Math Core

You now have exponential functions to describe change and logarithms to solve for time. This lesson puts them to work on real situations: populations, radioactive decay, cooling objects, and the spread of anything that eventually runs out of room to grow. Along the way you'll meet the logistic model, which fixes the biggest flaw of pure exponential growth.

Continuous exponential models

Any exponential model can be written as

A(t) = A_0 e^{kt},

where $A_0 = A(0)$ is the initial amount and $k$ is the continuous growth rate (or relative growth rate). If $k > 0$ the quantity grows; if $k < 0$ it decays. A model written as $A_0 b^t$ converts to this form with $k = \ln b$ , as you saw in the properties lesson.

What makes this form natural is that $k$ measures growth relative to size. At every instant, a quantity modeled by $A_0e^{kt}$ is changing at a rate of $k$ times its current amount. That's the signature of anything whose growth is fueled by what's already there: cells dividing, money earning interest on interest, atoms decaying independently of one another.

Doubling time and half-life

For $A = A_0e^{kt}$ :

If $k > 0$ , the doubling time is $T = \dfrac{\ln 2}{k}$ .
If $k < 0$ , the half-life is $h = \dfrac{\ln 2}{|k|}$ .

Neither depends on the starting amount. With a known half-life $h$ , you can also write $A = A_0\left(\tfrac{1}{2}\right)^{t/h}$ .

To see the doubling-time formula, set $A = 2A_0$ : then $e^{kT} = 2$ , so $kT = \ln 2$ .

Building a model from data

If you know the amount at two times, you can find $k$ and then predict anything else.

Worked example: Bacteria culture

A culture contains $500$ bacteria at the start of an experiment and $1400$ three hours later. Assume continuous exponential growth.

Find the model $N = N_0e^{kt}$ .
Predict the count after $5$ hours.
When will the culture reach $5000$ bacteria?

Solution.

$N_0 = 500$ . From $1400 = 500e^{3k}$ you get $e^{3k} = 2.8$ , so $k = \dfrac{\ln 2.8}{3} \approx 0.3432$ . The model is $N \approx 500e^{0.3432t}$ , a continuous growth rate of about $34.3\%$ per hour.
$N(5) = 500e^{5k}$ . Because $e^{3k} = 2.8$ exactly, $e^{5k} = 2.8^{5/3}$ , and $N(5) = 500(2.8)^{5/3} \approx 2781$ bacteria.
Solve $5000 = 500e^{kt}$ : $e^{kt} = 10$ , so $t = \dfrac{\ln 10}{k} = \dfrac{3\ln 10}{\ln 2.8} \approx 6.71$ hours.

Tip

Keep $k$ exact (as $\frac{\ln 2.8}{3}$ ) or in calculator memory. Rounding $k$ to $0.34$ in part 3 gives $6.77$ hours instead of $6.71$ . Small errors in a rate are magnified in the exponent.

Radioactive decay

Radioactive isotopes decay exponentially with a fixed half-life. Carbon-14, with a half-life of about $5730$ years, is used to date once-living material: while an organism is alive its carbon-14 level stays constant, and after death it decays.

Worked example: Carbon dating

A wooden tool contains $30\%$ of the carbon-14 found in living wood. About how old is it?

Solution. With $A = A_0\left(\tfrac{1}{2}\right)^{t/5730}$ and $A = 0.30A_0$ :

0.30 = \left(\tfrac{1}{2}\right)^{t/5730} \quad\Longrightarrow\quad \frac{t}{5730}\ln 0.5 = \ln 0.30 \quad\Longrightarrow\quad t = \frac{5730\ln 0.30}{\ln 0.5} \approx 9953.

The tool is roughly $10{,}000$ years old. Reasonableness check: after $1$ half-life $50\%$ remains, after $2$ half-lives ( $11{,}460$ years) $25\%$ remains. $30\%$ lies between, and so does $9953$ years.

Newton's law of cooling

A hot object cools quickly at first, then more slowly as it approaches room temperature. The difference between its temperature and the surroundings decays exponentially:

T(t) = T_s + (T_0 - T_s)e^{-kt},

where $T_s$ is the surrounding temperature, $T_0$ the initial temperature and $k > 0$ a constant for the object. The graph has horizontal asymptote $T = T_s$ .

Worked example: Cooling coffee

A cup of coffee at $90^\circ\text{C}$ is placed in a $20^\circ\text{C}$ room. After $5$ minutes it is $70^\circ\text{C}$ . How long until it cools to $45^\circ\text{C}$ ?

Solution. The model is $T = 20 + 70e^{-kt}$ . Use the data point:

70 = 20 + 70e^{-5k} \;\Longrightarrow\; e^{-5k} = \frac{5}{7} \;\Longrightarrow\; k = \frac{\ln(7/5)}{5} \approx 0.06729.

Now solve $45 = 20 + 70e^{-kt}$ :

e^{-kt} = \frac{25}{70} \;\Longrightarrow\; t = \frac{\ln(70/25)}{k} = \frac{5\ln 2.8}{\ln 1.4} \approx 15.3 \text{ minutes}.

The coffee's temperature approaches the room temperature, 20 °C.Open in grapher →

Logistic growth

Exponential growth cannot last forever. A population in a lake, a rumor in a school or sales of a new phone all eventually run into a limit. The logistic model starts out looking exponential, then bends and levels off at a maximum.

Definition

Logistic model

A logistic function has the form

P(t) = \frac{C}{1 + Ae^{-kt}}, \qquad C, A, k > 0.

The constant $C$ is the carrying capacity: $P(t) \to C$ as $t \to \infty$ . The initial value is $P(0) = \dfrac{C}{1 + A}$ .

Why does it level off? As $t$ grows, $e^{-kt} \to 0$ , so the denominator approaches $1$ and $P \to C$ . For very negative $t$ the denominator is huge and $P \to 0$ . The graph is an S-shaped curve with horizontal asymptotes $P = 0$ and $P = C$ .

Growth is fastest at the inflection point, where $P = \dfrac{C}{2}$ . Before that point the curve bends upward like an exponential; after it, growth slows as the population crowds its limit.

Worked example: Fish in a lake

A lake is stocked with fish, and the population after $t$ years is modeled by

P(t) = \frac{1200}{1 + 39e^{-0.4t}}.

How many fish were stocked, and what is the carrying capacity?
When does the population reach half the carrying capacity?

Solution.

$P(0) = \dfrac{1200}{1 + 39} = 30$ fish were stocked. The carrying capacity is $C = 1200$ fish.
Solve $P(t) = 600$ :

\frac{1200}{1 + 39e^{-0.4t}} = 600 \;\Longrightarrow\; 1 + 39e^{-0.4t} = 2 \;\Longrightarrow\; e^{-0.4t} = \frac{1}{39} \;\Longrightarrow\; t = \frac{\ln 39}{0.4} \approx 9.16 \text{ years}.

This is when the population is growing fastest.

The logistic model (solid) tracks the exponential 30e^(0.4t) (dashed) early on, then levels off at 1200.Open in grapher →

Common mistake

Don't use an exponential model outside the range where it makes sense. An exponential fit to the first few years of fish data would predict about $30e^{0.4 \cdot 25} \approx 660{,}000$ fish after $25$ years. The logistic model, which knows about the lake's limit, predicts just under $1200$ .

Choosing and checking a model

A quick way to test whether data are exponential: compute $\ln y$ for each data point. If $y = A_0e^{kt}$ , then

\ln y = \ln A_0 + kt,

which is linear in $t$ with slope $k$ . So exponential data look like a straight line when plotted on a logarithmic vertical axis (a semi-log plot). Logistic data look linear there at first, then flatten.

Situation	Model
Growth or decay by a constant percent, no limit in sight	$A_0e^{kt}$ or $A_0b^t$
Decay toward a nonzero level (cooling, warming)	$T_s + (T_0 - T_s)e^{-kt}$
Growth limited by a maximum (population, spread of news)	$\dfrac{C}{1 + Ae^{-kt}}$

Practice

Practice 1

A deer population in a state park is modeled by $P(t) = 250e^{0.03t}$ , where $t$ is years since 2020. Find $P(10)$ , rounded to the nearest whole number.