Why were logarithms invented? Before calculators, multiplying large numbers was tedious. But logarithms transform multiplication into addition, and addition is easy.
The key insight is that logarithms convert:
This makes logarithms incredibly powerful for simplifying expressions and solving equations.
Self-check: Can you answer these questions? If not, review the linked prerequisites first.
| Question | If you struggle... |
|---|---|
| What is $\ln(e^5)$? | Review Natural Logarithm |
| Simplify $e^a \cdot e^b$ using exponent rules | Review exponent laws: answer is $e^{a+b}$ |
| What is $\log_b x$ in terms of $b$ and $y$ when $b^y = x$? | Review Logarithm Definition |
| Simplify $(e^a)^r$ | Review exponent laws: answer is $e^{ar}$ |
For any base $b > 0$:
| Operation | Exponent Rule |
|---|---|
| Multiplication | $b^m \cdot b^n = b^{m+n}$ |
| Division | $\frac{b^m}{b^n} = b^{m-n}$ |
| Power of a power | $(b^m)^n = b^{mn}$ |
The log laws are the "mirror image" of these.
For base $b$:
For natural logs:
| Property | Value |
|---|---|
| Concept | Inverse Functions |
| Course | MATH162 |
| Section | Stewart 6.3 |
| Difficulty | Intermediate |
| Time | ~30 minutes |
For $b > 0$, $b \neq 1$, and $x, y > 0$:
$$\boxed{\text{Product Rule: } \log_b(xy) = \log_b x + \log_b y}$$
$$\boxed{\text{Quotient Rule: } \log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y}$$
$$\boxed{\text{Power Rule: } \log_b(x^r) = r \log_b x \quad \text{(for any real } r\text{)}}$$
Each log law follows from the corresponding exponent law. The proof below establishes the product rule for natural logs.
Proof of Product Rule: Let $a = \ln x$ and $b = \ln y$. Then $e^a = x$ and $e^b = y$.
We want to show $\ln(xy) = \ln x + \ln y = a + b$.
$$xy = e^a \cdot e^b = e^{a+b}$$
Taking $\ln$ of both sides: $$\ln(xy) = \ln(e^{a+b}) = a + b = \ln x + \ln y \quad \checkmark$$
The quotient and power rules follow similarly from $\frac{e^a}{e^b} = e^{a-b}$ and $(e^a)^r = e^{ar}$.
| Operation Inside Log | Becomes Outside Log | Example |
|---|---|---|
| Multiplication $xy$ | Addition $+$ | $\ln(2 \cdot 3) = \ln 2 + \ln 3$ |
| Division $\frac{x}{y}$ | Subtraction $-$ | $\ln\frac{5}{7} = \ln 5 - \ln 7$ |
| Power $x^r$ | Multiplication by $r$ | $\ln(x^3) = 3\ln x$ |
These are common errors: avoid them!
| Wrong | Right | Why |
|---|---|---|
| $\ln(x + y) = \ln x + \ln y$ | No simplification | Logs don't distribute over addition |
| $\ln(x - y) = \ln x - \ln y$ | No simplification | Logs don't distribute over subtraction |
| $\frac{\ln x}{\ln y} = \ln\frac{x}{y}$ | No relation | Division of logs $\neq$ log of quotient |
| $(\ln x)^r = r \ln x$ | $\ln(x^r) = r\ln x$ | Power rule needs the power inside the log |
| $\ln x \cdot \ln y = \ln(xy)$ | No relation | Product of logs $\neq$ log of product |
Memory aid: The log laws convert operations inside the logarithm to operations outside. They don't work in reverse on operations already outside.
Goal: Write a single logarithm as a sum/difference of simpler logarithms.
Strategy: Apply the laws from "inside out": start with the outermost operation.
Example: Expand $\ln \frac{x^2\sqrt{x^2+2}}{3x+1}$.
Step 1: The outermost operation is division, so apply the quotient rule: $$\ln \frac{x^2\sqrt{x^2+2}}{3x+1} = \ln(x^2\sqrt{x^2+2}) - \ln(3x+1)$$
Step 2: In the first term, we have multiplication, so apply the product rule: $$= \ln(x^2) + \ln(\sqrt{x^2+2}) - \ln(3x+1)$$
Step 3: Apply the power rule to each term with a power:
Final answer: $$\ln \frac{x^2\sqrt{x^2+2}}{3x+1} = 2\ln x + \frac{1}{2}\ln(x^2+2) - \ln(3x+1)$$
Goal: Combine multiple logarithms into a single logarithm.
Strategy: Apply the laws in reverse: work from "outside in."
Example: Express $\ln a + \frac{1}{2}\ln b$ as a single logarithm.
Step 1: Apply the power rule in reverse to handle the coefficient: $$\frac{1}{2}\ln b = \ln(b^{1/2}) = \ln\sqrt{b}$$
Step 2: Apply the product rule in reverse (addition → multiplication inside): $$\ln a + \ln\sqrt{b} = \ln(a\sqrt{b})$$
Final answer: $$\ln a + \frac{1}{2}\ln b = \ln(a\sqrt{b})$$
Example: Evaluate $\log_4 2 + \log_4 32$.
Method 1: Use the product rule: $$\log_4 2 + \log_4 32 = \log_4(2 \cdot 32) = \log_4 64 = 3$$ (since $4^3 = 64$)
Method 2: Evaluate each term separately:
| Mistake | Why It's Wrong | Correction |
|---|---|---|
| $\ln(x+y) = \ln x + \ln y$ | Addition inside ≠ addition outside | No simplification possible |
| $\ln(x-y) = \ln x - \ln y$ | Subtraction inside ≠ subtraction outside | No simplification possible |
| $2\ln x = \ln(2x)$ | Coefficient should become exponent | $2\ln x = \ln(x^2)$ |
| $\ln x^2 = (\ln x)^2$ | Different operations | $\ln x^2 = 2\ln x$ |
| Applying laws when $x$ or $y$ is negative | Domain restriction | All arguments must be positive |
Evaluate each expression.
(a) $\log_5 75 - \log_5 3$
(b) $\log_4 8 + \log_4 2$
Expand each expression using the laws of logarithms. Assume all variables represent positive quantities.
(a) $\log_b\left(\frac{m^5}{n^3}\right)$
(b) $\ln\left(\frac{7t^2}{t+4}\right)$
Expand completely: $\ln \frac{t^3 \sqrt{t+1}}{(t-2)^2}$
Express as a single logarithm:
(a) $\ln 6 + 3\ln 2$
(b) $\log_5 x + 2\log_5 y - \frac{1}{2}\log_5(x+y)$
Express as a single logarithm and simplify:
$$2\ln(t+1) - \ln(t^2+3t+2) + \ln(t+2)$$
(a) Prove the power rule: $\ln(x^r) = r\ln x$ for any real number $r$ and $x > 0$.
(b) Explain why $\ln(x + y) \neq \ln x + \ln y$ in general. Give a specific numerical counterexample.
Logarithms as Translators:
Think of logarithms as translating between two "languages":
| Exponential World | Log World |
|---|---|
| $xy$ | $\log x + \log y$ |
| $\frac{x}{y}$ | $\log x - \log y$ |
| $x^r$ | $r \cdot \log x$ |
The logarithm converts from exponential language to log language. The exponential function converts back.
Historical insight: Before calculators, people used log tables to multiply. To compute $347 \times 892$:
This converted hard multiplication into easy addition!
Looking back:
Looking ahead:
Key pattern: Whenever you see products, quotients, or powers inside a logarithm, you can potentially simplify using these laws. Conversely, when you need to combine separate logarithms, use the laws in reverse.
| Previous | Up | Next |
|---|---|---|
| Natural Logarithm | Skills Index | Derivatives of Logarithms |
Last updated: 2026-01-23