Chapter 4: Exponential and Logarithmic Functions

Precalculus · Updated February 2026 · 25 min read

Exponential and logarithmic functions are among the most important in all of mathematics. They model phenomena that grow or decay at rates proportional to their current value: population growth, radioactive decay, compound interest, sound intensity, and earthquake magnitude. In this chapter, we build up from the definition of exponential functions, introduce logarithms as their inverses, develop the algebraic properties that make logarithms so powerful, and then apply everything to solve real-world problems.

4.1 Exponential Functions

You have already encountered integer exponents: $2^3 = 8$, $5^{-1} = \frac{1}{5}$, and so on. An exponential function takes this idea and lets the exponent be any real number, creating a smooth, continuous curve that models growth and decay.

Definition: Exponential Function

An exponential function is a function of the form

$$f(x) = b^x$$

where $b > 0$ and $b \neq 1$. The constant $b$ is called the base.

Why do we require $b > 0$? If $b$ were negative, then $b^x$ would be undefined for many values of $x$ (for example, $(-2)^{1/2}$ is not a real number). And why $b \neq 1$? Because $1^x = 1$ for all $x$, which is just a constant function and not very interesting.

Key Properties of $f(x) = b^x$

Every exponential function $f(x) = b^x$ shares several fundamental properties:

Growth vs. Decay

The behavior of $f(x) = b^x$ depends on whether the base is greater than or less than $1$:

Use the interactive graph below to explore how changing the base $b$ affects the shape of the exponential curve. When $b > 1$ the curve rises to the right; when $0 < b < 1$ it falls to the right. Notice that regardless of the base, every curve passes through $(0, 1)$.

Interactive: Drag the slider to change the base $b$ and observe growth vs. decay.

The Natural Exponential Function

Among all possible bases, one stands out as the most important in mathematics: the number $e$, known as Euler's number. It is an irrational constant defined by the limit

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

The function $f(x) = e^x$ is called the natural exponential function. It arises naturally in calculus because it is the unique exponential function that is its own derivative: $\frac{d}{dx} e^x = e^x$. For this reason, $e^x$ appears throughout physics, engineering, finance, and biology whenever continuous growth or decay is modeled.

Worked Example 4.1

Evaluate the following without a calculator:

(a) $4^{3/2}$    (b) $27^{-2/3}$    (c) $e^0$

Part (a) Rewrite the fractional exponent: $4^{3/2} = \left(4^{1/2}\right)^3 = 2^3 = 8$.

Part (b) Apply the negative exponent and fractional exponent rules: $27^{-2/3} = \dfrac{1}{27^{2/3}} = \dfrac{1}{\left(27^{1/3}\right)^2} = \dfrac{1}{3^2} = \dfrac{1}{9}$.

Part (c) Any nonzero base raised to the zero power equals $1$: $e^0 = 1$.

Worked Example 4.2

A culture of bacteria doubles every 3 hours. If the initial population is 500, write an exponential function modeling the population after $t$ hours and find the population after 9 hours.

Step 1 Since the population doubles every 3 hours, the base is $2$ and the growth period is 3 hours. The model is:

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

Step 2 Evaluate at $t = 9$:

$$P(9) = 500 \cdot 2^{9/3} = 500 \cdot 2^3 = 500 \cdot 8 = 4000$$

After 9 hours, the population is 4,000 bacteria.

4.2 Logarithmic Functions

We have just seen that every exponential function $f(x) = b^x$ (with $b > 0, b \neq 1$) is one-to-one. This means it has an inverse function. That inverse is the logarithm.

Definition: Logarithmic Function

For $b > 0$ and $b \neq 1$, the logarithm base $b$ of $x$ is defined by

$$\log_b(x) = y \quad \iff \quad b^y = x$$

In words: $\log_b(x)$ is the exponent you must raise $b$ to in order to get $x$.

Think of the logarithm as answering a question. The expression $\log_2(8)$ asks: "2 raised to what power gives 8?" Since $2^3 = 8$, the answer is $\log_2(8) = 3$.

Common and Natural Logarithms

Two bases are used so frequently that they have special notation:

Properties of $g(x) = \log_b(x)$

Since $g(x) = \log_b(x)$ is the inverse of $f(x) = b^x$, its properties mirror those of the exponential:

The Inverse Relationship

Because $f(x) = b^x$ and $g(x) = \log_b(x)$ are inverses, they "undo" each other:

$$b^{\log_b(x)} = x \quad \text{for } x > 0 \qquad \text{and} \qquad \log_b(b^x) = x \quad \text{for all } x$$

Graphically, the curves $y = b^x$ and $y = \log_b(x)$ are reflections of each other across the line $y = x$. The interactive graph below demonstrates this. Adjust the slider to change the base and observe how both curves change together while always remaining symmetric about the diagonal.

Interactive: The exponential (blue) and its inverse logarithm (red) reflected across $y = x$ (dashed).

Worked Example 4.3

Evaluate each logarithm without a calculator:

(a) $\log_3(81)$    (b) $\log_{1/2}(8)$    (c) $\ln(e^5)$

Part (a) Ask: $3^y = 81$. Since $3^4 = 81$, we have $\log_3(81) = 4$.

Part (b) Ask: $\left(\frac{1}{2}\right)^y = 8$. Write $\frac{1}{2} = 2^{-1}$, so $(2^{-1})^y = 2^{-y} = 8 = 2^3$. Thus $-y = 3$, giving $y = -3$. So $\log_{1/2}(8) = -3$.

Part (c) By the inverse property $\ln(e^x) = x$, we get $\ln(e^5) = 5$.

Worked Example 4.4

Find the domain of $f(x) = \ln(3 - 2x)$.

Step 1 The argument of the natural logarithm must be strictly positive:

$$3 - 2x > 0$$

Step 2 Solve for $x$:

$$-2x > -3 \implies x < \frac{3}{2}$$

The domain is $\left(-\infty, \dfrac{3}{2}\right)$.

4.3 Properties of Logarithms

Logarithms were originally invented to simplify computation: they turn multiplication into addition, division into subtraction, and exponentiation into multiplication. These properties arise directly from the corresponding exponent rules and remain essential tools for solving equations and simplifying expressions.

Logarithm Properties

Let $b > 0$, $b \neq 1$, and suppose $M > 0$ and $N > 0$. Then:

1. Product Rule: $\quad \log_b(MN) = \log_b(M) + \log_b(N)$

2. Quotient Rule: $\quad \log_b\!\left(\dfrac{M}{N}\right) = \log_b(M) - \log_b(N)$

3. Power Rule: $\quad \log_b(M^n) = n \cdot \log_b(M)$

4. Change of Base: $\quad \log_b(x) = \dfrac{\log_a(x)}{\log_a(b)} = \dfrac{\ln(x)}{\ln(b)}$

These properties follow from the exponent laws. For instance, the product rule works because if $\log_b(M) = p$ and $\log_b(N) = q$, then $b^p = M$ and $b^q = N$, so $MN = b^p \cdot b^q = b^{p+q}$, which means $\log_b(MN) = p + q = \log_b(M) + \log_b(N)$.

Common Mistakes to Avoid:

$\log_b(M + N) \neq \log_b(M) + \log_b(N)$ — the product rule applies to $\log_b(M \cdot N)$, not $\log_b(M + N)$.

$\dfrac{\log_b(M)}{\log_b(N)} \neq \log_b\!\left(\dfrac{M}{N}\right)$ — the quotient rule applies when the logarithm is on the outside.

Worked Example 4.5 — Expanding a Logarithm

Expand $\log_2\!\left(\dfrac{x^3 \sqrt{y}}{z^4}\right)$ as a sum and difference of simpler logarithms.

Step 1 Apply the quotient rule:

$$\log_2\!\left(\frac{x^3 \sqrt{y}}{z^4}\right) = \log_2(x^3 \sqrt{y}) - \log_2(z^4)$$

Step 2 Apply the product rule to the first term:

$$= \log_2(x^3) + \log_2(\sqrt{y}) - \log_2(z^4)$$

Step 3 Apply the power rule (recall $\sqrt{y} = y^{1/2}$):

$$= 3\log_2(x) + \frac{1}{2}\log_2(y) - 4\log_2(z)$$

Worked Example 4.6 — Condensing Logarithms

Write $2\ln(x) - \frac{1}{3}\ln(y) + \ln(z)$ as a single logarithm.

Step 1 Apply the power rule in reverse:

$$= \ln(x^2) - \ln(y^{1/3}) + \ln(z)$$

Step 2 Combine using the product and quotient rules:

$$= \ln\!\left(\frac{x^2 \cdot z}{y^{1/3}}\right) = \ln\!\left(\frac{x^2 z}{\sqrt[3]{y}}\right)$$

Worked Example 4.7 — Change of Base

Evaluate $\log_5(20)$ to four decimal places using natural logarithms.

Step 1 Apply the change of base formula:

$$\log_5(20) = \frac{\ln(20)}{\ln(5)}$$

Step 2 Evaluate using a calculator:

$$= \frac{2.9957}{1.6094} \approx 1.8614$$

4.4 Solving Exponential and Logarithmic Equations

The key to solving exponential equations is to isolate the exponential expression and then apply a logarithm to both sides. The key to solving logarithmic equations is to rewrite the equation in exponential form or combine logarithms and then exponentiate. In either case, you must always check your solutions for extraneous values.

Strategy for Exponential Equations

  1. Isolate the exponential expression on one side of the equation.
  2. Take the logarithm (usually $\ln$ or $\log$) of both sides.
  3. Use the power rule to bring down the exponent.
  4. Solve for the variable.

Strategy for Logarithmic Equations

  1. Use logarithm properties to combine into a single logarithm if needed.
  2. Convert to exponential form: $\log_b(A) = C$ becomes $b^C = A$.
  3. Solve for the variable.
  4. Check that each solution makes the argument of every original logarithm positive (reject extraneous solutions).

Worked Example 4.8 — Solving an Exponential Equation

Solve $3^{2x-1} = 15$.

Step 1 Take the natural logarithm of both sides:

$$\ln(3^{2x-1}) = \ln(15)$$

Step 2 Apply the power rule:

$$(2x - 1)\ln(3) = \ln(15)$$

Step 3 Solve for $x$:

$$2x - 1 = \frac{\ln(15)}{\ln(3)} \implies 2x = 1 + \frac{\ln(15)}{\ln(3)} \implies x = \frac{1}{2}\!\left(1 + \frac{\ln(15)}{\ln(3)}\right)$$

Step 4 Evaluate numerically: $\dfrac{\ln(15)}{\ln(3)} \approx 2.4649$, so $x \approx \dfrac{1 + 2.4649}{2} \approx 1.7325$.

Worked Example 4.9 — Solving a Logarithmic Equation

Solve $\log_2(x + 3) + \log_2(x - 1) = 5$.

Step 1 Combine using the product rule:

$$\log_2\!\big((x + 3)(x - 1)\big) = 5$$

Step 2 Convert to exponential form:

$$(x + 3)(x - 1) = 2^5 = 32$$

Step 3 Expand and solve the quadratic:

$$x^2 + 2x - 3 = 32 \implies x^2 + 2x - 35 = 0 \implies (x + 7)(x - 5) = 0$$

So $x = -7$ or $x = 5$.

Step 4 Check for extraneous solutions. Substituting $x = -7$: we need $x + 3 = -4 > 0$ (false) and $x - 1 = -8 > 0$ (false). So $x = -7$ is extraneous.

Substituting $x = 5$: $x + 3 = 8 > 0$ and $x - 1 = 4 > 0$. Both arguments are positive, so $x = 5$ is valid.

The solution is $x = 5$.

Worked Example 4.10 — An Equation with $e$

Solve $e^{2x} - 5e^x + 6 = 0$.

Step 1 Recognize this as a quadratic in disguise. Let $u = e^x$, so the equation becomes:

$$u^2 - 5u + 6 = 0$$

Step 2 Factor:

$$(u - 2)(u - 3) = 0 \implies u = 2 \text{ or } u = 3$$

Step 3 Substitute back $u = e^x$:

$$e^x = 2 \implies x = \ln(2) \approx 0.6931$$

$$e^x = 3 \implies x = \ln(3) \approx 1.0986$$

Step 4 Both solutions are valid since $e^x > 0$ always holds. The solutions are $x = \ln(2)$ and $x = \ln(3)$.

4.5 Applications

Exponential and logarithmic functions model a wide variety of real-world phenomena. In this section, we focus on two of the most common: compound interest in finance and exponential decay in the sciences.

Compound Interest

When money earns interest on both the principal and the previously accumulated interest, the growth is exponential. The compound interest formula depends on how frequently the interest is compounded.

Compound Interest Formulas

Periodic compounding: If a principal $P$ is invested at an annual rate $r$ (as a decimal), compounded $n$ times per year for $t$ years, the accumulated amount is:

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

Continuous compounding: If interest is compounded continuously, the formula becomes:

$$A = Pe^{rt}$$

The continuous compounding formula is the limiting case as $n \to \infty$. It provides a slightly larger amount than any finite compounding frequency, but the difference is often small in practice.

Worked Example 4.11 — Compound Interest

You invest $\$5{,}000$ at $6\%$ annual interest. Find the amount after 10 years if interest is compounded (a) quarterly and (b) continuously.

Part (a) Here $P = 5000$, $r = 0.06$, $n = 4$ (quarterly), and $t = 10$:

$$A = 5000\!\left(1 + \frac{0.06}{4}\right)^{4 \cdot 10} = 5000(1.015)^{40}$$

Using a calculator: $(1.015)^{40} \approx 1.8140$, so

$$A \approx 5000 \times 1.8140 = \$9{,}070.09$$

Part (b) With continuous compounding:

$$A = 5000 \cdot e^{0.06 \times 10} = 5000 \cdot e^{0.6}$$

Since $e^{0.6} \approx 1.8221$:

$$A \approx 5000 \times 1.8221 = \$9{,}110.59$$

Continuous compounding yields about $\$40.50$ more than quarterly compounding over 10 years.

Exponential Growth and Decay

Many natural processes follow an exponential model. If a quantity $Q$ changes at a rate proportional to its current value, then:

$$Q(t) = Q_0 \cdot e^{kt}$$

where $Q_0$ is the initial quantity at $t = 0$ and $k$ is the growth/decay constant. If $k > 0$, the quantity grows; if $k < 0$, it decays.

Half-Life

In radioactive decay, the half-life $h$ is the time it takes for a substance to reduce to half its initial amount. The relationship between $k$ and $h$ is:

$$Q_0 \cdot e^{kh} = \frac{Q_0}{2} \implies e^{kh} = \frac{1}{2} \implies k = \frac{-\ln(2)}{h}$$

Worked Example 4.12 — Radioactive Decay

Carbon-14 has a half-life of approximately 5,730 years. A fossil contains 30% of the original Carbon-14. How old is the fossil?

Step 1 Find the decay constant $k$:

$$k = \frac{-\ln(2)}{5730} \approx \frac{-0.6931}{5730} \approx -0.0001210$$

Step 2 Set up the equation. We want $Q(t) = 0.30 \cdot Q_0$:

$$Q_0 \cdot e^{kt} = 0.30 \cdot Q_0 \implies e^{kt} = 0.30$$

Step 3 Take the natural logarithm:

$$kt = \ln(0.30) \implies t = \frac{\ln(0.30)}{k} = \frac{\ln(0.30)}{-0.0001210}$$

Step 4 Evaluate:

$$t = \frac{-1.2040}{-0.0001210} \approx 9{,}950 \text{ years}$$

The fossil is approximately 9,950 years old.

4.6 Practice Problems

Test your understanding with the following problems. Each has a hidden solution you can reveal after attempting the problem on your own.

Problem 1

Evaluate without a calculator: $\log_4(64)$.

Show Solution

We need $4^y = 64$. Since $4^3 = 64$, we have $\log_4(64) = 3$.

Problem 2

Simplify $\log_3(27) + \log_3\!\left(\frac{1}{9}\right)$.

Show Solution

$\log_3(27) = 3$ since $3^3 = 27$.

$\log_3\!\left(\frac{1}{9}\right) = \log_3(3^{-2}) = -2$.

Therefore $\log_3(27) + \log_3\!\left(\frac{1}{9}\right) = 3 + (-2) = 1$.

Alternatively, by the product rule: $\log_3\!\left(27 \cdot \frac{1}{9}\right) = \log_3(3) = 1$.

Problem 3

Expand completely: $\ln\!\left(\dfrac{x^2(x+1)^3}{e^x}\right)$.

Show Solution

Apply the quotient rule, then the product and power rules:

$$\ln\!\left(\frac{x^2(x+1)^3}{e^x}\right) = \ln(x^2) + \ln(x+1)^3 - \ln(e^x)$$

$$= 2\ln(x) + 3\ln(x+1) - x$$

(since $\ln(e^x) = x$).

Problem 4

Write as a single logarithm: $\frac{1}{2}\log(x) - 3\log(y) + \log(z)$.

Show Solution

Apply the power rule: $\log(x^{1/2}) - \log(y^3) + \log(z)$.

Combine: $\log\!\left(\dfrac{\sqrt{x} \cdot z}{y^3}\right)$.

Problem 5

Solve for $x$: $5^{x+2} = 125$.

Show Solution

Since $125 = 5^3$, we can write $5^{x+2} = 5^3$.

Equating exponents: $x + 2 = 3$, so $x = 1$.

Problem 6

Solve for $x$: $2 \cdot e^{3x} = 14$.

Show Solution

Divide both sides by 2: $e^{3x} = 7$.

Take the natural logarithm: $3x = \ln(7)$.

Solve: $x = \dfrac{\ln(7)}{3} \approx \dfrac{1.9459}{3} \approx 0.6486$.

Problem 7

Solve for $x$: $\log(x - 2) + \log(x + 1) = 1$.

Show Solution

Combine: $\log\!\big((x-2)(x+1)\big) = 1$.

Convert to exponential form: $(x-2)(x+1) = 10^1 = 10$.

Expand: $x^2 - x - 2 = 10 \implies x^2 - x - 12 = 0$.

Factor: $(x-4)(x+3) = 0 \implies x = 4$ or $x = -3$.

Check: $x = -3$ gives $\log(-5)$, which is undefined. Reject $x = -3$.

$x = 4$ gives $\log(2) + \log(5) = \log(10) = 1$. Valid.

Solution: $x = 4$.

Problem 8

Find the domain of $g(x) = \log_5(x^2 - 9)$.

Show Solution

We need the argument to be positive: $x^2 - 9 > 0$.

Factor: $(x-3)(x+3) > 0$.

By sign analysis, this holds when $x < -3$ or $x > 3$.

Domain: $(-\infty, -3) \cup (3, \infty)$.

Problem 9

You deposit $\$2{,}000$ in an account earning $4.5\%$ annual interest compounded monthly. How long until the balance reaches $\$3{,}000$?

Show Solution

Use $A = P\!\left(1 + \frac{r}{n}\right)^{nt}$ with $A = 3000$, $P = 2000$, $r = 0.045$, $n = 12$:

$$3000 = 2000\!\left(1 + \frac{0.045}{12}\right)^{12t}$$

$$1.5 = (1.00375)^{12t}$$

Take $\ln$ of both sides:

$$\ln(1.5) = 12t \cdot \ln(1.00375)$$

$$t = \frac{\ln(1.5)}{12 \cdot \ln(1.00375)} = \frac{0.4055}{12 \times 0.003743} \approx \frac{0.4055}{0.04492} \approx 9.03 \text{ years}$$

It takes approximately 9 years for the balance to reach $\$3{,}000$.

Problem 10

A radioactive substance has a half-life of 12 hours. If you start with 80 grams, how much remains after 40 hours?

Show Solution

Method 1 (using the half-life formula directly):

$$Q(t) = Q_0 \cdot \left(\frac{1}{2}\right)^{t/h} = 80 \cdot \left(\frac{1}{2}\right)^{40/12} = 80 \cdot \left(\frac{1}{2}\right)^{10/3}$$

$$= 80 \cdot 2^{-10/3} \approx 80 \cdot 0.09921 \approx 7.94 \text{ grams}$$

Method 2 (using $k$):

$k = \dfrac{-\ln 2}{12} \approx -0.05776$. Then:

$$Q(40) = 80 \cdot e^{-0.05776 \times 40} = 80 \cdot e^{-2.3105} \approx 80 \times 0.09921 \approx 7.94 \text{ grams}$$

Approximately 7.94 grams remain after 40 hours.

← Chapter 3: Polynomial & Rational Functions Chapter 5: Trigonometric Functions →