Exponential Functions

Model explosive growth and decay with exponents

You have probably heard someone say that something is “growing exponentially.” Maybe it was about the spread of a viral video, the growth of a savings account, or even the spread of an actual virus. People use this phrase to mean “really, really fast,” but there is something more specific going on. Exponential growth is not just fast growth; it is growth that accelerates. The more you have, the more you gain. The bigger something gets, the faster it gets even bigger.

You have likely experienced this yourself. If you have ever watched a small savings account slowly accumulate interest, you know that the first few years feel painfully slow. But leave that money alone long enough, and suddenly the interest is earning interest, and the growth starts to feel almost magical. That is exponential growth in action, and understanding it mathematically gives you power: the power to predict, to plan, and to make better decisions with money, time, and resources.

In this lesson, we move beyond the exponent rules you learned in Algebra 1. There, exponents were just a way to write repeated multiplication. Here, we let the exponent become a variable. That single change transforms exponents from a shorthand into a function, one that models some of the most dramatic phenomena in nature, finance, and science.

Core Concepts

Exponential vs. Linear and Quadratic Growth

To appreciate what makes exponential functions special, let us compare them to the functions you already know.

Linear functions grow by adding the same amount each time. If you earn $50 per day, your total after $x$ days is $f(x) = 50x$. Each day adds $50, no matter how many days have passed. The growth is steady and predictable.

Quadratic functions grow by adding increasingly larger amounts. The function $f(x) = x^2$ goes 1, 4, 9, 16, 25… The jumps between values get bigger (3, 5, 7, 9…), but they grow at a constant rate of increase.

Exponential functions grow by multiplying by the same factor each time. If a population doubles every year, you do not add the same number each year; you multiply by 2 each year. Starting with 100 individuals: 100, 200, 400, 800, 1600… The jumps themselves are doubling.

Here is a comparison to make this concrete:

$x$ Linear: $f(x) = 10x$ Quadratic: $f(x) = x^2$ Exponential: $f(x) = 2^x$
1 10 1 2
2 20 4 4
3 30 9 8
5 50 25 32
10 100 100 1,024
20 200 400 1,048,576

Notice how the exponential function starts slowly but eventually leaves the others far behind. By $x = 20$, the exponential value is over a million, while linear is at 200 and quadratic is at 400.

The General Form: $f(x) = ab^x$

An exponential function has the form:

$$f(x) = ab^x$$

where:

  • $a$ is the initial value (the value when $x = 0$, since $b^0 = 1$)
  • $b$ is the base (must be positive and not equal to 1)
  • $x$ is the exponent (this is what makes it exponential: the variable is in the exponent)

Why must $b > 0$ and $b \neq 1$? If $b$ were negative, we would get undefined or imaginary values for some inputs (like $(-2)^{1/2}$). If $b = 1$, we would just have $f(x) = a \cdot 1^x = a$, a boring horizontal line.

Growth vs. Decay

The value of the base $b$ determines whether the function grows or decays:

Growth ($b > 1$): When the base is greater than 1, the function increases as $x$ increases. Each time $x$ goes up by 1, the output multiplies by $b$. If $b = 2$, the function doubles. If $b = 1.05$, it increases by 5%.

Decay ($0 < b < 1$): When the base is between 0 and 1, the function decreases as $x$ increases. Each time $x$ goes up by 1, the output multiplies by a fraction, making it smaller. If $b = 0.5$, the function halves. If $b = 0.92$, it decreases by 8%.

Here is the key insight: a decay factor like 0.92 means “keep 92% of what you had,” which is the same as “lose 8%.” A growth factor like 1.05 means “keep 100% plus add 5% more.”

The Natural Base $e$

There is one base that appears so often in mathematics, science, and finance that it gets its own letter: $e$.

$$e \approx 2.71828…$$

Like $\pi$, the number $e$ is irrational; its decimal expansion goes on forever without repeating. But why is this particular number so special?

Imagine you put $1 in a bank that offers 100% annual interest (a fantasy, but bear with me). If they compound once at the end of the year, you get $2. If they compound twice (50% each time), you get $1.5 \times 1.5 = \$2.25$. If they compound monthly (about 8.33% each time), you get even more. As you compound more and more frequently, the amount approaches a limit:

$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$

This is why $e$ appears naturally whenever something grows (or decays) continuously. It is not a number mathematicians invented; it is a number that nature seems to prefer.

Transformations of Exponential Functions

Just like other functions, exponential functions can be shifted and stretched:

Vertical shift: $f(x) = b^x + k$ shifts the graph up (if $k > 0$) or down (if $k < 0$). The horizontal asymptote moves from $y = 0$ to $y = k$.

Horizontal shift: $f(x) = b^{x-h}$ shifts the graph right (if $h > 0$) or left (if $h < 0$).

Vertical stretch/compression: $f(x) = ab^x$ stretches vertically if $|a| > 1$ and compresses if $|a| < 1$. If $a < 0$, the graph reflects across the x-axis.

Reflection: $f(x) = b^{-x}$ reflects the graph across the y-axis. This turns growth into decay and vice versa: $2^{-x} = \left(\frac{1}{2}\right)^x$.

Compound Interest

One of the most practical applications of exponential functions is compound interest. When interest is compounded (meaning the interest earns interest), your money grows exponentially.

The compound interest formula is:

$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$

where:

  • $A$ = final amount
  • $P$ = principal (initial investment)
  • $r$ = annual interest rate (as a decimal)
  • $n$ = number of times interest is compounded per year
  • $t$ = time in years

Notice the structure: $\left(1 + \frac{r}{n}\right)$ is the growth factor per compounding period, and $nt$ is the total number of compounding periods.

Continuous Compounding

What happens if interest is compounded not monthly, not daily, not hourly, but continuously? As the number of compounding periods approaches infinity, we get:

$$A = Pe^{rt}$$

This formula uses $e$ because continuous compounding is exactly the kind of situation where $e$ arises naturally. In practice, continuous compounding does not give you much more money than daily compounding, but the formula is simpler and appears throughout physics, biology, and engineering.

Exponential Growth and Decay Models

Many real-world situations follow exponential patterns and can be modeled with:

Exponential growth: $A(t) = A_0 \cdot b^t$ or $A(t) = A_0 e^{kt}$ (where $k > 0$)

Exponential decay: $A(t) = A_0 \cdot b^t$ (where $0 < b < 1$) or $A(t) = A_0 e^{kt}$ (where $k < 0$)

Two special measurements describe how fast exponential processes occur:

Doubling time: The time it takes for a quantity to double. If something doubles every $T$ years, the model is $A(t) = A_0 \cdot 2^{t/T}$.

Half-life: The time it takes for a quantity to become half its original amount. If something has a half-life of $T$ years, the model is $A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{t/T}$.

Notation and Terminology

Term Meaning Example
Exponential function A function of the form $f(x) = ab^x$ $f(x) = 2 \cdot 3^x$
Growth factor The base $b$ when $b > 1$ 1.05 means 5% growth per period
Decay factor The base $b$ when $0 < b < 1$ 0.92 means 8% decay per period
$e$ The natural base, $\approx 2.71828…$ Used in continuous growth models
Half-life Time for a quantity to halve Radioactive decay, drug metabolism
Doubling time Time for a quantity to double Population growth, investments
Asymptote A line the graph approaches but never reaches For $y = 2^x$, the asymptote is $y = 0$

Examples

Example 1: Graphing a Basic Exponential Function

Graph $f(x) = 2^x$ and identify its key features.

Step 1: Create a table of values.

$x$ $f(x) = 2^x$
-2 $2^{-2} = \frac{1}{4} = 0.25$
-1 $2^{-1} = \frac{1}{2} = 0.5$
0 $2^0 = 1$
1 $2^1 = 2$
2 $2^2 = 4$
3 $2^3 = 8$

Step 2: Identify key features.

  • y-intercept: $(0, 1)$ because $2^0 = 1$
  • Horizontal asymptote: $y = 0$ (the x-axis). As $x \to -\infty$, the function approaches 0 but never reaches it.
  • Domain: All real numbers $(-\infty, \infty)$
  • Range: All positive numbers $(0, \infty)$
  • Behavior: Increasing (growth function since $b = 2 > 1$)
  • The graph passes through $(1, 2)$ since $2^1 = 2$

Answer: The graph rises from left to right, passing through $(0, 1)$, approaching but never touching the x-axis on the left, and increasing rapidly on the right.

Example 2: Identifying Growth or Decay

Determine whether $f(x) = 0.7^x$ represents exponential growth or decay.

Step 1: Identify the base.

The function is in the form $f(x) = ab^x$ where $a = 1$ and $b = 0.7$.

Step 2: Check the value of $b$.

Since $0 < 0.7 < 1$, this is an exponential decay function.

Step 3: Interpret what this means.

Each time $x$ increases by 1, the output is multiplied by 0.7. This means the function retains 70% of its previous value (or loses 30%) with each unit increase in $x$.

Answer: This is exponential decay. The function decreases as $x$ increases.

Example 3: Writing an Exponential Equation from a Description

Write an exponential function for the following scenario: An investment starts at $500 and grows at 8% per year.

Step 1: Identify the initial value.

The investment starts at $500, so $a = 500$.

Step 2: Determine the growth factor.

Growing at 8% per year means the investment keeps 100% and gains an additional 8%, so the growth factor is: $$b = 1 + 0.08 = 1.08$$

Step 3: Write the function.

Let $t$ represent the number of years. The exponential function is: $$A(t) = 500 \cdot (1.08)^t$$

Step 4: Verify with a quick check.

After 1 year: $A(1) = 500 \cdot 1.08 = 540$. This is indeed $500 plus 8% of $500 ($40).

Answer: $A(t) = 500(1.08)^t$

Example 4: Compound Interest Calculation

You invest $1000 at 6% annual interest, compounded quarterly, for 5 years. How much will you have?

Step 1: Identify the values for the compound interest formula.

  • $P = 1000$ (principal)
  • $r = 0.06$ (6% as a decimal)
  • $n = 4$ (compounded quarterly means 4 times per year)
  • $t = 5$ (years)

Step 2: Apply the compound interest formula.

$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$

$$A = 1000\left(1 + \frac{0.06}{4}\right)^{4 \cdot 5}$$

Step 3: Simplify inside the parentheses.

$$A = 1000\left(1 + 0.015\right)^{20}$$

$$A = 1000(1.015)^{20}$$

Step 4: Calculate.

$(1.015)^{20} \approx 1.34686…$

$$A \approx 1000 \times 1.34686 = 1346.86$$

Answer: After 5 years, you will have approximately $1,346.86.

Example 5: Finding Time to Reach a Target Population

A population of 10,000 grows at 3% per year. When will it reach 25,000?

Step 1: Set up the exponential equation.

With initial population 10,000 and growth rate 3%: $$P(t) = 10000 \cdot (1.03)^t$$

We want to find $t$ when $P(t) = 25000$: $$25000 = 10000 \cdot (1.03)^t$$

Step 2: Isolate the exponential expression.

Divide both sides by 10000: $$2.5 = (1.03)^t$$

Step 3: Take the natural logarithm of both sides.

$$\ln(2.5) = \ln\left((1.03)^t\right)$$

Using the logarithm power rule, $\ln(a^b) = b \cdot \ln(a)$: $$\ln(2.5) = t \cdot \ln(1.03)$$

Step 4: Solve for $t$.

$$t = \frac{\ln(2.5)}{\ln(1.03)}$$

$$t = \frac{0.9163…}{0.02956…} \approx 31.0$$

Answer: The population will reach 25,000 in approximately 31 years.

Example 6: Carbon Dating with Half-Life

Carbon-14 has a half-life of 5730 years. If a sample contains 40% of its original Carbon-14, how old is the sample?

Step 1: Set up the half-life model.

The amount remaining after time $t$ with half-life $T$ is: $$A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{t/T}$$

With $T = 5730$ years and the remaining amount being 40% of the original ($A = 0.40 A_0$): $$0.40 A_0 = A_0 \cdot \left(\frac{1}{2}\right)^{t/5730}$$

Step 2: Simplify by dividing both sides by $A_0$.

$$0.40 = \left(\frac{1}{2}\right)^{t/5730}$$

Step 3: Take the natural logarithm of both sides.

$$\ln(0.40) = \ln\left(\left(\frac{1}{2}\right)^{t/5730}\right)$$

$$\ln(0.40) = \frac{t}{5730} \cdot \ln\left(\frac{1}{2}\right)$$

Step 4: Solve for $t$.

$$t = \frac{\ln(0.40)}{\ln(0.5)} \cdot 5730$$

$$t = \frac{-0.9163…}{-0.6931…} \cdot 5730$$

$$t \approx 1.322 \cdot 5730 \approx 7576$$

Answer: The sample is approximately 7,576 years old.

Key Properties and Rules

Property Description Example
y-intercept Always $(0, a)$ for $f(x) = ab^x$ $f(x) = 5 \cdot 2^x$ has y-intercept $(0, 5)$
Horizontal asymptote $y = 0$ for basic form; $y = k$ for $f(x) = ab^x + k$ $f(x) = 2^x + 3$ has asymptote $y = 3$
Domain All real numbers $(-\infty, \infty)$
Range $(0, \infty)$ for $a > 0$; $(-\infty, 0)$ for $a < 0$ Always positive (or always negative)
Growth condition $b > 1$ $f(x) = 3^x$ increases
Decay condition $0 < b < 1$ $f(x) = 0.5^x$ decreases
One-to-one Passes horizontal line test Every output comes from exactly one input

Converting between growth rate and growth factor:

  • Growth factor from percentage: If something grows by $r%$, the growth factor is $b = 1 + \frac{r}{100}$
  • Decay factor from percentage: If something decays by $r%$, the decay factor is $b = 1 - \frac{r}{100}$

Useful logarithm relationships for solving exponential equations:

  • $\ln(e^x) = x$
  • $e^{\ln(x)} = x$
  • $\log_b(b^x) = x$
  • $b^{\log_b(x)} = x$

Real-World Applications

Population Growth

Populations often grow exponentially when resources are abundant. If a city grows at 2.5% per year:

$$P(t) = P_0 \cdot (1.025)^t$$

This model explains why population growth can seem manageable for decades and then suddenly feel overwhelming. The actual number of people added each year keeps increasing, even though the percentage stays the same.

Compound Interest and Investments

Your retirement account, the national debt, and credit card balances all grow exponentially. The compound interest formula:

$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$

This is why financial advisors stress starting to save early. Someone who invests $5,000 at age 25 will have more at retirement than someone who invests $10,000 at age 45, assuming the same interest rate.

Radioactive Decay

Radioactive isotopes decay exponentially, each with a characteristic half-life. This predictable decay allows scientists to date ancient artifacts (Carbon-14), treat cancer (Iodine-131), and generate nuclear power (Uranium-235).

$$A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{t/T}$$

Bacteria and Virus Spread

In ideal conditions, bacteria divide and double at regular intervals. A single bacterium can become billions in a day. Similarly, early stages of disease outbreaks often show exponential growth in cases, which is why early intervention is so critical.

Depreciation

Cars, computers, and most equipment lose value exponentially. A car that depreciates 15% per year follows:

$$V(t) = V_0 \cdot (0.85)^t$$

This model explains why a new car loses so much value the moment you drive it off the lot. The first year’s 15% loss is the biggest absolute dollar amount; subsequent years’ depreciation is 15% of an already-smaller value.

Self-Test Problems

Problem 1: Identify whether $f(x) = 1.15^x$ represents growth or decay, and interpret the rate.

Show Answer

Since $b = 1.15 > 1$, this is exponential growth. The function grows by 15% for each unit increase in $x$ (because $1.15 = 1 + 0.15$).

Problem 2: Write an exponential function for a quantity that starts at 800 and decreases by 12% per year.

Show Answer

Initial value: $a = 800$

Decay factor: $b = 1 - 0.12 = 0.88$ (keeping 88% means losing 12%)

Function: $A(t) = 800 \cdot (0.88)^t$

Problem 3: Calculate the value of $2,500 invested at 4% annual interest, compounded monthly, after 3 years.

Show Answer

Using $A = P\left(1 + \frac{r}{n}\right)^{nt}$:

$A = 2500\left(1 + \frac{0.04}{12}\right)^{12 \cdot 3}$

$A = 2500(1.00333…)^{36}$

$A = 2500 \times 1.1273… \approx \$2,818.31$

Problem 4: What is the horizontal asymptote of $g(x) = 3 \cdot 2^x - 5$?

Show Answer

For a function of the form $f(x) = ab^x + k$, the horizontal asymptote is $y = k$.

Therefore, the horizontal asymptote of $g(x) = 3 \cdot 2^x - 5$ is $y = -5$.

Problem 5: A radioactive substance has a half-life of 20 days. If you start with 100 grams, how much remains after 60 days?

Show Answer

Using the half-life formula: $A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{t/T}$

$A(60) = 100 \cdot \left(\frac{1}{2}\right)^{60/20}$

$A(60) = 100 \cdot \left(\frac{1}{2}\right)^{3}$

$A(60) = 100 \cdot \frac{1}{8} = 12.5$ grams

Alternatively: After 20 days: 50g. After 40 days: 25g. After 60 days: 12.5g.

Problem 6: How long will it take for an investment to double if it earns 5% interest compounded annually?

Show Answer

We need $2P = P(1.05)^t$, which simplifies to $2 = (1.05)^t$.

Taking the natural log of both sides: $\ln(2) = t \cdot \ln(1.05)$

$t = \frac{\ln(2)}{\ln(1.05)} = \frac{0.6931…}{0.04879…} \approx 14.2$ years

(Note: The “Rule of 72” approximation gives $72 \div 5 = 14.4$ years, which is close!)

Problem 7: A population of bacteria doubles every 4 hours. If there are initially 500 bacteria, how many will there be after 1 day (24 hours)?

Show Answer

Using the doubling model: $P(t) = P_0 \cdot 2^{t/T}$ where $T = 4$ hours.

$P(24) = 500 \cdot 2^{24/4}$

$P(24) = 500 \cdot 2^6$

$P(24) = 500 \cdot 64 = 32,000$ bacteria

Summary

  • Exponential functions have the form $f(x) = ab^x$, where the variable is in the exponent. This is fundamentally different from polynomial functions where the variable is in the base.

  • Growth vs. decay depends on the base: if $b > 1$, the function grows; if $0 < b < 1$, the function decays. The growth/decay factor $b$ tells you what fraction of the previous value you have after one unit of time.

  • The natural base $e \approx 2.718…$ appears naturally in continuous growth situations. It is to exponential functions what $\pi$ is to circles: a fundamental constant that emerges from the mathematics itself.

  • Key features of exponential graphs include: passing through $(0, a)$, having a horizontal asymptote at $y = 0$ (or $y = k$ if shifted), domain of all real numbers, and range of all positive numbers (for $a > 0$).

  • Compound interest follows the formula $A = P\left(1 + \frac{r}{n}\right)^{nt}$, and continuous compounding simplifies to $A = Pe^{rt}$.

  • Half-life and doubling time are convenient ways to describe exponential processes: $A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{t/T}$ for half-life; $A(t) = A_0 \cdot 2^{t/T}$ for doubling time.

  • To solve exponential equations for the variable in the exponent, use logarithms. The key step is taking $\ln$ (or $\log$) of both sides and using the power rule: $\ln(b^x) = x \cdot \ln(b)$.

  • Exponential models appear everywhere: population dynamics, radioactive decay, interest calculations, disease spread, depreciation, and countless other phenomena where the rate of change is proportional to the current amount.