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Exponential Functions

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Textbook Reference

Primary source OpenStax Calculus Volume 1, Section 1.2: “Basic Classes of Functions”
Direct link https://openstax.org/books/calculus-volume-1/pages/1-2-basic-classes-of-functions
Supplementary OpenStax Precalculus 2e, Section 6.1: “Exponential Functions”
Supplementary link https://openstax.org/books/precalculus-2e/pages/6-1-exponential-functions

Both sources are free and openly licensed.


Try This First

A single bacterium divides into two every hour.

Fill in the table of bacteria counts:

Hours $0$ $1$ $2$ $3$ $4$ $5$ $10$
Count $1$ ? ? ? ? ? ?

Before computing the full table, answer these two questions:

  1. How does each entry compare to the previous entry?
  2. How does this differ from a linear model where, say, 2 new bacteria are added each hour?

Keep your answers. The key concepts below will explain the structural difference.


Key idea

In the bacteria example, the count doubles each hour: each output is a fixed multiple of the previous output. This is the defining property of an exponential function. The input $x$ appears as an exponent, not as a base. This is the crucial structural difference between an exponential function $b^x$ (variable in the exponent) and a power function $x^n$ (variable in the base). These two functions grow at completely different rates, and they model completely different situations.


Prerequisite Check

Before this lesson, make sure you can do all of the following:

If any of these is uncertain, review Power Functions and exponent rules before continuing.


Quick Reference

Definition. An exponential function has the form $$ f(x) = b^x $$ where the base $b > 0$ and $b \neq 1$, and the exponent $x$ is the variable.

Feature $b > 1$ (growth) $0 < b < 1$ (decay)
Domain All real numbers All real numbers
Range $(0, +\infty)$ $(0, +\infty)$
$f(0)$ $1$ $1$
Behavior as $x \to +\infty$ $f(x) \to +\infty$ $f(x) \to 0$
Behavior as $x \to -\infty$ $f(x) \to 0$ $f(x) \to +\infty$
Horizontal asymptote $y = 0$ $y = 0$

Key exponent laws (for $b > 0$):

Law Formula
Product rule $b^{x+y} = b^x \cdot b^y$
Quotient rule $b^{x-y} = b^x / b^y$
Power of a power $(b^x)^y = b^{xy}$
Negative exponent $b^{-x} = 1/b^x$
Zero exponent $b^0 = 1$

The natural base. $e \approx 2.71828\ldots$ is defined as $e = \displaystyle\lim_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n$. The natural exponential function $f(x) = e^x$ is the exponential function with this special base.


Key Concepts

1. The Exponential Function as a Process

An exponential function $f(x) = b^x$ describes a process where each unit increase in the input multiplies the output by the same factor $b$.

Compare linear and exponential growth:

Hours Linear: $2t + 1$ Exponential: $2^t$
$0$ $1$ $1$
$1$ $3$ $2$
$2$ $5$ $4$
$3$ $7$ $8$
$4$ $9$ $16$
$10$ $21$ $1024$

In the linear model, the output increases by 2 each hour (constant addition). In the exponential model, the output doubles each hour (constant multiplication). By $t = 10$, the linear model gives $21$ and the exponential model gives $1024$ -- nearly 50 times larger.

Two representations, one idea.

Algebraic: $f(t) = 2^t$ satisfies $f(t+1) = 2^{t+1} = 2 \cdot 2^t = 2 \cdot f(t)$. Each unit increase in $t$ multiplies the output by 2.

Graphical: The graph of $2^t$ curves upward steeply. The graph of a linear function is a straight line. For large $t$, the exponential graph rises far above the linear graph.

Translation prompt: Using the table, compute the ratio $f(t+1)/f(t)$ for each row of the exponential column. Confirm it is always 2. Then compute the difference $f(t+1) - f(t)$ for the linear column. Confirm it is always 2. Write one sentence connecting the constant ratio to the word “exponential” and the constant difference to the word “linear”.


2. Growth vs. Decay

Growth ($b > 1$). The output increases as $x$ increases. The graph rises steeply to the right and flattens toward $y = 0$ on the left.

Decay ($0 < b < 1$). The output decreases as $x$ increases. The graph falls toward $y = 0$ on the right. Note that $0 < b < 1$ is equivalent to writing $f(x) = (1/b)^{-x} = c^{-x}$ for some $c = 1/b > 1$.

Predict-then-check. For $f(x) = (1/2)^x = 2^{-x}$, predict: is $f(3)$ greater than or less than $f(0)$?

Prediction: $b = 1/2 < 1$, so this is a decay function. As $x$ increases, $f(x)$ decreases. So $f(3) < f(0)$.

Check: $f(0) = 1$. $f(3) = (1/2)^3 = 1/8 = 0.125$. Yes, $0.125 < 1$. Prediction correct.

Both growth and decay have the same range $(0, +\infty)$ and the same $y$-intercept $f(0) = 1$.

The horizontal asymptote $y = 0$ appears on the left side for growth functions (as $x \to -\infty$) and on the right side for decay functions (as $x \to +\infty$).


3. The Natural Base $e$

The number $e \approx 2.71828$ is irrational and transcendental (not a root of any polynomial with rational coefficients). Despite its seemingly unusual definition, $e$ appears naturally in any situation where a quantity grows continuously at a fixed rate.

Where $e$ comes from. If \$1 is invested at 100% annual interest, the amount after 1 year depends on how frequently interest is compounded:

Compounding Formula Amount after 1 year
Annually $(1 + 1)^1$ $\$2.00$
Monthly $(1 + 1/12)^{12}$ $\$2.613$
Daily $(1 + 1/365)^{365}$ $\$2.7146$
Hourly $(1 + 1/8760)^{8760}$ $\$2.7181$
Continuously $\lim_{n \to \infty}(1+1/n)^n$ $e \approx 2.71828$

As the compounding becomes continuous (instantaneous), the limit is $e$.

The natural exponential $e^x$ is the unique exponential function equal to its own derivative. This property makes $e^x$ indispensable in calculus. You will see this in Chapter 3.

Ask why: Why is continuous compounding the most natural base? Because many physical processes (radioactive decay, population growth in an unlimited environment, heat dissipation) change continuously and at a rate proportional to the current amount. The mathematical model for “grows at a rate proportional to its current size” is exactly $f(x) = Ce^{kx}$.


4. Exponent Laws in Practice

The exponent laws are not arbitrary rules. They follow from the meaning of exponentiation.

Product rule: $b^{x+y} = b^x \cdot b^y$.

Why: $b^{x+y}$ means multiply $b$ by itself $(x+y)$ times. This is the same as multiplying $b$ by itself $x$ times, then $y$ more times: $b^x \cdot b^y$.

Example. Simplify $e^{2x} \cdot e^{3x+1}$: $$ e^{2x} \cdot e^{3x+1} = e^{2x + 3x + 1} = e^{5x+1}. $$

Example. Write $e^{-x}$ in an alternative form: $$ e^{-x} = \frac{1}{e^x}. $$

Predict-then-check. Simplify $\dfrac{4^{2x}}{4^{x-1}}$ and evaluate at $x = 1$.

Prediction: $\dfrac{4^{2x}}{4^{x-1}} = 4^{2x-(x-1)} = 4^{x+1}$. At $x = 1$: $4^{2} = 16$.

Check: $\dfrac{4^2}{4^0} = \dfrac{16}{1} = 16$. Correct.


5. Properties of the Graph

Every exponential function $f(x) = b^x$ (with $b > 0$, $b \neq 1$) has these properties:

  1. Domain: all real numbers.
  2. Range: $(0, +\infty)$. The output is always positive (never zero, never negative).
  3. $y$-intercept: $(0, 1)$, since $b^0 = 1$ for any $b > 0$.
  4. Horizontal asymptote: $y = 0$.
  5. One-to-one: the function is strictly monotone (always increasing for $b > 1$, always decreasing for $0 < b < 1$), so it has an inverse. The inverse of $b^x$ is the logarithm $\log_b x$.

Ask why: Why is the range $(0, +\infty)$, not $[0, +\infty)$? Because $b^x > 0$ for all real $x$ -- the exponential function never reaches zero. Zero would require $x = -\infty$, which is not a real number. The asymptote $y = 0$ is approached but never reached.


Named Misconception

Misconception ID: action-view-of-function

A power function like $x^2$ has a variable base and a constant exponent. An exponential function like $2^x$ has a constant base and a variable exponent. Students sometimes treat both as “just using exponents” and confuse their behavior.

Why this breaks. At $x = 10$: $10^2 = 100$ (power function) and $2^{10} = 1024$ (exponential). At $x = 20$: $20^2 = 400$ and $2^{20} = 1{,}048{,}576$. The exponential grows orders of magnitude faster than the power function for large inputs.

The confusion often appears in problem setup: a student asked to model “a quantity that doubles every year” may write $f(t) = 2t$ (linear) or $f(t) = t^2$ (power function) instead of $f(t) = 2^t$ (exponential). The phrase “doubles every year” means the output is multiplied by 2 for each unit increase in $t$, which is the product rule for exponentials: $f(t+1) = 2 \cdot f(t)$, satisfied by $f(t) = C \cdot 2^t$.

Correct question to ask yourself: Is the exponent fixed (power function: variable base) or is the base fixed (exponential: variable exponent)? These are structurally different functions with different growth rates and different graphs.


Worked Example

Problem. A culture of yeast starts with 50 cells and doubles every 3 hours.

(a) Write an exponential model for the number of cells $N(t)$ after $t$ hours. (b) How many cells are present after 9 hours? (c) After how many hours will there be more than 5000 cells? (Estimate without logarithms; confirm with the model.)

Step 1: Predict. Since the count doubles every 3 hours (not every 1 hour), the model is $N(t) = 50 \cdot 2^{t/3}$.

Why $t/3$: when $t = 3$, $2^{3/3} = 2^1 = 2$, so $N(3) = 50 \cdot 2 = 100$. Doubles in 3 hours. When $t = 6$, $N(6) = 50 \cdot 4 = 200$. Doubles again. Consistent.

Step 2: Evaluate at $t = 9$.

$$N(9) = 50 \cdot 2^{9/3} = 50 \cdot 2^3 = 50 \cdot 8 = 400 \text{ cells.}$$

Step 3: Estimate when $N > 5000$.

$N = 5000$ means $50 \cdot 2^{t/3} = 5000$, so $2^{t/3} = 100$.

Estimate: $2^6 = 64$ and $2^7 = 128$. So $t/3$ is between 6 and 7, meaning $t$ is between 18 and 21 hours.

More precisely, $2^{6.6} \approx 97.6$ and $2^{6.65} \approx 100.5$. So $t/3 \approx 6.64$, giving $t \approx 19.9$ hours. After approximately 20 hours, there are more than 5000 cells.

Step 4: Check. $N(18) = 50 \cdot 2^6 = 50 \cdot 64 = 3200 < 5000$. $N(21) = 50 \cdot 2^7 = 50 \cdot 128 = 6400 > 5000$. The threshold is between 18 and 21 hours. Consistent.


Common Errors

Error Example Correction
Confusing $b^x$ and $x^b$ “Doubling means $f(t) = t^2$” Doubling each unit: $f(t) = 2^t$; the base is the multiplier, not the exponent
Forgetting the $y$-intercept is always 1 “The graph of $3^x$ starts at 0” $3^0 = 1$; the $y$-intercept is always $(0, 1)$ for $b^x$
Assuming the range includes 0 “The exponential can equal 0 for very negative $x$” $b^x > 0$ for all real $x$; the asymptote $y = 0$ is never reached
Misapplying exponent laws $2^x \cdot 3^x = 6^{2x}$ $2^x \cdot 3^x = (2 \cdot 3)^x = 6^x$; the exponent stays $x$, not $2x$
Writing decay as a negative function “Decay means $f(x) = -b^x$” Decay means $0 < b < 1$, or write $f(x) = b^{-x}$ for $b > 1$; output is still positive

Common Misconceptions

Common misconception

$b^x$ and $x^b$ are the same kind of function because both involve a base and an exponent.

This is the action-view-of-function error applied to exponential vs power notation. The two expressions have the variable in entirely different positions: in $b^x$ (an exponential function) the base is fixed and the exponent varies; in $x^b$ (a power function) the base varies and the exponent is fixed. For $b = 2$: $2^x$ doubles at each unit step (growth is multiplicative), while $x^2$ grows as a parabola (growth is polynomial). They differ in long-run behavior, graph shape, and domain. Treating “anything with an exponent” as the same kind of function leads to incorrect limit and derivative reasoning.


Leveled Practice

Level 1 -- Direct Application

Problem 1. Evaluate each without a calculator.

(a) $5^0$ (b) $3^{-2}$ (c) $\left(\dfrac{1}{2}\right)^4$ (d) $e^0$

Show answer

(a) $5^0 = 1$ (any positive base to the zero power is 1).

(b) $3^{-2} = 1/3^2 = 1/9$.

(c) $(1/2)^4 = 1/16$.

(d) $e^0 = 1$.


Problem 2. Simplify each expression.

(a) $e^{3x} \cdot e^{x+2}$ (b) $\dfrac{2^{5x}}{2^{x+1}}$ (c) $(e^{x})^4$

Show answer

(a) $e^{3x} \cdot e^{x+2} = e^{3x + x + 2} = e^{4x+2}$.

(b) $\dfrac{2^{5x}}{2^{x+1}} = 2^{5x - (x+1)} = 2^{4x-1}$.

(c) $(e^x)^4 = e^{4x}$.


Problem 3. For $f(x) = 3^x$:

(a) Is $f$ increasing or decreasing? (b) What is the $y$-intercept? (c) What is the horizontal asymptote? (d) Is $f(x)$ ever zero or negative? Explain.

Show answer

(a) Increasing ($b = 3 > 1$).

(b) $f(0) = 3^0 = 1$. The $y$-intercept is $(0, 1)$.

(c) Horizontal asymptote: $y = 0$ (approached as $x \to -\infty$).

(d) Never. $3^x > 0$ for all real $x$. The range is $(0, +\infty)$.


Level 2 -- Multiple Representations

Problem 4. The table below shows values of a function $f$.

$x$ $0$ $1$ $2$ $3$ $4$
$f(x)$ $4$ $12$ $36$ $108$ $324$

(a) Is this function linear or exponential? Explain. (b) Find a formula for $f(x)$. (c) Predict $f(6)$.

Show answer

(a) Compute ratios: $12/4 = 3$, $36/12 = 3$, $108/36 = 3$, $324/108 = 3$. Constant ratio of 3 per unit increase in $x$: exponential.

(Differences: $12-4 = 8$, $36-12 = 24$, $108-36 = 72$. Not constant, so not linear.)

(b) $f(0) = 4$ and each step multiplies by 3: $f(x) = 4 \cdot 3^x$.

(c) $f(6) = 4 \cdot 3^6 = 4 \cdot 729 = 2916$.


Problem 5. Sketch rough graphs of $y = 2^x$, $y = (1/2)^x$, and $y = e^x$ on the same axes.

(a) At $x = 0$, what do all three graphs have in common? (b) For $x > 0$, rank the three functions from smallest to largest output. (c) For $x < 0$, rank them from smallest to largest output.

Show answer

(a) All three equal 1 at $x = 0$.

(b) For $x > 0$: $2^x < e^x < $ ... wait, $e \approx 2.718 > 2$, so $e^x > 2^x$. Also $(1/2)^x = 2^{-x} < 1 < 2^x < e^x$ for $x > 0$.

Order: $(1/2)^x < 2^x < e^x$.

(c) For $x < 0$: $2^x < 1$ and $e^x < 1$ (both decaying), but $(1/2)^x = 2^{-x}$ with $-x > 0$, so $(1/2)^x = 2^{-x} > 1$.

At $x = -2$: $(1/2)^{-2} = 4$, $2^{-2} = 0.25$, $e^{-2} \approx 0.135$.

Order: $e^x < 2^x < (1/2)^x$.


Level 3 -- Extension

Problem 6 (Low Floor, High Ceiling). Prove that $e^x$ grows faster than any polynomial for large $x$. Specifically, show that for any positive integer $n$, $$\lim_{x \to \infty} \frac{x^n}{e^x} = 0.$$

You may use the fact that $e^x \geq 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^{n+1}}{(n+1)!}$ for all $x > 0$.

Show answer

Since $e^x \geq \dfrac{x^{n+1}}{(n+1)!}$ for all $x > 0$:

$$0 \leq \frac{x^n}{e^x} \leq \frac{x^n}{x^{n+1}/(n+1)!} = \frac{(n+1)!}{x}.$$

As $x \to \infty$, the right side $\dfrac{(n+1)!}{x} \to 0$ (it is $(n+1)!$ divided by an increasing quantity).

By the Squeeze Theorem, $\dfrac{x^n}{e^x} \to 0$ as $x \to \infty$.

Conclusion: $e^x$ grows faster than $x^n$ for any fixed positive integer $n$. Equivalently, the ratio $x^n / e^x \to 0$: the exponential eventually dominates any polynomial, no matter how high the degree.

This argument extends to all polynomials (not just $x^n$) because a polynomial of degree $n$ is at most $C \cdot x^n$ for some constant $C$ for large $x$, and $C \cdot x^n / e^x = C \cdot (x^n / e^x) \to 0$.


Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:


Mental Model

An exponential function $b^x$ is a repeated multiplication machine. Each time the input increases by 1, the output is multiplied by $b$. For growth ($b > 1$), each multiplication makes the output larger than the last -- and it makes it larger by a factor of $b$, so the outputs grow without bound and at an accelerating rate. For decay ($0 < b < 1$), each multiplication makes the output smaller, approaching zero but never reaching it.

Compare with a linear function: each time the input increases by 1, the output is increased by $m$ (added, not multiplied). Addition produces linear growth; multiplication produces exponential growth. This is why, for large inputs, exponential growth eventually dominates any linear or polynomial growth.

The number $e$ is the base that arises naturally when a quantity grows or decays continuously at a rate proportional to its current amount. It is in this sense the “natural” base -- not because of any human convention, but because of the mathematics of continuous change.


Connections

Within Chapter 1

Toward Later Calculus

Audience Notes

For students who find the base $e$ mysterious: Begin with the compound interest story. The number $e$ is simply what happens when interest is compounded continuously -- as the compounding period shrinks to zero, the amount approaches $e$ times the principal. There is nothing mysterious; it is a limit.

For students who want to go further: The number $e$ is transcendental, meaning it cannot be the root of any polynomial with rational coefficients. This was proved by Hermite in 1873. The proof uses techniques from complex analysis. The transcendence of $e$ implies that no algebraic formula can express $e$ using just fractions and roots.


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