentage each time period, it follows an exponential model. Populations, investments, a car’s value, a cooling cup of hot chocolate, and medicine in your bloodstream all behave this way. This page shows how to build the equation from a description and use it to answer questions.

Key ideas

The growth and decay model

$$A(t) = A_0(1 + r)^t \qquad \text{or} \qquad A(t) = A_0(1 - r)^t$$

• $A_0$ is the initial amount (the value when $t = 0$).
• $r$ is the rate per time period, as a decimal: $3\% = 0.03$.
• $1 + r$ is the growth factor; $1 - r$ is the decay factor.
• $t$ is the number of time periods.

A $3\%$ increase means multiplying by $1.03$ each period. A $15\%$ decrease means multiplying by $0.85$, because $85\%$ is left.

Doubling and half-life

When you know how long it takes to double or halve, use base $2$ or $\tfrac{1}{2}$:

$$A(t) = A_0 (2)^{\frac{t}{D}} \qquad\qquad A(t) = A_0 \left(\tfrac{1}{2}\right)^{\frac{t}{H}}$$

Here $D$ is the doubling time and $H$ is the half-life. The exponent $\tfrac{t}{H}$ counts how many half-lives have passed.

An $80$ mg dose with a half-life of $5$ hours: $A(t) = 80\left(\tfrac{1}{2}\right)^{\frac{t}{5}}$.

Domain and range in context

A model only makes sense for realistic values. Usually $t \ge 0$, and the amount stays positive: a decaying quantity gets close to $0$ but never reaches it in the model.

Solving for time

In this course, find an unknown time by rewriting both sides with the same base (when the numbers allow it), by guess and check with a calculator, or by reading a graph. (A faster method, using logarithms, comes in Grade 12.)

Worked examples

Example 1: Population growth

A town of $12\,000$ people grows by $3\%$ per year. Write a model, and estimate the population after $10$ years.

Solution. $A_0 = 12\,000$ and the growth factor is $1.03$:

$$P(t) = 12\,000(1.03)^t$$ $$P(10) = 12\,000(1.03)^{10} \approx 16\,127$$

The population will be about $16\,127$ people.

Example 2: Depreciation

A car costs $28 000 and loses $15\%$ of its value each year. Find its value after $5$ years.

Solution. The decay factor is $1 - 0.15 = 0.85$:

$$V(t) = 28\,000(0.85)^t$$ $$V(5) = 28\,000(0.85)^5 \approx 12\,423.75$$

After $5$ years, the car is worth about $12 423.75.

Example 3: Half-life

A patient takes an $80$ mg dose of a medicine with a half-life of $5$ hours. How much is left after $12$ hours? When will $10$ mg be left?

Solution.

$$A(t) = 80\left(\tfrac{1}{2}\right)^{\frac{t}{5}}$$ $$A(12) = 80\left(\tfrac{1}{2}\right)^{2.4} \approx 15.16$$

About $15.16$ mg is left after $12$ hours.

For $10$ mg: $80 \to 40 \to 20 \to 10$ takes three halvings. Algebraically, $\left(\tfrac{1}{2}\right)^{\frac{t}{5}} = \tfrac{10}{80} = \left(\tfrac{1}{2}\right)^3$, so $\tfrac{t}{5} = 3$ and $t = 15$ hours. The graph above shows the same thing.

Example 4: Doubling

A culture starts with $500$ bacteria and doubles every $20$ minutes. How many are there after $2$ hours? When will there be $8000$?

Solution. Measure time in minutes, so $D = 20$:

$$N(t) = 500(2)^{\frac{t}{20}}$$

After $2$ hours ($120$ minutes): $N(120) = 500(2)^6 = 500(64) = 32\,000$.

For $8000$: $2^{\frac{t}{20}} = \tfrac{8000}{500} = 16 = 2^4$, so $\tfrac{t}{20} = 4$ and $t = 80$ minutes.

Common mistakes

Using the percentage as the factor. A $3\%$ increase is a factor of $1.03$, not $1.3$ (that would be $30\%$) and not $0.03$.

Using the rate instead of what’s left for decay. Losing $15\%$ means multiplying by $0.85$, not $0.15$.

Mixing time units. If the half-life is in hours, $t$ must be in hours too. In Example 4, $2$ hours had to become $120$ minutes.

Multiplying before applying the exponent. $12\,000(1.03)^{10}$ means $(1.03)^{10}$ first, then times $12\,000$. Don’t compute $(12\,000 \times 1.03)^{10}$.

Treating it like linear growth. Growing $3\%$ a year doesn’t mean adding the same number of people each year. Each year’s increase is $3\%$ of a bigger population.

Rounding too early. Keep full calculator values until the end, then round.

Practice

1. (Warm-up) Does each model show growth or decay? By what percentage, or how often does it halve?

• (a) $A = 500(1.08)^t$
• (b) $A = 200(0.75)^t$
• (c) $A = 50\left(\tfrac{1}{2}\right)^{\frac{t}{3}}$

Solution

(a) Growth of $8\%$ per period.

(b) Decay of $25\%$ per period.

(c) Decay: the amount halves every $3$ time units.

2. (Warm-up) $1500 is invested and grows by $6\%$ each year. Write a model for its value after $t$ years.

Solution

$A(t) = 1500(1.06)^t$

3. (Warm-up) What is the initial value of $N(t) = 350(0.9)^t$, and what percentage is lost each period?

Solution

Initial value $350$. The factor $0.9$ means $10\%$ is lost each period.

4. (Core) A village of $4500$ people is shrinking by $2\%$ per year. Estimate its population after $8$ years.

Solution

$$P(8) = 4500(0.98)^8 \approx 3828$$

About $3828$ people.

5. (Core) A ball is dropped from $3$ m. Each bounce reaches $70\%$ of the previous height. Write a model for the height after the $n$th bounce, and find the height after the $4$th bounce, to the nearest centimetre.

Solution

$$h(n) = 3(0.7)^n$$$$h(4) = 3(0.7)^4 = 3(0.2401) = 0.7203$$

About $0.72$ m, or $72$ cm.

6. (Core) Iodine-131 has a half-life of about $8$ days. A hospital has a $200$ mg sample. How much is left after $24$ days? After $30$ days (to two decimal places)?

Solution

$$A(t) = 200\left(\tfrac{1}{2}\right)^{\frac{t}{8}}$$

$24$ days is $3$ half-lives: $A(24) = 200\left(\tfrac{1}{8}\right) = 25$ mg.

$A(30) = 200\left(\tfrac{1}{2}\right)^{3.75} \approx 14.87$ mg.

7. (Core) A savings account’s balance is shown below. Show that it’s growing exponentially, write a model, and predict the balance after $6$ years.

Year $t$	$0$	$1$	$2$	$3$
Balance ($)	$2000$	$2200$	$2420$	$2662$

Solution

The ratios are $\tfrac{2200}{2000} = \tfrac{2420}{2200} = \tfrac{2662}{2420} = 1.1$, all the same, so it’s exponential with a growth rate of $10\%$.

$$A(t) = 2000(1.1)^t, \qquad A(6) = 2000(1.1)^6 \approx 3543.12$$

About $3543.12 after $6$ years.

8. (Challenge) Money invested at $5\%$ per year grows by a factor of $1.05^t$. Use guess and check to find how many years it takes to double.

Solution

Look for $1.05^t \approx 2$:

• $1.05^{14} \approx 1.980$ (not quite double)
• $1.05^{15} \approx 2.079$ (more than double)

So it takes between $14$ and $15$ years: the money has more than doubled after $15$ full years. (A more precise answer is about $14.2$ years.)

9. (Challenge) For the car in Example 2, $V(t) = 28\,000(0.85)^t$, state a reasonable domain and range, and explain why the model never gives a value of $0.

Solution

Domain $\{t \in \mathbb{R} \mid t \ge 0\}$ (time since purchase). Range $\{V \in \mathbb{R} \mid 0 \lt V \le 28\,000\}$.

Each year the value is multiplied by $0.85$, which shrinks it but never makes it $0$: $85\%$ of a positive number is still positive. The graph approaches the asymptote $V = 0$ without reaching it. (In real life, the car might eventually be sold for scrap, so the model only works for a reasonable number of years.)