Suppose you know that $2^x = 8$. What is $x$? You can see that $x = 3$ because $2^3 = 8$. But what if you know that $2^x = 10$? What power of 2 gives you 10?
This is exactly the question that logarithms answer. The logarithm is the operation that "undoes" exponentiation: it tells you what exponent was used.
The key insight: If $b^y = x$, then $y$ is called "the logarithm base $b$ of $x$," written $\log_b x = y$.
In other words: The logarithm is the exponent.
Self-check: Can you answer these questions? If not, review the linked prerequisites first.
| Question | If you struggle... |
|---|---|
| What does $f^{-1}$ mean? How is it related to $f$? | Review inverse functions |
| Explain why $f(x) = b^x$ (with $b > 1$) passes the horizontal line test | Review one-to-one functions |
| Evaluate $3^4$, $2^{-3}$, and $25^{1/2}$ | Review exponent rules |
| If $f(f^{-1}(x)) = x$, what does this tell you? | This is the cancellation property of inverses |
If $f$ is one-to-one, its inverse function $f^{-1}$ satisfies:
The graph of $f^{-1}$ is the reflection of the graph of $f$ about the line $y = x$.
For $b > 0$ and $b \neq 1$, the function $f(x) = b^x$ is either:
Either way, it passes the horizontal line test, so it is one-to-one and therefore has an inverse function.
| Property | Value |
|---|---|
| Concept | Inverse Functions |
| Course | MATH162 |
| Section | Stewart 6.3 |
| Difficulty | Foundational |
| Time | ~25 minutes |
If $b > 0$ and $b \neq 1$, the logarithm base $b$ of $x$ is defined by:
$$\boxed{\log_b x = y \quad \Longleftrightarrow \quad b^y = x}$$
In words: $\log_b x$ is the exponent to which $b$ must be raised to get $x$.
Example: $\log_3 81 = 4$ because $3^4 = 81$.
| Restriction | Reason |
|---|---|
| $b > 0$ | Negative bases create complex number issues (what is $(-2)^{1/2}$?) |
| $b \neq 1$ | If $b = 1$, then $1^y = 1$ for all $y$; the function isn't one-to-one |
| $x > 0$ | Since $b^y > 0$ for all $y$, logarithms can only output positive numbers |
The logarithmic function $f(x) = \log_b x$ is the inverse of the exponential function $g(x) = b^x$.
This gives us the cancellation equations:
$$\boxed{\log_b(b^x) = x \quad \text{for every } x \in \mathbb{R}}$$
$$\boxed{b^{\log_b x} = x \quad \text{for every } x > 0}$$
Think of it this way:
| Function | Domain | Range |
|---|---|---|
| $f(x) = b^x$ | $\mathbb{R}$ (all real numbers) | $(0, \infty)$ |
| $f^{-1}(x) = \log_b x$ | $(0, \infty)$ | $\mathbb{R}$ (all real numbers) |
Notice how the domain and range swap. This always happens with inverse functions.
y y
| y = b^x | y = log_b(x)
| / | ___
| / | ___/
| / | ___/
+----/------→ x +--/----------→ x
| / | |
| / | (vertical asymptote
|/ | at x = 0)For $b > 1$:
$$\lim_{x \to \infty} \log_b x = \infty \qquad \text{(grows without bound, but slowly)}$$
$$\lim_{x \to 0^+} \log_b x = -\infty \qquad \text{(vertical asymptote at } x = 0\text{)}$$
Physical interpretation: If $b = 10$, then $\log_{10}(1{,}000{,}000) = 6$. To get a logarithm of just 6, the input had to be a million. Logarithms grow very slowly.
Strategy: Ask yourself "What power of the base gives me the argument?"
Examples:
| Expression | Question | Answer | Because |
|---|---|---|---|
| $\log_3 81$ | $3^{?} = 81$ | $4$ | $3^4 = 81$ |
| $\log_{25} 5$ | $25^{?} = 5$ | $\frac{1}{2}$ | $25^{1/2} = \sqrt{25} = 5$ |
| $\log_{10} 0.001$ | $10^{?} = 0.001$ | $-3$ | $10^{-3} = \frac{1}{1000} = 0.001$ |
| $\log_2 1$ | $2^{?} = 1$ | $0$ | $2^0 = 1$ |
| $\log_b b$ | $b^{?} = b$ | $1$ | $b^1 = b$ |
For any valid base $b$:
$$\log_b 1 = 0 \qquad \text{(because } b^0 = 1\text{)}$$
$$\log_b b = 1 \qquad \text{(because } b^1 = b\text{)}$$
Key features (for $b > 1$):
The graph is the reflection of $y = b^x$ about the line $y = x$.
| Mistake | Why It's Wrong | Correction |
|---|---|---|
| $\log_b 0 = 0$ | There is no power of $b$ that equals 0 | $\log_b 0$ is undefined |
| $\log_b(-5)$ exists | $b^y$ is always positive | Log of negative numbers is undefined |
| Confusing $\log_b x$ with $\frac{\log x}{b}$ | Notation issue | $\log_b x$ means "log base $b$ of $x$" |
| $\log_b(x + y) = \log_b x + \log_b y$ | Logs don't distribute over addition | This is wrong; see Laws of Logarithms |
Evaluate each logarithm.
(a) $\log_2 32$ (b) $\log_5 125$ (c) $\log_4 2$
Evaluate each logarithm.
(a) $\log_3 \frac{1}{27}$ (b) $\log_{16} 8$ (c) $\log_{100} 10$
Simplify each expression.
(a) $\log_7(7^{12})$ (b) $5^{\log_5 19}$ (c) $\log_2(2^x \cdot 2^{3x})$
If $\log_8 x = \frac{2}{3}$, find $x$.
(a) Explain why $\log_b(b^x) = x$ for all real $x$, but $b^{\log_b x} = x$ only for $x > 0$.
(b) Without a calculator, determine which is larger: $\log_2 10$ or $\log_3 10$. Justify your answer.
The Logarithm as a "Power Finder":
Think of $\log_b$ as asking a question: "What power?"
The exponential $b^x$ takes a power and produces a number. The logarithm $\log_b x$ takes a number and produces a power.
They undo each other.
Looking back:
Looking ahead:
Real-world applications:
| Previous | Up | Next |
|---|---|---|
| Exponential Functions | Skills Index | The Natural Logarithm |
Last updated: 2026-01-23