Exponential growth and decay

Many quantities change at a rate that is proportional to their size. A population of ${2000}$ rabbits has twice as many births as a population of ${1000}$. A large sample of a radioactive element loses more atoms each second than a small one. Such quantities grow or decay exponentially. This page shows how to model them and answer questions about them.

The exponential growth model

${n(t)=n_0e^{rt}}$

Here ${n(t)}$ is the size at time ${t}$, and ${n_0=n(0)}$ is the starting size. The number ${r}$ is the relative growth rate: the growth per year (or per hour) as a fraction of the current size. For example, ${r=0.02}$ means ${2\%}$ per year.

If ${r>0}$, the quantity grows. If ${r<0}$, it decays.

Example 1: A town had ${20{,}000}$ people in ${2020}$. Its population grows at a relative rate of ${2\%}$ per year.

(a) Find the population in ${2030}$.

(b) When will the population reach ${30{,}000}$?

Solution:

Let ${t}$ be the number of years after ${2020}$. Then ${n_0=20{,}000}$ and ${r=0.02}$:

${n(t)=20{,}000\,e^{0.02t}}$

(a) The year ${2030}$ is ${t=10}$:

${n(10)=20{,}000\,e^{0.2}\approx 24{,}428}$

(b) Set ${n(t)=30{,}000}$, divide by ${20{,}000}$, and take ${\ln}$ (see exponential and logarithmic equations):

$\begin{align*}e^{0.02t}&=1.5\\0.02t&=\ln 1.5\\t&=\dfrac{\ln 1.5}{0.02}\approx 20.3\end{align*}$

The population reaches ${30{,}000}$ about ${20}$ years after ${2020}$, in the year ${2040}$.

Often the rate ${r}$ is not given. Then we find it from two measurements.

Example 2: A culture starts with ${100}$ bacteria. After ${3}$ hours there are ${450}$. Assume exponential growth. Find the relative growth rate, and predict the number after ${5}$ hours.

Solution:

The model is ${n(t)=100e^{rt}}$, with ${t}$ in hours. Use ${n(3)=450}$ to find ${r}$:

$\begin{align*}100e^{3r}&=450\\e^{3r}&=4.5\\3r&=\ln 4.5\\r&=\dfrac{\ln 4.5}{3}\approx 0.5014\end{align*}$

So the culture grows at about ${50\%}$ per hour. After ${5}$ hours:

${n(5)=100e^{0.5014\cdot 5}\approx 1227}$

There will be about ${1227}$ bacteria.

Tip: Keep the exact value ${r=\dfrac{\ln 4.5}{3}}$ in your calculator. Rounding ${r}$ too early changes the final answer.

Radioactive decay and half-life

A radioactive element slowly turns into other elements. The time it takes for half of a sample to decay is its half-life, ${h}$. After one half-life, half of the sample is left. After two half-lives, a quarter is left. So after ${t}$ years, ${\dfrac{t}{h}}$ half-lives have passed, and

${m(t)=m_0\left(\dfrac{1}{2}\right)^{t/h}}$

Here ${m_0}$ is the starting mass. The same model can be written as ${m(t)=m_0e^{-rt}}$, with the decay rate ${r=\dfrac{\ln 2}{h}}$.

16003200480064002550(1600, 25)(3200, 12.5)tm
A ${50}$ mg sample of radium-226. Every ${1600}$ years, half of what is left decays.

Example 3: Radium-226 has a half-life of ${1600}$ years. A sample has a mass of ${50}$ mg.

(a) How much is left after ${500}$ years?

(b) When will only ${10}$ mg be left?

Solution:

The model is $m(t)=50\left(\dfrac{1}{2}\right)^{t/1600}$.

(a) Put ${t=500}$:

$m(500)=50\left(\dfrac{1}{2}\right)^{500/1600}\approx 40.3$ mg

(b) Set ${m(t)=10}$, and divide both sides by ${50}$:

$\left(\dfrac{1}{2}\right)^{t/1600}=\dfrac{1}{5}$

Take ${\ln}$ of both sides, and use the power law:

$\begin{align*}\dfrac{t}{1600}\ln\dfrac{1}{2}&=\ln\dfrac{1}{5}\\t&=1600\cdot\dfrac{\ln(1/5)}{\ln(1/2)}\approx 3715\end{align*}$

Only ${10}$ mg is left after about ${3715}$ years.

Example 4: Living things contain carbon-14, which has a half-life of ${5730}$ years. After a plant or animal dies, its carbon-14 decays. An old bone has ${30\%}$ of the carbon-14 it had when the animal was alive. How old is the bone?

Solution:

The amount left is ${0.30}$ times the starting amount:

${\left(\dfrac{1}{2}\right)^{t/5730}=0.30}$

Take ${\ln}$ of both sides, and solve for ${t}$:

$\begin{align*}\dfrac{t}{5730}\ln 0.5&=\ln 0.30\\t&=5730\cdot\dfrac{\ln 0.30}{\ln 0.5}\approx 9953\end{align*}$

The bone is about ${10{,}000}$ years old. This method is called carbon dating.

Newton's law of cooling

A hot cup of coffee cools down to room temperature. The difference between its temperature and the room temperature decays exponentially:

${T(t)=T_s+D_0e^{-kt}}$

Here ${T_s}$ is the temperature of the surroundings (the room), ${D_0}$ is the starting temperature difference, and ${k>0}$ is a constant that depends on the object.

Example 5: A cup of coffee at ${90}$°C is placed in a room at ${20}$°C. After ${10}$ minutes, the coffee is at ${60}$°C. When will it be at ${30}$°C?

Solution:

Here ${T_s=20}$, and the starting difference is ${D_0=90-20=70}$. So

${T(t)=20+70e^{-kt}}$.

Use ${T(10)=60}$ to find ${k}$:

$\begin{align*}20+70e^{-10k}&=60\\e^{-10k}&=\dfrac{40}{70}=\dfrac{4}{7}\\-10k&=\ln\dfrac{4}{7}\\k&=\dfrac{\ln(7/4)}{10}\approx 0.0560\end{align*}$

Now set ${T(t)=30}$:

$\begin{align*}20+70e^{-kt}&=30\\e^{-kt}&=\dfrac{1}{7}\\-kt&=\ln\dfrac{1}{7}=-\ln 7\\t&=\dfrac{\ln 7}{k}\approx 34.8\end{align*}$

The coffee reaches ${30}$°C after about ${35}$ minutes.

Summary

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