**Contents**:

## Exponential Growth

**Exponential growth** is an increase in some quantity that follows the relationship

**N(t) = A e ^{(kt)}**

where *A* and *k* are positive real-valued constants.

Before diving further into the mathematics, let’s look at a graphical example of exponential growth. This plot assumes that A = 3 and k = 1.

The function’s initial value at t=0 is A=3. The variable *k* is the growth constant. The larger the value of *k*, the faster the growth will occur.

## Differential Equation

The exponential behavior explored above is the solution to the differential equation below:

**dN/dt = kN**

The differential equation states that exponential change in a population is directly proportional to its size. Initially, the small population (3 in the above graph) is growing at a relatively slow rate. However, as the population grows, the growth rate increases rapidly.

## Exponential Growth: Example Problems

Exponential growth can be found in a range of natural phenomena, from the growth of bacterial populations to the speed of computer processors.

Problem 1: A colony of bacteria doubles its population every 4 hours. If the colony originally has ten bacteria, how large will the colony be 24 hours later?

Solution: Since the colony has an original population of 10, then A=10. Knowing that the population will be 20 four hours later, we can solve for the growth constant.

- N(t) = A e
^{(kt)} - 20 = 10 e
^{(k*4 hours)} - ln(2) = (4 hours)*k
- k =
**0.173 /hours**

Then, the growth constant can be used to determine the population’s size one day later.

- N(24 hours) = 10 e
^{(0.173 /hours * 24 hours)} - N(24 hours) =
**635**

Amazingly, the original handful of bacteria will blossom into a colony of nearly a thousand in one day’s time. That’s the power of exponential growth.

Problem 2: A client deposits $100 in a savings account at the local bank. The account grows by 1% interest, compounded annually. What will the value of the account be after ten years?

Solution: The initial size of the account is $100, so A=100. The account’s value will be $101 after one year, due to the interest. Knowing this, we can calculate the growth constant.

N(t) = A e^{(kt)}

101 = 100 e^{(k*1 year)}

ln(1.01) = (1 year)*k

k = **0.00995 /years**

To find the value of the account at ten years, t=10.

N(10 years) = $100 e^{(0.00995/years * 10 years)}

N(10 years) = **$110.46**

## Exponential Decay

**Exponential decay** is a decrease in a quantity that follows the mathematical relationship. In many ways you can think of it as the opposite of exponential growth: where exponential growth goes up, exponential growth goes down.

**N(t) = A e ^{(-kt)}**

where A and k are positive, real-valued constants.

The following graph shows exponential decay, where A = 5 and k = 1.

The function’s initial value at t=0 is A=5. *k* is a variable that represents the decay constant. The larger the value of *k*, the faster the decay will occur.

## Half-Life in Exponential Decay

The **half-life** is the time after which half of the original population has decayed. From the language of our original exponential decay equation, the half-life is the time at which the population’s size is A/2. Then, by plugging this value into our equation, we arrive at an expression for the half-life:

- A/2 = A e
^{(-kt)} - ½ = e
^{(-kt)} - ln(0.5) = -kt
**t = -ln(0.5)/k**

Problem 1: The half-life of carbon-14 is 5,730 years. What is its decay constant?

Solution: To solve, we can use the equation for half-life.

- t = -ln(0.5)/k
- 5730 = 0.693/k
- k =
**1.21 * 10**^{(-4)}/years

## Differential Equation

The exponential behavior explored above is the solution to the differential equation:

**dN/dt = -kN**

The differential equation states that exponential change in a population is directly proportional to its size. For a large population, the decay is rapid. In contrast, as the population shrinks in size, the rate of decay becomes slower.

**Radioactivity** is the most common natural example of exponential decay. Over time, an unstable atom will eject particles from its nucleus. As these particles discharge, less radioactive material remains. Carbon dating, cancer therapies, and x-ray machines all involve radioactivity.

Problem 2: A wooden tool recovered from an archaeological site contains 65% of its original composition of carbon-14. Approximately how long ago was the wooden tool created?

Solution: Because carbon-14 decays due to radioactivity, we can use the exponential decay equation.

N(t) = A e^{(-kt)}.

From Problem 1, the decay constant for carbon-14 is 1.21 * 10^{(-4)} /years. Additionally, if the original population of carbon-14 is A, then N(t) = 0.65A.

0.65A = A e^{(-1.21 * 10(-4) t) }

ln(0.65) = -1.21 * 10^{(-4)} t

t = **3560 years ago**

## Exponential Growth and Decay: References

Matthews, John A. “Exponential Growth.” 2014: 387–387. Print.

“Exponential Decay.” The Penguin Dictionary of Physics. United Kingdom, Penguin, 2000.

**CITE THIS AS:**

**Stephanie Glen**. "Exponential Growth & Decay: Simple Definition, Step by Step Examples" From

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