Exponential Functions

Understand explosive growth and rapid decay

Have you ever wondered why a viral video can go from a few hundred views to millions seemingly overnight? Or why financial advisors insist that starting to save in your twenties makes such a dramatic difference compared to waiting until your thirties? Or why a single bacterium can turn into billions in just a day? The answer to all of these questions lies in exponential functions.

Exponential functions describe situations where quantities grow (or shrink) by a consistent percentage over equal time intervals. Unlike linear growth, where you add the same amount each time, exponential growth multiplies by the same factor each time. This difference might seem subtle, but it leads to dramatically different outcomes. A quantity that doubles every hour does not just grow twice as fast as something that adds a fixed amount each hour; given enough time, it completely dwarfs it.

In this chapter, we are going to explore these powerful functions, understand their properties, and see how they model everything from population growth to radioactive decay to the interest accumulating in your bank account.

Core Concepts

What Is an Exponential Function?

An exponential function is a function where the variable appears in the exponent rather than the base. The general form is:

$$f(x) = a \cdot b^x$$

Where:

  • $a$ is the initial value (the value of the function when $x = 0$)
  • $b$ is the base, which must satisfy $b > 0$ and $b \neq 1$
  • $x$ is the exponent (the independent variable)

Why the restrictions on $b$? If $b$ were negative, we would get complex numbers for some values of $x$ (like $(-2)^{1/2}$). If $b$ were 0, the function would just be 0 for all positive $x$. And if $b$ were 1, we would have $f(x) = a \cdot 1^x = a$, which is just a constant, not particularly interesting.

Growth vs. Decay

The behavior of an exponential function depends entirely on the value of the base $b$:

Exponential Growth ($b > 1$): When the base is greater than 1, each time $x$ increases by 1, the function value multiplies by $b$. The function increases without bound as $x$ gets larger.

$$f(x) = 2^x \quad \text{doubles each time } x \text{ increases by 1}$$

Exponential Decay ($0 < b < 1$): When the base is between 0 and 1, each time $x$ increases by 1, the function value multiplies by a number less than 1, making it smaller. The function approaches 0 as $x$ gets larger.

$$f(x) = \left(\frac{1}{2}\right)^x \quad \text{halves each time } x \text{ increases by 1}$$

Here is a helpful way to remember: if you are multiplying by something greater than 1, you are getting bigger. If you are multiplying by something between 0 and 1, you are getting smaller.

The Natural Base $e$

Among all possible bases for exponential functions, one number holds a special place in mathematics: the number $e$, approximately equal to 2.71828. This might seem like an arbitrary choice, but $e$ appears naturally in countless mathematical and scientific contexts.

The number $e$ can be defined in several equivalent ways:

$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828…$$

Think of it this way: if you invested $1 at 100% interest for one year, compounded more and more frequently, you would approach exactly $e$ dollars.

The function $f(x) = e^x$ is called the natural exponential function. It has a remarkable property: its rate of change at any point equals its value at that point. This makes it incredibly useful in calculus and differential equations. For now, just know that $e$ is a special number that appears throughout advanced mathematics, much like $\pi$ appears in geometry.

Notation and Terminology

Term Symbol Meaning
Exponential function $f(x) = a \cdot b^x$ A function with the variable in the exponent
Base $b$ The number being raised to a power (must be positive, not 1)
Initial value $a$ The function value when $x = 0$; also called the coefficient
Natural base $e$ Approximately 2.71828…; the base of natural exponentials
Growth factor $b$ (when $b > 1$) The multiplier for each unit increase in $x$
Decay factor $b$ (when $0 < b < 1$) The multiplier for each unit increase in $x$
Asymptote horizontal line $y = 0$ The line the graph approaches but never touches
Compound interest $A = P(1 + r/n)^{nt}$ Interest calculated on principal and accumulated interest
Continuous compounding $A = Pe^{rt}$ Interest compounded infinitely often

Graphing Exponential Functions

The Basic Shape

All exponential functions $f(x) = b^x$ (where $b > 0$ and $b \neq 1$) share some common features:

  1. Domain: All real numbers (you can plug in any value for $x$)
  2. Range: All positive real numbers (the output is always positive)
  3. $y$-intercept: The point $(0, 1)$ since $b^0 = 1$
  4. Horizontal asymptote: The $x$-axis ($y = 0$). The graph gets closer and closer to this line but never touches it.
  5. No $x$-intercept: Since the output is always positive, the graph never crosses the $x$-axis.

For $f(x) = a \cdot b^x$, the $y$-intercept shifts to $(0, a)$, and the horizontal asymptote remains at $y = 0$ (assuming $a > 0$).

Transformations

Just like other functions, exponential functions can be transformed:

Vertical shifts: $f(x) = b^x + k$

  • Shifts the graph up by $k$ units (if $k > 0$) or down (if $k < 0$)
  • The horizontal asymptote moves to $y = k$

Horizontal shifts: $f(x) = b^{x - h}$

  • Shifts the graph right by $h$ units (if $h > 0$) or left (if $h < 0$)

Vertical stretches and reflections: $f(x) = a \cdot b^x$

  • If $|a| > 1$, the graph is stretched vertically
  • If $0 < |a| < 1$, the graph is compressed vertically
  • If $a < 0$, the graph is reflected across the $x$-axis

Horizontal reflections: $f(x) = b^{-x}$

  • Reflects the graph across the $y$-axis
  • Note: $b^{-x} = \left(\frac{1}{b}\right)^x$, so reflection converts growth to decay and vice versa
Example 1: Evaluating Exponential Functions

Evaluate $f(x) = 3 \cdot 2^x$ at $x = -1$, $x = 0$, $x = 1$, and $x = 4$.

For $x = -1$: $$f(-1) = 3 \cdot 2^{-1} = 3 \cdot \frac{1}{2} = \frac{3}{2} = 1.5$$

For $x = 0$: $$f(0) = 3 \cdot 2^0 = 3 \cdot 1 = 3$$

For $x = 1$: $$f(1) = 3 \cdot 2^1 = 3 \cdot 2 = 6$$

For $x = 4$: $$f(4) = 3 \cdot 2^4 = 3 \cdot 16 = 48$$

Notice how each time $x$ increases by 1, the output doubles. The initial value (when $x = 0$) is 3.

Example 2: Identifying Growth vs. Decay

Classify each function as exponential growth or decay, and identify the initial value:

(a) $f(x) = 5 \cdot (1.08)^x$

(b) $g(x) = 100 \cdot (0.75)^x$

(c) $h(x) = 2^{-x}$

Solutions:

(a) Since $b = 1.08 > 1$, this is exponential growth. The initial value is $a = 5$.

(b) Since $b = 0.75$ and $0 < 0.75 < 1$, this is exponential decay. The initial value is $a = 100$.

(c) We can rewrite this as $h(x) = \left(\frac{1}{2}\right)^x$. Since $\frac{1}{2} < 1$, this is exponential decay. The initial value is $a = 1$.

Example 3: Graphing with Transformations

Describe the transformations applied to $f(x) = 2^x$ to obtain $g(x) = 2^{x-3} + 1$, and identify the new horizontal asymptote and $y$-intercept.

Transformations:

  • The $(x - 3)$ in the exponent shifts the graph 3 units to the right.
  • The $+1$ outside shifts the graph 1 unit up.

Horizontal asymptote: The original asymptote $y = 0$ shifts up by 1, so the new asymptote is $y = 1$.

$y$-intercept: Set $x = 0$: $$g(0) = 2^{0-3} + 1 = 2^{-3} + 1 = \frac{1}{8} + 1 = \frac{9}{8} = 1.125$$

The $y$-intercept is $(0, 1.125)$.

Compound Interest

One of the most practical applications of exponential functions is compound interest. When you deposit money in a savings account, you earn interest not just on your original deposit, but also on the interest you have already earned. This is called compound interest.

The Compound Interest Formula

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

Where:

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

Common compounding frequencies:

  • Annually: $n = 1$
  • Semi-annually: $n = 2$
  • Quarterly: $n = 4$
  • Monthly: $n = 12$
  • Daily: $n = 365$

Continuous Compounding

What happens if we compound interest not just daily, but every hour? Every second? Every instant? As the compounding frequency approaches infinity, we get continuous compounding:

$$A = Pe^{rt}$$

Where $e \approx 2.71828$ is the natural base. This formula gives the maximum amount your money can grow to for a given interest rate and time period.

Example 4: Comparing Compounding Frequencies

You invest $5,000 at 6% annual interest for 10 years. Calculate the final amount if interest is compounded (a) annually, (b) monthly, and (c) continuously.

Given: $P = 5000$, $r = 0.06$, $t = 10$

(a) Annually ($n = 1$): $$A = 5000\left(1 + \frac{0.06}{1}\right)^{1 \cdot 10} = 5000(1.06)^{10}$$ $$A = 5000 \times 1.7908… \approx \$8,954.24$$

(b) Monthly ($n = 12$): $$A = 5000\left(1 + \frac{0.06}{12}\right)^{12 \cdot 10} = 5000(1.005)^{120}$$ $$A = 5000 \times 1.8194… \approx \$9,096.98$$

(c) Continuously: $$A = 5000 \cdot e^{0.06 \times 10} = 5000 \cdot e^{0.6}$$ $$A = 5000 \times 1.8221… \approx \$9,110.59$$

Comparison: More frequent compounding yields more money, but with diminishing returns. The difference between monthly and continuous compounding is only about $14 over 10 years.

Example 5: Finding Time to Double Your Money

How long does it take for an investment to double if it earns 8% annual interest compounded continuously?

Setting up the equation: We want to find $t$ when $A = 2P$ (the amount doubles the principal).

$$2P = Pe^{0.08t}$$

Solving: Divide both sides by $P$: $$2 = e^{0.08t}$$

To solve for $t$, we need to “undo” the exponential. We take the natural logarithm of both sides: $$\ln(2) = 0.08t$$

$$t = \frac{\ln(2)}{0.08} = \frac{0.6931…}{0.08} \approx 8.66 \text{ years}$$

Answer: It takes approximately 8.66 years (or about 8 years and 8 months) to double your money at 8% continuous interest.

The Rule of 72: A quick approximation for doubling time is to divide 72 by the interest rate percentage. For 8%: $72 \div 8 = 9$ years. This is close to our exact answer of 8.66 years.

Solving Exponential Equations

Sometimes you need to find what exponent makes an equation true. The simplest method works when you can express both sides of the equation with the same base.

The Same-Base Method

If $b^m = b^n$, then $m = n$ (assuming $b > 0$ and $b \neq 1$).

This works because exponential functions are one-to-one: each input gives a unique output, and each output comes from a unique input.

Example 6: Solving Exponential Equations (Same Base)

Solve each equation:

(a) $2^{3x} = 32$

Step 1: Express 32 as a power of 2. $$32 = 2^5$$

Step 2: Rewrite the equation with the same base on both sides: $$2^{3x} = 2^5$$

Step 3: Since the bases are equal, the exponents must be equal: $$3x = 5$$ $$x = \frac{5}{3}$$

(b) $9^{x+1} = 27$

Step 1: Express both sides as powers of 3. $$9 = 3^2 \quad \text{and} \quad 27 = 3^3$$

Step 2: Rewrite: $$(3^2)^{x+1} = 3^3$$

Step 3: Use the power rule $(a^m)^n = a^{mn}$: $$3^{2(x+1)} = 3^3$$ $$3^{2x+2} = 3^3$$

Step 4: Set exponents equal: $$2x + 2 = 3$$ $$2x = 1$$ $$x = \frac{1}{2}$$

(c) $4^x = 8^{x-1}$

Step 1: Express both bases as powers of 2. $$4 = 2^2 \quad \text{and} \quad 8 = 2^3$$

Step 2: Rewrite: $$(2^2)^x = (2^3)^{x-1}$$ $$2^{2x} = 2^{3(x-1)}$$ $$2^{2x} = 2^{3x-3}$$

Step 3: Set exponents equal: $$2x = 3x - 3$$ $$-x = -3$$ $$x = 3$$

Check: $4^3 = 64$ and $8^{3-1} = 8^2 = 64$. It works!

Key Properties and Rules

Properties of Exponential Functions $f(x) = a \cdot b^x$ (where $a > 0$, $b > 0$, $b \neq 1$)

Property Description
Domain All real numbers $(-\infty, \infty)$
Range All positive real numbers $(0, \infty)$
$y$-intercept $(0, a)$
$x$-intercept None (the function is always positive)
Horizontal asymptote $y = 0$
Behavior as $x \to \infty$ If $b > 1$: $f(x) \to \infty$; if $0 < b < 1$: $f(x) \to 0$
Behavior as $x \to -\infty$ If $b > 1$: $f(x) \to 0$; if $0 < b < 1$: $f(x) \to \infty$
One-to-one Yes; passes both horizontal and vertical line tests

Exponential Rules (Review)

These rules from your study of exponents remain essential:

Rule Formula
Product of powers $b^m \cdot b^n = b^{m+n}$
Quotient of powers $\frac{b^m}{b^n} = b^{m-n}$
Power of a power $(b^m)^n = b^{mn}$
Negative exponent $b^{-n} = \frac{1}{b^n}$
Zero exponent $b^0 = 1$
Fractional exponent $b^{m/n} = \sqrt[n]{b^m}$

Key Relationships

  • $b^{-x} = \left(\frac{1}{b}\right)^x$ (reflecting across the $y$-axis converts growth to decay)
  • If $f(x) = b^x$, then $f(x+1) = b \cdot f(x)$ (each unit increase multiplies by $b$)
  • The natural exponential $e^x$ has a special property: its derivative equals itself (important in calculus)

Real-World Applications

Population Growth

When a population grows without constraints, it often follows an exponential model:

$$P(t) = P_0 \cdot e^{rt}$$

Where $P_0$ is the initial population, $r$ is the growth rate, and $t$ is time.

If a city of 100,000 people grows at 3% per year: $$P(t) = 100000 \cdot e^{0.03t}$$

After 25 years: $P(25) = 100000 \cdot e^{0.75} \approx 211,700$ people.

Radioactive Decay

Radioactive substances decay exponentially. The half-life is the time it takes for half the material to decay.

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

Where $A_0$ is the initial amount and $h$ is the half-life.

Carbon-14 has a half-life of about 5,730 years. If an artifact contains 25% of its original C-14, you can calculate its age: $$0.25 = \left(\frac{1}{2}\right)^{t/5730}$$ $$\frac{1}{4} = \left(\frac{1}{2}\right)^{t/5730}$$ $$\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{t/5730}$$ $$2 = \frac{t}{5730}$$ $$t = 11,460 \text{ years}$$

Medicine and Pharmacology

Drug concentrations in the bloodstream typically decay exponentially as the body metabolizes the drug:

$$C(t) = C_0 \cdot e^{-kt}$$

Where $k$ is the elimination rate constant. This is why medications often need to be taken at regular intervals to maintain therapeutic levels.

Viral Spread and Social Media

Early in an outbreak or when content goes viral, growth often appears exponential. If each infected person (or each person who shares) reaches a certain number of new people, the spread follows:

$$N(t) = N_0 \cdot b^t$$

This is why early intervention in disease outbreaks is so critical; exponential growth starts slow but accelerates dramatically.

Cooling and Heating

Newton’s Law of Cooling states that the temperature difference between an object and its surroundings decreases exponentially:

$$T(t) = T_s + (T_0 - T_s) \cdot e^{-kt}$$

Where $T_s$ is the surrounding temperature and $T_0$ is the initial temperature of the object.

Self-Test Problems

Problem 1: Evaluate $f(x) = 4 \cdot 3^x$ at $x = -2$, $x = 0$, and $x = 2$.

Show Answer

$f(-2) = 4 \cdot 3^{-2} = 4 \cdot \frac{1}{9} = \frac{4}{9}$

$f(0) = 4 \cdot 3^0 = 4 \cdot 1 = 4$

$f(2) = 4 \cdot 3^2 = 4 \cdot 9 = 36$

Problem 2: Classify $g(x) = 200 \cdot (0.85)^x$ as growth or decay, identify the initial value, and describe what happens to $g(x)$ as $x \to \infty$.

Show Answer

Since $b = 0.85$ and $0 < 0.85 < 1$, this is exponential decay.

The initial value is $a = 200$ (the value when $x = 0$).

As $x \to \infty$, $g(x) \to 0$. The function approaches but never reaches zero.

Problem 3: Solve the equation $5^{2x-1} = 125$.

Show Answer

Express 125 as a power of 5: $125 = 5^3$

Rewrite: $5^{2x-1} = 5^3$

Set exponents equal: $2x - 1 = 3$

Solve: $2x = 4$, so $x = 2$

Check: $5^{2(2)-1} = 5^3 = 125$. Correct!

Problem 4: You invest $2,500 at 4.5% annual interest compounded quarterly. How much will you have after 6 years?

Show Answer

Use the compound interest formula with $P = 2500$, $r = 0.045$, $n = 4$, $t = 6$:

$$A = 2500\left(1 + \frac{0.045}{4}\right)^{4 \cdot 6}$$ $$A = 2500(1.01125)^{24}$$ $$A = 2500 \times 1.3080…$$ $$A \approx \$3,270.09$$

Problem 5: A bacteria culture starts with 500 bacteria and doubles every 3 hours. Write an exponential function for the population $P(t)$ after $t$ hours, and find the population after 12 hours.

Show Answer

Since the population doubles every 3 hours, we can write:

$$P(t) = 500 \cdot 2^{t/3}$$

The exponent $t/3$ counts how many 3-hour periods have passed.

After 12 hours: $$P(12) = 500 \cdot 2^{12/3} = 500 \cdot 2^4 = 500 \cdot 16 = 8000$$

There will be 8,000 bacteria after 12 hours.

Problem 6: Describe the transformations applied to $f(x) = e^x$ to obtain $g(x) = -2e^{x+1} + 3$, and state the horizontal asymptote of $g$.

Show Answer

Starting from $f(x) = e^x$, the transformations are:

  1. Horizontal shift left 1 unit (from $x+1$ in the exponent)
  2. Vertical stretch by factor of 2 (from the coefficient 2)
  3. Reflection across the $x$-axis (from the negative sign)
  4. Vertical shift up 3 units (from the $+3$)

The horizontal asymptote of the original function $y = 0$ is first unaffected by the horizontal shift, stretch, and reflection, but then shifts up by 3.

Horizontal asymptote: $y = 3$

Summary

  • An exponential function has the form $f(x) = a \cdot b^x$, where $b > 0$ and $b \neq 1$. The variable is in the exponent, not the base.

  • Exponential growth occurs when $b > 1$; the function increases as $x$ increases. Exponential decay occurs when $0 < b < 1$; the function decreases toward zero as $x$ increases.

  • The natural base $e \approx 2.71828$ is a special irrational number that appears throughout mathematics. The function $f(x) = e^x$ is called the natural exponential function.

  • All exponential functions $f(x) = b^x$ pass through $(0, 1)$, have a horizontal asymptote at $y = 0$, have domain all real numbers, and have range all positive real numbers.

  • Transformations work the same way as with other functions: $f(x) = a \cdot b^{x-h} + k$ shifts horizontally by $h$, vertically by $k$, and stretches or reflects based on $a$.

  • Compound interest is modeled by $A = P(1 + r/n)^{nt}$, where $P$ is principal, $r$ is annual rate, $n$ is compounding frequency, and $t$ is time in years.

  • Continuous compounding uses $A = Pe^{rt}$ and represents the theoretical maximum growth for a given interest rate.

  • To solve exponential equations with the same-base method, express both sides as powers of the same base, then set the exponents equal.

  • Exponential functions model many real-world phenomena: population growth, radioactive decay, drug metabolism, viral spread, compound interest, and temperature change.

  • The key insight of exponential growth is that it accelerates over time. What starts as barely noticeable change becomes dramatic very quickly, which is why understanding exponential functions matters for everything from personal finance to public health.