Exponential Growth and Decay

Learning goals

Multiply by the growth factor $b = 1 + r$ each step
Distinguish the rate from the factor, since $r = b - 1$
Read growth, standstill or decay off $b$
Divide the rate and multiply the time when compounding $n$ times
Chain percent changes by multiplying factors
Take a logarithm for doubling time or half-life

Adding an amount versus multiplying by a factor

Start two accounts with 1000 dollars each. Account L gains a flat 100 dollars every year. Account E gains $10\%$ every year. Track them side by side.

Year	Account L, plus 100 dollars a year	Account E, plus $10\%$ a year
$0$	$1000$	$1000$
$1$	$1100$	$1100$
$2$	$1200$	$1210$
$3$	$1300$	$1331$
$4$	$1400$	$1464.10$
$5$	$1500$	$1610.51$

They tie after the first year, and then they separate. Account L adds $100$ every single year, because a flat amount does not care how much is already there. Account E adds $100$ , then $110$ , then $121$ , then $133.10$ . The percentage never changed. What changed is the balance the percentage is taken from.

That is the whole distinction, and it is worth stating in one line:

A fixed amount added each step gives linear change.
A fixed percentage added each step gives exponential change.

Everything else in this lesson follows from turning that second sentence into algebra.

Where the model comes from

Take a quantity and increase it by a rate $r$ , written as a decimal, so $10\%$ means $r = 0.10$ . The increase is $r$ times the current amount, and the new amount is the old amount plus that increase:

A_{\text{new}} \;=\; A_{\text{old}} + r \cdot A_{\text{old}} \;=\; A_{\text{old}}(1 + r).

Factoring out $A_{\text{old}}$ is the move that makes the whole subject work. One step of growth is one multiplication, by the number $1 + r$ . That number gets its own name.

The growth rate is $r$ , the fraction added per step. The growth factor is $b = 1 + r$ , the number you multiply by. They are not the same number, and confusing them is the single most common error in this topic. A rate of $r = 0.07$ goes with a factor of $b = 1.07$ . Running it the other way, $r = b - 1$ , so a factor of $b = 0.85$ means $r = -0.15$ , a $15\%$ decrease.

After $t$ steps, $A(t) = P(1+r)^t$ #

Let $P$ be the starting amount, and suppose every step multiplies the current amount by the same factor $b = 1 + r$ . Write $A(t)$ for the amount after $t$ steps.

At the start no step has happened, so $A(0) = P$ , which agrees with $P b^0 = P \cdot 1 = P$ . Now suppose the formula $A(t) = P b^t$ holds after some particular whole number of steps $t$ . The next step does one thing only: it multiplies by $b$ . So

A(t+1) = A(t) \cdot b = P b^t \cdot b = P b^{t+1},

which is the same formula with $t+1$ in place of $t$ . The formula is therefore true at step $0$ , and whenever it is true at one step it is true at the next. So the formula is true at every whole number of steps. Restoring $b = 1 + r$ gives $A(t) = P(1+r)^t$ .

The converse also holds, and it is worth checking, because it says the two descriptions really are the same description. Suppose a quantity follows $A(t) = P b^t$ with $P > 0$ and $b > 0$ . From one step to the next it changes by

\frac{A(t+1) - A(t)}{A(t)} = \frac{P b^{t+1} - P b^{t}}{P b^{t}} = b - 1,

which does not depend on $t$ . So the quantity changes by the same fraction, the same percentage, at every step. Constant percent change per step and the model $P b^t$ each force the other.

One honest caveat about the exponent. The argument above pins the formula down at whole numbers of steps, since those are the only steps that were taken. To read the model at times between steps we ask for something more. The extra demand is that the growth over a stretch of time depend only on the length of that stretch, and not on when it starts. That is exactly the rule $b^{t+s} = b^t \cdot b^s$ , which the exponential function already obeys for every real exponent, so $A(t) = P b^t$ is the extension that keeps the property we care about. Half a step then multiplies by $b^{1/2}$ , and the rational exponents of the previous chapter give that a meaning.

Growth, decay, and reading the factor

The factor $b$ decides the entire character of the model. Take a positive starting amount $P > 0$ and a positive factor $b$ . Then exactly one of three things happens:

If $b > 1$ , each step multiplies by more than $1$ , so the amount grows without bound.
If $b = 1$ , each step multiplies by $1$ , so the amount never changes.
If $0 < b < 1$ , each step multiplies by less than $1$ , so the amount decays toward zero.

Decay never actually reaches zero. Multiplying a positive number by $0.7$ leaves a smaller positive number, forever, which is why the decay curve below flattens against the horizontal axis without touching it.

Both curves start at the same amount, 100. Multiplying by 1.3 at every step bends the curve upward and it climbs without bound. Multiplying by 0.7 at every step drives the curve down toward zero, which it approaches but never reaches.

Reading a factor out of a sentence takes care. “Grows by $30\%$ ” keeps the original and adds $30\%$ more, so the factor is $1 + 0.30 = 1.30$ . “Falls by $30\%$ ” keeps $70\%$ of the original, so the factor is $1 - 0.30 = 0.70$ . “Doubles” is a $100\%$ increase, factor $2$ . “Halves” is a $50\%$ decrease, factor $\tfrac{1}{2}$ .

Now watch what happens when you chain two of them. A price rises $30\%$ and then falls $30\%$ . Many people expect to land back where they started. Multiply the factors:

1.30 \times 0.70 = 0.91,

a $9\%$ net loss. The two $30\%$ s are percentages of different amounts, so they do not cancel. Percent changes never add; the factors multiply. This is not a quirk of the numbers $1.3$ and $0.7$ : a rise of $r$ followed by a fall of $r$ always gives $(1+r)(1-r) = 1 - r^2$ . For every nonzero $r$ , that value $1 - r^2$ is less than $1$ .

Check your understanding

A town's population falls by $12\%$ each year. If the population starts at $P$ , which model gives the population after $t$ years?

P(0.12)^t

P(1.12)^t

P(0.88)^t

P(-0.12)^t

Compounding more than once a year

Banks rarely wait a whole year to pay interest. A rate quoted as ” $6\%$ per year, compounded monthly” means the year is cut into $12$ equal periods and each period pays one twelfth of the annual rate. Nothing new is needed: apply the one-step rule with the period rate.

With an annual rate $r$ compounded $n$ times a year, each period grows by $r/n$ , so each period multiplies by $1 + \frac{r}{n}$ . Over $t$ years there are $nt$ such periods, and $nt$ multiplications give an exponent of $nt$ :

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

Divide the rate, multiply the time. Setting $n = 1$ recovers $A = P(1+r)^t$ , so the annual formula is just the special case with one period a year. Compounding more often pays a little more, because the interest you earn early starts earning interest itself sooner.

Worked example 1 2000 dollars at $6\%$ per year, compounded monthly, for 5 years

Identify the four ingredients. The principal is $P = 2000$ dollars, the annual rate is $r = 0.06$ , the number of periods per year is $n = 12$ , and the time is $t = 5$ years.

The monthly rate is $r/n$ and the number of periods is $nt$ :

\frac{r}{n} = \frac{0.06}{12} = 0.005, \qquad nt = 12 \cdot 5 = 60.

So the balance multiplies by $1.005$ sixty times:

A = 2000(1.005)^{60} \approx 2000(1.34885) \approx 2697.70.

The account holds about $2697.70$ dollars. Compare that with the same rate compounded once a year, $2000(1.06)^5 \approx 2676.45$ dollars. Monthly compounding earns about $21$ dollars more over the five years, which is the interest that the interest earned.

Check your understanding

Which expression gives the balance on 500 dollars after $3$ years at an annual rate of $8\%$ compounded quarterly?

500(1.08)^{3}

500(1.02)^{12}

500(1.02)^{3}

500(1.08)^{12}

Doubling time and half-life

So far the unknown has always been the amount. The more interesting question runs the other way: how long until the amount reaches some target? That puts the unknown in the exponent, and the logarithm is precisely the tool built to bring an exponent down.

The doubling time is the time it takes for a growing quantity to reach twice its size. Set $A(t) = 2P$ and solve:

P b^t = 2P \quad \Longrightarrow \quad b^t = 2.

Look at what just happened: $P$ cancelled. The doubling time does not depend on the starting amount at all. A colony of $10$ bacteria and a colony of $10$ million bacteria with the same growth factor take exactly the same time to double. Taking the logarithm of both sides and using the power law $\log(b^t) = t\log b$ ,

t \log b = \log 2 \quad \Longrightarrow \quad t = \frac{\log 2}{\log b}.

The half-life is the mirror image: the time for a decaying quantity to fall to half its size. The same cancellation happens, leaving $b^t = \tfrac{1}{2}$ , so

t = \frac{\log \frac{1}{2}}{\log b}.

Here $\log \frac{1}{2}$ is negative, and for decay $b < 1$ makes $\log b$ negative too, so the quotient of two negatives is a positive time, as it must be.

Because the doubling time is the same everywhere along the curve, it gives a second, often tidier way to write the model. If the doubling time is $d$ , then the per-unit factor $b$ satisfies $b^d = 2$ , so $b = 2^{1/d}$ and therefore $b^t = 2^{t/d}$ . The same reasoning with a half-life $h$ gives $b = \left(\tfrac{1}{2}\right)^{1/h}$ . In short:

A(t) = P \cdot 2^{t/d} \qquad \text{and} \qquad A(t) = P\left(\tfrac{1}{2}\right)^{t/h}.

Sanity-check the second one at $t = h$ : the exponent is $1$ and the amount is exactly $\tfrac{P}{2}$ , which is what “half-life” means.

A quantity that starts at 100 and doubles every 3 years. Each 3-year stretch multiplies the amount by 2, whether it is climbing from 100 to 200 or from 400 to 800. The doubling time never changes, no matter how large the quantity has already grown.

Worked example 2 A population grows $4\%$ a year. How long until it doubles?

The growth factor is $b = 1 + 0.04 = 1.04$ . Doubling means the amount reaches $2P$ , and the starting population $P$ cancels straight away, leaving

(1.04)^t = 2.

The unknown is stuck in the exponent, so take the logarithm of both sides and use the power law to bring $t$ down in front:

t \log 1.04 = \log 2 \quad \Longrightarrow \quad t = \frac{\log 2}{\log 1.04}.

Evaluating, $\log 2 \approx 0.30103$ and $\log 1.04 \approx 0.01703$ , so

t \approx \frac{0.30103}{0.01703} \approx 17.7 \text{ years}.

Check it: $(1.04)^{17.7} \approx 2.00$ , so the population really has doubled. Notice that the answer never used the starting population. So a village of $300$ and a city of $3$ million both take about $17.7$ years to double at $4\%$ a year.

Worked example 3 A $60$ -gram sample has a half-life of $12$ years

Because the amount halves every $12$ years, use the half-life form of the model with $P = 60$ and $h = 12$ :

A(t) = 60\left(\tfrac{1}{2}\right)^{t/12}.

How much remains after $30$ years? Substitute $t = 30$ , so the exponent is $\tfrac{30}{12} = 2.5$ , meaning two and a half half-lives have passed:

A(30) = 60\left(\tfrac{1}{2}\right)^{2.5} = \frac{60}{2^{5/2}} = \frac{60}{4\sqrt{2}} \approx 10.61 \text{ grams}.

When does the sample fall to $10$ grams? Set $A(t) = 10$ and divide by $60$ first, which is the same cancellation as before:

\left(\tfrac{1}{2}\right)^{t/12} = \frac{10}{60} = \frac{1}{6}.

Take logarithms and use the power law:

\frac{t}{12}\log \tfrac{1}{2} = \log \tfrac{1}{6} \quad \Longrightarrow \quad \frac{t}{12} = \frac{\log \frac{1}{6}}{\log \frac{1}{2}} = \frac{\log 6}{\log 2} \approx \frac{0.77815}{0.30103} \approx 2.585.

Multiplying by $12$ gives $t \approx 31.0$ years. That agrees with part one: after $30$ years there were still $10.61$ grams, a little more than $10$ , so the sample should cross $10$ grams a little after year $30$ .

Check your understanding

A substance loses $20\%$ of its mass every hour. Which equation determines its half-life $h$ , in hours?

(0.2)^h = 0.5

(1.2)^h = 0.5

(0.8)^h = 2

(0.8)^h = 0.5

Building the model from two data points

In practice nobody hands you the rate. You measure the quantity twice and work the rate out. Suppose you know the starting amount $P = A(0)$ and one later reading $A(t_1)$ . Then

P b^{t_1} = A(t_1) \quad \Longrightarrow \quad b^{t_1} = \frac{A(t_1)}{P} \quad \Longrightarrow \quad b = \left(\frac{A(t_1)}{P}\right)^{1/t_1},

using the rational exponents from the previous chapter to undo the power $t_1$ . If neither reading is at time zero, divide one by the other and the unknown $P$ cancels: from $A(t_2)/A(t_1) = b^{t_2 - t_1}$ you get the factor the same way.

Worked example 4 A colony of $500$ grows to $800$ in $6$ years

Assume exponential growth, so $A(t) = 500\,b^{t}$ for an unknown annual factor $b$ . The reading at $t = 6$ gives an equation:

500\,b^{6} = 800 \quad \Longrightarrow \quad b^{6} = \frac{800}{500} = \frac{8}{5} = 1.6.

Undo the sixth power with a sixth root, which is the exponent $\tfrac{1}{6}$ :

b = 1.6^{1/6}.

To get a decimal, take logarithms: $\log b = \tfrac{1}{6}\log 1.6 \approx \tfrac{1}{6}(0.20412) \approx 0.03402$ , so $b \approx 10^{0.03402} \approx 1.0815$ . The colony grows about $8.15\%$ a year, which is the factor minus one, not the factor itself.

Check the model against the data: $500(1.0815)^6 \approx 500(1.600) \approx 800$ , as required.

There is also a log-free way to write the same model. The colony multiplies by $\tfrac{8}{5}$ every $6$ years, so exactly as with doubling time,

A(t) = 500\left(\tfrac{8}{5}\right)^{t/6}.

Both forms give the same numbers; the second one just refuses to round anything.

Common mistakes

Using the rate where the factor belongs. “Grows $6\%$ a year” gives the factor $1.06$ , not $0.06$ . The model $A = P(0.06)^t$ describes something that loses $94\%$ of itself every year.
Adding percent changes. A $30\%$ rise followed by a $30\%$ fall is not a wash. Multiply the factors: $1.30 \times 0.70 = 0.91$ , a $9\%$ net loss. Percents are taken from different amounts, so they do not cancel; only the factors combine, and they combine by multiplying.
Mishandling the compounding period. With $n$ periods a year the rate is divided and the time is multiplied: $P\left(1 + \tfrac{r}{n}\right)^{nt}$ . Writing $P\left(1 + \tfrac{r}{n}\right)^{t}$ or $P(1+r)^{nt}$ changes the answer.
Putting the half-life in the wrong place. A half-life of $h$ gives $\left(\tfrac{1}{2}\right)^{t/h}$ , not $\left(\tfrac{1}{2}\right)^{ht}$ . Test any candidate at $t = h$ , where the amount must come out to exactly half.
Thinking the doubling time depends on the starting amount. Solve $Pb^t = 2P$ and the $P$ cancels before you do anything else. Starting bigger does not slow the doubling down.
Dividing instead of taking a logarithm. From $b^t = 2$ you cannot write $t = 2/b$ . The unknown is an exponent, so take a log of both sides: $t\log b = \log 2$ .

Key takeaways

One step of growth is one multiplication. Increasing by a rate $r$ multiplies by the growth factor $b = 1 + r$ , so $t$ steps multiply by $b^t$ and $A(t) = P(1+r)^t$ .
The rate and the factor are different numbers: $r = 0.07$ pairs with $b = 1.07$ , and a factor of $b = 0.85$ means a $15\%$ decrease, since $r = b - 1$ .
With $P > 0$ the model grows when $b > 1$ , stays put when $b = 1$ , and decays toward zero (never reaching it) when $0 < b < 1$ .
Compounding $n$ times a year divides the rate and multiplies the time: $A = P\left(1 + \tfrac{r}{n}\right)^{nt}$ .
Percent changes chain by multiplying factors, so a $30\%$ gain followed by a $30\%$ loss leaves $1 - (0.3)^2 = 0.91$ of the original, not all of it.
Time questions put the unknown in the exponent, so take a logarithm. The doubling time is $t = \dfrac{\log 2}{\log b}$ and the half-life is $t = \dfrac{\log \frac{1}{2}}{\log b}$ , and neither depends on the starting amount $P$ .
A doubling time $d$ or a half-life $h$ rewrites the model as $A = P \cdot 2^{t/d}$ or $A = P\left(\tfrac{1}{2}\right)^{t/h}$ , and two data points pin down the factor through $b = \left(\tfrac{A_1}{P}\right)^{1/t_1}$ .

Exponential Growth and Decay

Learning goals

Adding an amount versus multiplying by a factor

Where the model comes from

Growth, decay, and reading the factor

Compounding more than once a year

Doubling time and half-life

Building the model from two data points

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)