Exponential Growth And Decay: Formula & Examples

What Is Exponential Growth And Decay?

Exponential growth and decay describes any quantity whose rate of change is proportional to how much of it there is right now. Twice as much substance decays twice as fast; twice as large a population breeds twice as many offspring per year. That single sentence becomes a differential equation:

$$\frac{dy}{dt} = ky$$

Here $y$ is the amount at time $t$, and $k$ is the constant of proportionality. The sign of $k$ decides everything:

• Growth ($k > 0$). The rate is positive, so $y$ increases, and it increases faster the larger it gets. This is the runaway behaviour of unchecked populations and continuously compounded money.

• Decay ($k < 0$). The rate is negative, so $y$ decreases, and it shrinks more slowly as it gets smaller. This is radioactive material, cooling coffee, and a drug clearing the bloodstream.

The equation is a separable differential equation, and solving it (next section) gives one clean formula that covers both cases:

$$y = y_0 e^{kt}$$

where $y_0 = y(0)$ is the value at $t = 0$. Every worked example, half-life, and doubling time in this article comes straight out of that one result.

How Do You Solve dy/dt = ky By Separation Of Variables?

The equation $\frac{dy}{dt} = ky$ is first-order and separable, so the method is to gather every $y$ on one side and every $t$ on the other, then integrate. This is the standard route for any first-order ordinary differential equation of this shape.

Step 1: Separate the variables. Divide both sides by $y$ and multiply by $dt$:

$$\frac{1}{y},dy = k,dt$$

Step 2: Integrate both sides. The left side integrates to a natural logarithm; the right side is a constant times $t$. Each side carries a constant of integration, which we combine into a single constant $C_1$ on the right:

$$\int \frac{1}{y},dy = \int k,dt \quad\Longrightarrow\quad \ln\lvert y \rvert = kt + C_1$$

Step 3: Exponentiate to free $y$. Raise $e$ to both sides:

$$\lvert y \rvert = e^{kt + C_1} = e^{C_1},e^{kt}$$

The factor $e^{C_1}$ is just a positive constant. Absorbing the sign that comes from dropping the absolute value, write it as a single constant $C$:

$$y = C e^{kt}$$

Step 4: Fix the constant with the initial value. At $t = 0$, $e^{kt} = e^0 = 1$, so $y(0) = C$. The constant is simply the starting amount, which we call $y_0$. This is an initial value problem: the differential equation gives the family of curves, and the starting value picks out one.

$$\boxed{,y = y_0 e^{kt},}$$

How Do You Verify The Solution By Substitution?

A solution to a differential equation is only trustworthy once you put it back in and check that both sides agree. Take $y = y_0 e^{kt}$ and differentiate it with respect to $t$. Because $y_0$ is a constant and the derivative of $e^{kt}$ is $k e^{kt}$ (see derivatives of exponential functions):

$$\frac{dy}{dt} = y_0 \cdot k e^{kt} = k\left(y_0 e^{kt}\right)$$

The bracket is exactly $y$, so

$$\frac{dy}{dt} = k y$$

which is the original equation. Check the initial condition too: $y(0) = y_0 e^{0} = y_0$. Both the equation and the starting value are satisfied, so $y = y_0 e^{kt}$ is confirmed.

This back-substitution is the geometric heart of the model. The curve's slope at every point equals $k$ times its own height, so a tall curve climbs steeply and a short one barely moves. That is what "rate proportional to amount" looks like on a graph, and it is why a direction field for $\frac{dy}{dt} = ky$ shows arrows that steepen as you move away from the horizontal axis.

What Are Half-Life And Doubling Time?

Half-life and doubling time turn the abstract constant $k$ into a number you can feel. Both answer the question, "how long until the amount changes by a fixed factor?" and the surprising result is that the answer does not depend on how much you started with.

Doubling time (growth, $k > 0$). Find the time $t_d$ at which $y$ reaches $2y_0$:

$$2y_0 = y_0 e^{k t_d} ;\Longrightarrow; 2 = e^{k t_d} ;\Longrightarrow; \ln 2 = k t_d ;\Longrightarrow; t_d = \frac{\ln 2}{k}$$

Half-life (decay, $k < 0$). Write $k = -\lambda$ with $\lambda > 0$ so the arithmetic stays positive, and find the time $t_h$ at which $y$ falls to $\tfrac{1}{2}y_0$:

$$\tfrac{1}{2}y_0 = y_0 e^{-\lambda t_h} ;\Longrightarrow; \tfrac{1}{2} = e^{-\lambda t_h} ;\Longrightarrow; -\ln 2 = -\lambda t_h ;\Longrightarrow; t_h = \frac{\ln 2}{\lambda} = \frac{\ln 2}{\lvert k \rvert}$$

The starting amount $y_0$ cancels in both derivations. That is why a physicist can quote "the half-life of carbon-14 is 5,730 years" without ever saying how much carbon there is: every sample halves in the same time.

Table: The one model, read as growth and as decay.

How Do You Use Exponential Growth And Decay? Worked Examples

Each example below is fully stepped, and every decimal is rounded to four places.

Example 1: A growing bacteria culture (growth).

A culture obeys $\frac{dy}{dt} = 0.2y$ with $y(0) = 100$ cells, where $t$ is in hours. Find the population after 10 hours and the doubling time.

The solution is $y = 100 e^{0.2t}$. At $t = 10$:

$$y(10) = 100 e^{0.2 \times 10} = 100 e^{2} = 100 (7.3891) = 738.9056 \text{ cells}$$

The doubling time is

$$t_d = \frac{\ln 2}{0.2} = \frac{0.6931}{0.2} = 3.4657 \text{ hours}$$

Check: at $t = 3.4657$, $y = 100 e^{0.2 \times 3.4657} = 100 e^{0.6931} = 100(2) = 200$, exactly double the start.

Final answer: about $738.9056$ cells after 10 hours; doubling time $3.4657$ hours.

Example 2: Carbon-14 dating (decay).

Carbon-14 has a half-life of 5,730 years. A wooden artefact contains 25% of the carbon-14 it had when the tree was alive. How old is it?

First find the decay constant from the half-life, using $t_h = \frac{\ln 2}{\lambda}$:

$$\lambda = \frac{\ln 2}{5730} = \frac{0.6931}{5730} = 0.0001210 \text{ per year}$$

Now solve $\frac{y}{y_0} = 0.25 = e^{-\lambda t}$ for $t$:

$$\ln(0.25) = -\lambda t ;\Longrightarrow; t = \frac{-\ln(0.25)}{\lambda} = \frac{1.3863}{0.0001210} = 11{,}460 \text{ years}$$

This matches intuition: 25% is one half of one half, so the sample has passed through exactly two half-lives, $2 \times 5730 = 11{,}460$ years.

Final answer: the artefact is about $11{,}460$ years old.

Example 3: Continuously compounded interest (growth).

An amount $P =$1{,}000$is invested at 5$\frac{dA}{dt} = 0.05A$. Find the balance after 10 years.

Continuous compounding is exactly the growth model with $k = r = 0.05$, giving $A = P e^{rt}$:

$$A(10) = 1000, e^{0.05 \times 10} = 1000, e^{0.5} = 1000(1.6487) = 1648.7213$$

Final answer: about $1{,}648.7213$, which rounds to $1{,}648.72$.

How Does Newton's Law Of Cooling Fit The Model?

Newton's law of cooling is the most useful variant of exponential decay, and it shows what happens when a quantity decays toward a value other than zero. A hot object does not cool toward $0$ degrees; it cools toward the temperature of the room around it. If $T$ is the object's temperature and $T_s$ is the surrounding (ambient) temperature, the law says the rate of cooling is proportional to the temperature difference:

$$\frac{dT}{dt} = k\left(T - T_s\right)$$

This is not quite $\frac{dy}{dt} = ky$, but a single substitution makes it so. Let $u = T - T_s$ be the temperature gap. Since $T_s$ is constant, $\frac{du}{dt} = \frac{dT}{dt}$, and the equation becomes $\frac{du}{dt} = k u$, the pure model. Its solution is $u = u_0 e^{kt}$, and substituting $u = T - T_s$ back gives

$$T = T_s + \left(T_0 - T_s\right) e^{kt}$$

where $T_0$ is the starting temperature. Verify by substitution: differentiating gives $\frac{dT}{dt} = k\left(T_0 - T_s\right)e^{kt}$, and the right side $k(T - T_s) = k\left(T_0 - T_s\right)e^{kt}$ matches, so the solution is correct.

Example 4: Cooling coffee.

Coffee at $90^\circ\text{C}$ sits in a $20^\circ\text{C}$ room and cools to $60^\circ\text{C}$ after 10 minutes. What is its temperature after 20 minutes?

Here $T_s = 20$ and $T_0 = 90$, so $T = 20 + 70 e^{kt}$. Use the 10-minute reading to find $k$:

$$60 = 20 + 70 e^{10k} ;\Longrightarrow; e^{10k} = \frac{40}{70} = \frac{4}{7}$$

At $t = 20$ the factor is $e^{20k} = \left(e^{10k}\right)^2 = \left(\tfrac{4}{7}\right)^2 = \tfrac{16}{49}$, so no logarithm is even needed:

$$T(20) = 20 + 70 \cdot \frac{16}{49} = 20 + 22.8571 = 42.8571^\circ\text{C}$$

Final answer: about $42.8571^\circ\text{C}$ after 20 minutes, still well above the $20^\circ\text{C}$ room it is heading toward.

Why Does Exponential Growth And Decay Work?

The model looks almost too simple to describe uranium, bank balances, and cooling tea at once. The reason it does is that all three share one mechanism: the change in the quantity is fed by the quantity itself.

• The rate is a fixed fraction of the whole. Each radioactive atom has the same chance of decaying per second, so with twice as many atoms you get twice as many decays. Each dollar earns the same interest per year, so twice the balance earns twice the interest. "Rate proportional to amount" is the honest description of both, and that is precisely $\frac{dy}{dt} = ky$.

• The exponential is the only function that is its own derivative. Up to the constant $k$, $e^{kt}$ is the unique curve whose slope everywhere equals a fixed multiple of its height. So the moment a process satisfies "rate proportional to amount," its graph has no choice but to be an exponential. For more on this curve on its own, see exponential functions.

• Decay never reaches zero, and growth never levels off. Because $e^{kt}$ is positive for every $t$, decaying quantities approach the axis as an asymptote without touching it, and growing ones climb forever. Real populations eventually run out of food and level off, which is why biologists switch to the logistic model once resources bite; the pure exponential is the early-stage, unlimited-resource picture.

Who Discovered Exponential Growth And Decay?

The exponential curve and the physical laws that ride on it arrived from three very different rooms: an economist worrying about food, a chemist splitting atoms, and a physicist watching things cool.

Two threads set the stage and one closed it:

• Thomas Malthus (1766–1834, England) argued in 1798 that population grows in a "geometric ratio," the discrete cousin of $\frac{dy}{dt} = ky$, an idea that later shaped Darwin's thinking on natural selection.

• Leonhard Euler (1707–1783, Switzerland) gave the constant $e \approx 2.71828$ its central role, making $e^{kt}$ the natural language of continuous growth. His constant is why continuous compounding, not yearly compounding, is the clean limiting case.

Where Is Exponential Growth And Decay Used In The Real World?

One equation quietly runs the clocks and counters of several sciences at once.

• Biology and epidemiology: early-stage bacterial cultures and the opening phase of an epidemic grow exponentially while resources and susceptible hosts are still plentiful, which is why early case counts double on a fixed schedule.

• Nuclear physics and archaeology: radioactive decay dates rocks, fossils, and artefacts through carbon-14 and other isotopes, and sets the storage timelines for nuclear waste.

• Finance: continuously compounded interest and continuously growing dividends are modelled by $A = P e^{rt}$, the backbone of present-value and option-pricing formulas.

• Medicine and pharmacology: most drugs clear the bloodstream by first-order kinetics, so a medicine's "half-life" tells a doctor how to space the doses.

• Forensics: Newton's law of cooling estimates a time of death from how far a body's temperature has fallen toward room temperature.

From a coroner's report to a carbon date to a compound-interest table, the underlying arithmetic is the same first-order equation, which is why exponential growth and decay is one of the most reused ideas a calculus student ever meets.

What Are The Most Common Mistakes With Exponential Growth And Decay?

These four errors account for most lost marks on the topic, and each matches a question real students ask on r/calculus, r/learnmath, and AP review guides such as Albert.io.

Getting the sign of $k$ wrong, or dropping the absolute value in half-life.

Where it slips in:

A student models decay with a positive $k$, so the "decaying" quantity grows, or computes a half-life as $\frac{\ln 2}{k}$ with a negative $k$ and reports a negative time.

Don't do this:

Do not carry a raw negative $k$ into the half-life formula, and do not assume $k > 0$ for a shrinking quantity.

The correct way:

Decide the sign from the physics first: $k > 0$ for growth, $k < 0$ for decay. For half-life use $t_h = \frac{\ln 2}{\lvert k \rvert}$, so the time always comes out positive.

Claiming exponential decay reaches zero.

Where it slips in:

A student says a radioactive sample "will be completely gone" after some finite time, or that cooling coffee "reaches room temperature" at a specific minute.

Don't do this:

Do not treat the asymptote as a value the curve lands on. Because $e^{kt} > 0$ for every $t$, $y = y_0 e^{kt}$ is never actually zero.

The correct way:

Say the quantity approaches zero (or, for cooling, approaches the ambient temperature $T_s$) as a limit. It gets arbitrarily close but never exactly arrives.

Forgetting the ambient shift in Newton's law of cooling.

Where it slips in:

A student writes the coffee's temperature as $T = T_0 e^{kt}$, sending it toward $0^\circ$ instead of toward the room.

Don't do this:

Do not apply the bare model $y = y_0 e^{kt}$ to a temperature. The quantity that decays exponentially is the gap $T - T_s$, not $T$ itself.

The correct way:

Substitute $u = T - T_s$, solve $u = u_0 e^{kt}$, then add $T_s$ back: $T = T_s + (T_0 - T_s)e^{kt}$.

Confusing exponential growth with logistic growth.

Where it slips in:

A student uses $y = y_0 e^{kt}$ to predict a population years into the future and forecasts an impossible, unbounded number.

Don't do this:

Do not assume "rate proportional to amount" holds once resources run short. The pure model has no ceiling.

The correct way:

Use $\frac{dy}{dt} = ky$ only for the early, unlimited-resource phase. When a carrying capacity $M$ matters, switch to the logistic equation $\frac{dy}{dt} = ky\left(1 - \frac{y}{M}\right)$, whose growth slows as $y$ nears $M$.

Practice Problems On Exponential Growth And Decay

Work each one, then check against the answer. Answers are verified, with decimals to four places.

1. Solve $\frac{dy}{dt} = 3y$ with $y(0) = 5$.
   (Answer: $y = 5e^{3t}$.)

2. A population follows $\frac{dy}{dt} = 0.1y$ with $y(0) = 200$. Find $y(5)$.
   (Answer: $200e^{0.5} = 329.7443$.)

3. Find the doubling time for $k = 0.04$ per year.
   (Answer: $\frac{\ln 2}{0.04} = 17.3287$ years.)

4. A substance has a half-life of 8 hours. Find its decay constant $\lambda$.
   (Answer: $\lambda = \frac{\ln 2}{8} = 0.0866$ per hour.)

5. A sample decays to 12.5% of its original amount. How many half-lives have passed?
   (Answer: $0.125 = \left(\tfrac{1}{2}\right)^3$, so 3 half-lives.)

6. Coffee at $80^\circ\text{C}$ in a $25^\circ\text{C}$ room cools to $55^\circ\text{C}$ in 6 minutes. Write its cooling equation.
   (Answer: $T = 25 + 55e^{kt}$ with $e^{6k} = \frac{30}{55} = \frac{6}{11}$.)

Where Should You Go Next After Exponential Growth And Decay?

This model is your first real differential equation, and several natural doors open from here.

1. Separable differential equations. Master the general method behind the trick used here, then apply it to equations that are not $\frac{dy}{dt} = ky$.

2. Initial value problems. Go deeper on how a starting condition selects one curve from the family of solutions.

3. Ordinary differential equations. See where this fits in the wider theory of first- and second-order equations.

If your child is meeting exponential models for the first time, a live Bhanzu trainer teaches them from the "rate proportional to amount" idea up, so the equation and its solution feel like one story, in the Bhanzu math program.