Laws of Logarithms

7 min read

Reference: Stewart §6.3

Multiplication Becomes Addition

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:

Products → Sums
Quotients → Differences
Powers → Products

This makes logarithms incredibly powerful for simplifying expressions and solving equations.

Prerequisite Map

Before You Start

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 $e^{ar}$

Refresh: Exponent Laws

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.

Refresh: Cancellation Equations

For base $b$:

$\log_b(b^x) = x$ for all real $x$
$b^{\log_b x} = x$ for all $x > 0$

For natural logs:

$\ln(e^x) = x$ for all real $x$
$e^{\ln x} = x$ for all $x > 0$

Quick Reference

Property	Value
Concept	Inverse Functions
Course	MATH162
Section	Stewart 6.3
Difficulty	Intermediate
Time	~30 minutes

Key Concepts

The Three Laws of Logarithms

For $b > 0$, $b \neq 1$, and $x, y > 0$:

$$\boxed{\text{Product Rule: } \log_b(xy) = \log_b x + \log_b y}$$

$$\boxed{\text{Quotient Rule: } \log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y}$$

$$\boxed{\text{Power Rule: } \log_b(x^r) = r \log_b x \quad \text{(for any real } r\text{)}}$$

Why These Laws Work

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 $(e^a)^r = e^{ar}$.

The Laws in Table Form

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 $r$	$\ln(x^3) = 3\ln x$

What the Laws Do NOT Say

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.

Expanding Logarithmic Expressions

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:

$\ln(x^2) = 2\ln x$
$\ln(\sqrt{x^2+2}) = \ln((x^2+2)^{1/2}) = \frac{1}{2}\ln(x^2+2)$

Final answer: $$\ln \frac{x^2\sqrt{x^2+2}}{3x+1} = 2\ln x + \frac{1}{2}\ln(x^2+2) - \ln(3x+1)$$

Condensing Logarithmic Expressions

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: $$\ln a + \frac{1}{2}\ln b = \ln(a\sqrt{b})$$

Using Laws to Evaluate Logarithms

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:

$\log_4 2 = \frac{1}{2}$ (since $4^{1/2} = 2$)
$\log_4 32 = \frac{5}{2}$ (since $4^{5/2} = (2^2)^{5/2} = 2^5 = 32$)
Sum: $\frac{1}{2} + \frac{5}{2} = 3$ ✓

Common Pitfalls

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

Practice Problems

Level 1 Basic Law Application

Evaluate each expression.

(a) $\log_5 75 - \log_5 3$

(b) $\log_4 8 + \log_4 2$

Thought Process

(a) Subtraction of logs becomes division inside: $\log_5 75 - \log_5 3 = \log_5\frac{75}{3}$.

(b) Addition of logs becomes multiplication inside: $\log_4 8 + \log_4 2 = \log_4(8 \cdot 2)$.

Show Answer

(a) Using the quotient rule: $$\log_5 75 - \log_5 3 = \log_5\frac{75}{3} = \log_5 25 = \boxed{2}$$ (since $5^2 = 25$)

(b) Using the product rule: $$\log_4 8 + \log_4 2 = \log_4(8 \cdot 2) = \log_4 16 = \boxed{2}$$ (since $4^2 = 16$)

Level 2 Expanding Logarithms

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) $\ln\left(\frac{7t^2}{t+4}\right)$

Thought Process

(a) First apply the quotient rule to separate numerator from denominator, then apply the power rule to each.

(b) First apply the quotient rule, then split the numerator using the product rule, and finally apply the power rule.

Show Answer

(a) Apply quotient rule, then power rule: $$\log_b\left(\frac{m^5}{n^3}\right) = \log_b(m^5) - \log_b(n^3) = \boxed{5\log_b m - 3\log_b n}$$

(b) Apply quotient rule, then product rule and power rule: $$\ln\left(\frac{7t^2}{t+4}\right) = \ln(7t^2) - \ln(t+4) = \ln 7 + \ln(t^2) - \ln(t+4)$$ $$= \boxed{\ln 7 + 2\ln t - \ln(t+4)}$$

Level 3 Expanding Complex Expressions

Expand completely: $\ln \frac{t^3 \sqrt{t+1}}{(t-2)^2}$

Thought Process

Step by step:

Apply quotient rule: numerator minus denominator
The numerator is a product, so apply the product rule
Apply the power rule to $t^3$, $\sqrt{t+1} = (t+1)^{1/2}$, and $(t-2)^2$

Show Answer

Step 1: Apply quotient rule: $$\ln \frac{t^3 \sqrt{t+1}}{(t-2)^2} = \ln(t^3 \sqrt{t+1}) - \ln((t-2)^2)$$

Step 2: Apply product rule to the numerator: $$= \ln(t^3) + \ln(\sqrt{t+1}) - \ln((t-2)^2)$$

Step 3: Apply power rule to each term:

$\ln(t^3) = 3\ln t$
$\ln(\sqrt{t+1}) = \ln((t+1)^{1/2}) = \frac{1}{2}\ln(t+1)$
$\ln((t-2)^2) = 2\ln(t-2)$

Final answer: $$\ln \frac{t^3 \sqrt{t+1}}{(t-2)^2} = \boxed{3\ln t + \frac{1}{2}\ln(t+1) - 2\ln(t-2)}$$

Level 3 Condensing Logarithms

Express as a single logarithm:

(a) $\ln 6 + 3\ln 2$

(b) $\log_5 x + 2\log_5 y - \frac{1}{2}\log_5(x+y)$

Thought Process

(a) First convert $3\ln 2$ to $\ln(2^3)$ using the power rule, then combine using the product rule.

(b) Convert each coefficient to an exponent, then combine: addition → multiplication inside, subtraction → division inside.

Show Answer

(a) Apply power rule to handle coefficient: $$3\ln 2 = \ln(2^3) = \ln 8$$

Apply product rule: $$\ln 6 + \ln 8 = \ln(6 \cdot 8) = \boxed{\ln 48}$$

(b) Apply power rule to handle coefficients:

$2\log_5 y = \log_5(y^2)$
$\frac{1}{2}\log_5(x+y) = \log_5((x+y)^{1/2}) = \log_5\sqrt{x+y}$

Combine using product and quotient rules: $$\log_5 x + \log_5(y^2) - \log_5\sqrt{x+y} = \log_5\frac{xy^2}{\sqrt{x+y}}$$

$$= \boxed{\log_5\frac{xy^2}{(x+y)^{1/2}}}$$

Level 4 Complex Condensing with Factoring

Express as a single logarithm and simplify:

$$2\ln(t+1) - \ln(t^2+3t+2) + \ln(t+2)$$

Thought Process

Convert coefficients to exponents using the power rule
Factor the quadratic $t^2+3t+2 = (t+1)(t+2)$
Combine using product/quotient rules
Look for cancellation: this is where the insight happens!

Show Answer

Step 1: Apply power rule to the first term: $$\ln(t+1)^2 - \ln(t^2+3t+2) + \ln(t+2)$$

Step 2: Factor the quadratic: $$t^2 + 3t + 2 = (t+1)(t+2)$$

So we have: $$\ln(t+1)^2 - \ln[(t+1)(t+2)] + \ln(t+2)$$

Step 3: Combine using product and quotient rules: $$= \ln\frac{(t+1)^2 \cdot (t+2)}{(t+1)(t+2)}$$

Step 4: Simplify by canceling common factors: $$= \ln\frac{(t+1)^2 (t+2)}{(t+1)(t+2)} = \ln(t+1)$$

Final answer: $$= \boxed{\ln(t+1)}$$

Note: The dramatic simplification happens because the factored quadratic contains both $(t+1)$ and $(t+2)$, which cancel with terms in the numerator.

Level 5 Conceptual: Why the Laws Work

(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.

Thought Process

(a) Start with the identity $x = e^{\ln x}$, raise both sides to power $r$, then take $\ln$ of both sides.

(b) Choose simple positive values for $x$ and $y$, compute both sides, and show they’re different.

Show Answer

(a) Proof of the Power Rule:

Start with the cancellation equation: $x = e^{\ln x}$ for $x > 0$.

Raise both sides to the power $r$: $$x^r = (e^{\ln x})^r$$

Apply the exponent law $(e^a)^r = e^{ar}$: $$x^r = e^{r \ln x}$$

Take $\ln$ of both sides: $$\ln(x^r) = \ln(e^{r \ln x})$$

Apply the cancellation equation $\ln(e^y) = y$: $$\ln(x^r) = r \ln x \quad \blacksquare$$

(b) Why $\ln(x+y) \neq \ln x + \ln y$:

The product rule says: $\ln(xy) = \ln x + \ln y$.

If $\ln(x+y)$ also equaled $\ln x + \ln y$, then we’d have $\ln(xy) = \ln(x+y)$, which would mean $xy = x + y$ for all positive $x, y$. This is false.

Counterexample: Let $x = y = 2$.

Left side: $\ln(2 + 2) = \ln 4 \approx 1.386$
Right side: $\ln 2 + \ln 2 = 2\ln 2 \approx 1.386$

Wait, these are equal! That is because $\ln 4 = \ln(2^2) = 2\ln 2$. Try $x = 1, y = 2$:

Left side: $\ln(1 + 2) = \ln 3 \approx 1.099$
Right side: $\ln 1 + \ln 2 = 0 + 0.693 = 0.693$

These are different: $1.099 \neq 0.693$ ✓

The underlying reason: The laws connect multiplication/division with addition/subtraction. Addition inside doesn’t have a corresponding outside operation for logs.

Common Misconceptions

Common misconception

$\ln(x + y) = \ln x + \ln y$, extending the product rule to sums.

This is the multiplicative-not-additive error. The product rule $\ln(xy) = \ln x + \ln y$ applies to a product inside the logarithm, not a sum. There is no simplification for $\ln(x + y)$. A numerical check shows the error: $\ln(1 + 2) = \ln 3 \approx 1.099$, while $\ln 1 + \ln 2 = 0 + 0.693 = 0.693$; these are not equal.

Mastery Checklist

I can state all three laws: product, quotient, and power rules
I can expand a logarithm of a product into a sum of logarithms
I can expand a logarithm of a quotient into a difference of logarithms
I can bring exponents out front as coefficients
I can condense sums/differences of logs into a single logarithm
I know that $\ln(x+y) \neq \ln x + \ln y$ (and why)
I can use log laws to simplify expressions before evaluating
I can factor expressions inside logarithms to enable simplification

Mental Model

Logarithms as Translators:

Think of logarithms as translating between two “languages”:

Exponential language: multiplication, division, powers
Logarithmic language: addition, subtraction, multiplication by scalars

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 $347 \times 892$:

Look up $\log(347) \approx 2.540$ and $\log(892) \approx 2.950$
Add: $2.540 + 2.950 = 5.490$
Look up what number has log $5.490$: antilog gives $\approx 309,000$

This converted hard multiplication into easy addition!

Connections

Looking back:

Logarithm Definition: The laws follow from the definition and exponent rules
Natural Logarithm: $\ln$ is the most common log in calculus
Exponent laws: Each log law mirrors an exponent law

Looking ahead:

Derivatives (Section 6.4): The power rule $\ln(x^r) = r\ln x$ is used in logarithmic differentiation
Solving equations: Log laws help isolate variables in exponential equations
Integration: Recognizing $\ln$ patterns is key to integrating rational functions

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.

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Last updated: 2026-01-23