Exponential growth and decay

Exponential models describe quantities whose change over equal intervals is proportional to the amount present. They are used for compound interest, population growth, radioactive decay, depreciation and many other processes.

The essential distinction is:

$$\text{linear change: add the same amount}, \qquad \text{exponential change: multiply by the same factor}.$$

Prerequisites

You should be able to:

• use the laws of indices;
• work with exponential functions;
• use logarithms and their laws;
• rearrange equations and convert percentages to decimals;
• substitute values with consistent units.

Recognising exponential change

Suppose a population increases by $8\%$ each year. After one year, $100\%+8\%=108\%$ of the previous population remains, so the annual multiplier is

$$1+\frac{8}{100}=1.08.$$

If its initial value is $P_0$, then

$$P_1=P_0(1.08),$$ $$P_2=P_0(1.08)^2,$$

and after $n$ years,

$$\boxed{P_n=P_0(1.08)^n}.$$

The repeated multiplication creates an exponential model.

For a percentage rate $r\%$ per interval:

$$\boxed{\text{growth multiplier}=1+\frac{r}{100}},$$ $$\boxed{\text{decay multiplier}=1-\frac{r}{100}}.$$

For ordinary decay, $0<r<100$, so the multiplier lies between $0$ and $1$.

Description	Multiplier
increases by $6\%$	$1.06$
decreases by $6\%$	$0.94$
becomes $115\%$ of its value	$1.15$
loses one quarter of its value	$0.75$
doubles	$2$
halves	$0.5$

The percentage acts on the current amount

A decrease of $10\%$ followed by an increase of $10\%$ does not return to the starting value:

$$0.9\times1.1=0.99.$$

The final value is $99\%$ of the original value.

The discrete model $Q=Ab^t$

A quantity multiplied by a constant factor $b$ in each unit interval can be modelled by

$$\boxed{Q=Ab^t},$$

where:

• $A$ is the value when $t=0$;
• $b$ is the multiplier per unit of time;
• $b>1$ gives growth;
• $0<b<1$ gives decay.

Although repeated multiplication initially gives integer values of $t$, the formula $Ab^t$ also defines values between the recorded times. Whether that interpolation is sensible depends on the context.

Worked example 1: compound growth

A colony contains $2400$ cells and grows by $12\%$ every hour. Find the model and the predicted population after $7$ hours.

The initial value is $A=2400$. The hourly multiplier is

$$b=1+\frac{12}{100}=1.12.$$

Therefore

$$P=2400(1.12)^t.$$

At $t=7$,

$$P=2400(1.12)^7 \approx5305.63.$$

A cell count must be a whole number, so the model predicts approximately

$$\boxed{5306\text{ cells}}.$$

Do not round intermediate values. Doing so compounds the rounding error as well as the growth.

Worked example 2: depreciation

A machine is bought for $\pounds 36\,000$ and loses $18\%$ of its value each year.

The annual multiplier is

$$1-0.18=0.82.$$

Hence its value after $t$ years is

$$V=36\,000(0.82)^t.$$

After $5$ years,

$$V=36\,000(0.82)^5 \approx13\,346.63.$$

The predicted value is

$$\boxed{\pounds 13\,346.63}.$$

The machine loses $18\%$ of its current value each year, not $18\%$ of $\pounds36\,000$ each year. The latter would be linear depreciation.

Worked example 3: find the percentage rate

An investment grows from $\pounds 5000$ to $\pounds 6749.29$ in $4$ years at a constant annual percentage rate. Find the rate.

Let the annual multiplier be $b$. Then

$$5000b^4=6749.29.$$

Divide by $5000$:

$$b^4=1.349858.$$

Take the positive fourth root because a financial multiplier is positive:

$$b=(1.349858)^{1/4}\approx1.077884.$$

Thus the percentage growth rate is

$$(b-1)\times100\%\approx7.7884\%.$$

So the annual rate is approximately

$$\boxed{7.79\%}.$$

Multiplier is not percentage rate

If $b=1.08$, the rate is $8\%$, not $108\%$. If $b=0.92$, the decay rate is $8\%$, not $92\%$.

The continuous model $Q=Ae^{kt}$

Many processes are modelled as changing continuously rather than once per fixed interval. The standard form is

$$\boxed{Q=Ae^{kt}},$$

where $A=Q(0)$ and $k$ is a constant.

• If $k>0$, then $Q$ grows.
• If $k<0$, then $Q$ decays.
• If $k=0$, then $Q=A$ is constant.

Since $e^{kt}>0$ for every real $t$, the model never becomes negative. If $A>0$ and $k<0$, then $Q$ approaches zero but never reaches zero in finite time.

The units of $kt$ must cancel because an exponent has no units. Therefore, if $t$ is measured in years, $k$ has units $\text{year}^{-1}$.

Why $e$ appears

For

$$Q=Ae^{kt},$$

differentiation gives

$$\frac{dQ}{dt}=kAe^{kt}=kQ.$$

Thus the instantaneous rate of change is proportional to the amount present. This is the defining feature of continuous exponential growth or decay.

The full construction and solution of such models is developed in differential equation models and separable differential equations.

Worked example 4: determine $A$ and $k$

A population is modelled by $P=Ae^{kt}$, where $t$ is measured in years. Initially the population is $3200$, and after $6$ years it is $5000$. Find the model.

Use the initial condition:

$$P(0)=Ae^0=A=3200.$$

So

$$P=3200e^{kt}.$$

Use $P(6)=5000$:

$$5000=3200e^{6k}.$$

Divide by $3200$:

$$e^{6k}=\frac{25}{16}.$$

Take natural logarithms:

$$6k=\ln\left(\frac{25}{16}\right).$$

Hence

$$k=\frac16\ln\left(\frac{25}{16}\right) \approx0.07438.$$

The model may be written exactly as

$$\boxed{P=3200e^{\frac{t}{6}\ln(25/16)}}.$$

Equivalently,

$$\boxed{P=3200\left(\frac{25}{16}\right)^{t/6}}.$$

The second form makes the six year multiplier $25/16$ visible.

Worked example 5: find when a threshold is crossed

The mass of a substance is modelled by

$$m=80e^{-0.035t},$$

where $m$ is measured in grams and $t$ in days. Find when the mass first falls below $30$ g.

First solve the boundary equation:

$$80e^{-0.035t}=30.$$

Divide by $80$:

$$e^{-0.035t}=0.375.$$

Take natural logarithms:

$$-0.035t=\ln(0.375).$$

Therefore

$$t=\frac{\ln(0.375)}{-0.035} \approx28.02.$$

Because the model decreases continuously, the mass is below $30$ g when

$$\boxed{t>28.02\text{ days, approximately}}.$$

If observations are made only at the end of each whole day, the first such observation would be on day $29$. Always interpret the time in context.

Converting between $Ab^t$ and $Ae^{kt}$

The two forms are algebraically equivalent when the same time unit is used:

$$b^t=e^{t\ln b}.$$

Therefore

$$\boxed{k=\ln b} \qquad\text{and}\qquad \boxed{b=e^k}.$$

Worked example 6: convert an annual rate

A quantity increases by $5\%$ per year, so

$$Q=A(1.05)^t.$$

Since $1.05^t=e^{t\ln(1.05)}$,

$$Q=Ae^{kt}, \qquad k=\ln(1.05)\approx0.04879.$$

The equivalent continuous constant is approximately $0.04879\text{ year}^{-1}$.

It is tempting to use $k=0.05$, but that gives

$$e^{0.05}\approx1.05127,$$

which is about $5.127\%$ growth per year, not exactly $5\%$.

Small rates

For small $r$, $\ln(1+r)\approx r$, where $r$ is a decimal. This explains why $5\%$ and $k=0.05$ are close, but they are not identical.

Doubling time and half life

For $Q=Ae^{kt}$ with $k>0$, the doubling time $T_d$ satisfies

$$Ae^{kT_d}=2A.$$

Cancel $A$ and take logarithms:

$$kT_d=\ln2.$$

Thus

$$\boxed{T_d=\frac{\ln2}{k}}.$$

For decay, it is often clearer to write $Q=Ae^{-\lambda t}$ with $\lambda>0$. The half life $T_{1/2}$ satisfies

$$Ae^{-\lambda T_{1/2}}=\frac{A}{2},$$

so

$$\boxed{T_{1/2}=\frac{\ln2}{\lambda}}.$$

If the model is written as $Ae^{kt}$ with $k<0$, then

$$\boxed{T_{1/2}=-\frac{\ln2}{k}}.$$

Doubling time and half life do not depend on the initial amount $A$.

Worked example 7: use a half life

A radioactive sample has initial mass $120$ mg and half life $8$ days. Find a model and the mass after $30$ days.

Using $m=120e^{-\lambda t}$,

$$\lambda=\frac{\ln2}{8}.$$

Hence

$$m=120e^{-(\ln2)t/8}.$$

Since $e^{-(\ln2)t/8}=2^{-t/8}$, an equally useful form is

$$m=120\left(\frac12\right)^{t/8}.$$

At $t=30$,

$$m=120\left(\frac12\right)^{30/8} \approx8.919.$$

Therefore

$$\boxed{m\approx8.92\text{ mg}}.$$

The exponent $30/8=3.75$ means that $3.75$ half lives have elapsed.

Worked example 8: recover the half life from data

A substance decays from $250$ g to $190$ g in $12$ hours. Assuming exponential decay, find its half life.

Let

$$m=250e^{-\lambda t}.$$

At $t=12$,

$$190=250e^{-12\lambda}.$$

Therefore

$$e^{-12\lambda}=0.76,$$

so

$$-12\lambda=\ln(0.76)$$

and

$$\lambda=-\frac{\ln(0.76)}{12}.$$

The half life is

$$T_{1/2}=\frac{\ln2}{\lambda} =\frac{12\ln2}{-\ln(0.76)} \approx30.31.$$

Thus the predicted half life is

$$\boxed{30.3\text{ hours, approximately}}.$$

Shifted exponential models

Not every exponentially changing quantity approaches zero. A model of the form

$$\boxed{Q=L+Ae^{kt}}$$

has horizontal asymptote $Q=L$ when $k<0$. The difference from the limiting value is exponential:

$$Q-L=Ae^{kt}.$$

This is useful for temperature approaching room temperature, drug concentration approaching a background level, or a quantity approaching a long term equilibrium.

Worked example 9: cooling towards room temperature

A drink at $90^\circ\text{C}$ is placed in a room at $20^\circ\text{C}$. Its excess temperature above room temperature halves every $15$ minutes. Find its temperature after $40$ minutes.

It is the excess temperature, not the actual temperature, that decays exponentially. Initially,

$$90-20=70.$$

Therefore

$$T-20=70\left(\frac12\right)^{t/15},$$

so

$$T=20+70\left(\frac12\right)^{t/15}.$$

At $t=40$,

$$T=20+70\left(\frac12\right)^{40/15} \approx31.02.$$

Thus

$$\boxed{T\approx31.0^\circ\text{C}}.$$

The incorrect model $90(1/2)^{t/15}$ tends to $0^\circ\text{C}$. That ignores the room temperature and predicts physically inappropriate long term behaviour.

Comparing linear and exponential models

Suppose two accounts each begin with $\pounds1000$.

• Account L gains $\pounds50$ each year: $L=1000+50t$.
• Account E gains $5\%$ each year: $E=1000(1.05)^t$.

After one year both contain $\pounds1050$. After ten years,

$$L=1000+50(10)=1500,$$

whereas

$$E=1000(1.05)^{10}\approx1628.89.$$

The linear model has constant first differences. The exponential model has a constant ratio:

$$\frac{E(t+1)}{E(t)}=1.05.$$

For equally spaced data, nearly constant ratios suggest an exponential model. Real data contain noise, so the ratios will rarely be exactly constant. Linearising exponential data gives a more systematic test.

Interpreting and criticising a model

An exponential formula is an assumption about reality, not a guarantee.

State the variables and units

Writing $P=500e^{0.2t}$ is incomplete unless the meanings and units of $P$ and $t$ are known. Changing the unit of time changes the numerical value of the growth constant.

For example, if $k=0.12\text{ year}^{-1}$ and time in months is $u=12t$, then

$$e^{0.12t}=e^{0.12(u/12)}=e^{0.01u}.$$

The monthly constant is $0.01\text{ month}^{-1}$.

Check initial and long term behaviour

For $Q=Ae^{kt}$:

$$Q(0)=A.$$

If $A>0$ and $k<0$, then $Q\to0$ as $t\to\infty$. If the real quantity should approach a nonzero limit, a shifted model may be needed.

Avoid unjustified extrapolation

Population growth may slow as resources become scarce. An asset cannot usually lose the same percentage forever while the simple model remains commercially meaningful. A medicine may be eliminated by several biological processes rather than one constant proportional rate.

A model can fit observed data well and still give poor predictions far outside the observed interval.

Continuous quantities and counts

A model may predict $317.6$ people or bacteria. This is not literally possible. The decimal is the output of a continuous approximation and should be rounded appropriately when interpreted, but not during intermediate calculation.

Common misconceptions

Confusing a rate with a multiplier

A $23\%$ decrease gives multiplier

$$1-0.23=0.77,$$

not $0.23$ and not $1.23$.

Subtracting the same percentage of the original value

Repeated percentage change acts on the current value. If $Q$ falls by $15\%$ per year, use

$$Q=Q_0(0.85)^t,$$

not $Q=Q_0-0.15Q_0t$.

Reading $k$ as the exact percentage per unit time

In $Q=Ae^{kt}$, the multiplier over one time unit is $e^k$. The exact percentage change is

$$100(e^k-1)\%.$$

Losing the sign in a decay calculation

Both forms below describe decay:

$$Ae^{kt}\quad\text{with }k<0,$$ $$Ae^{-\lambda t}\quad\text{with }\lambda>0.$$

State the convention. A positive exponent with a positive constant describes growth.

Rounding the constant too early

If $k=\dfrac{\ln(5/3)}{8}$, retain this exact value or store the full calculator value. A rounded $k$ is then multiplied by $t$ in the exponent, so its error can become noticeable.

Self check

1. A car worth $\pounds24\,000$ depreciates by $14\%$ each year. Write a model and find its value after $3$ years.
2. A culture is modelled by $N=600e^{0.18t}$, where $t$ is in hours. Find its doubling time.
3. A sample has mass $48$ g initially and $30$ g after $10$ days. Assuming $m=Ae^{kt}$, find $A$ and $k$.
4. A medicine has half life $6$ hours. What fraction of the initial amount remains after $15$ hours?
5. A temperature is modelled by $T=18+62e^{-0.08t}$. State the initial temperature and the limiting temperature.
6. Explain why $P=2000(1.04)^t$ and $P=2000e^{kt}$ describe the same model when $k=\ln(1.04)$.

Answers

1. The multiplier is $0.86$, so

   $$V=24\,000(0.86)^t.$$

   Hence

   $$V(3)=24\,000(0.86)^3=\boxed{\pounds15\,902.78}$$

   to the nearest penny.

2. Set $e^{0.18t}=2$:

   $$t=\frac{\ln2}{0.18}\approx\boxed{3.85\text{ hours}}.$$

3. Since $m(0)=A$, $A=48$. Also

   $$30=48e^{10k},$$

   so

   $$k=\frac1{10}\ln\left(\frac{30}{48}\right) =\boxed{\frac1{10}\ln\left(\frac58\right)} \approx-0.04700\text{ day}^{-1}.$$

4. The number of half lives is $15/6=2.5$, so the fraction is

   $$\left(\frac12\right)^{2.5} =\boxed{\frac{1}{4\sqrt2}} \approx0.1768.$$

5. Initially,

   $$T(0)=18+62=\boxed{80}.$$

   As $t\to\infty$, $e^{-0.08t}\to0$, so $T\to\boxed{18}$.

6. Using $a^x=e^{x\ln a}$,

   $$2000(1.04)^t =2000e^{t\ln(1.04)} =2000e^{kt}.$$

Next steps

• Practise solving for time and unknown parameters in exponential and logarithmic equations.
• Learn how logarithms turn exponential data into a straight line in linearising exponential and power relationships.
• Build rate based models in constructing differential equation models.
• Solve proportional growth and decay equations in separable differential equations.