Teacher Guide: Product Rule for Logarithms

Learn why the logarithm of a product equals the sum of the logarithms, and use it to expand, condense and evaluate logs.

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10 questions on Logarithms. Students join with a name, you see everyone's score.

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Learning Objectives

State the product rule for logarithms and the conditions on the base and the arguments
Explain the rule using the exponent rule $b^m \cdot b^n = b^{m+n}$
Expand logarithms of products into sums of logarithms
Condense sums of logarithms with the same base into a single logarithm
Approximate logarithms of products from known values such as $\log 2 \approx 0.301$

Prerequisites

• Definition of a Logarithm: $\log_b y = x$ means $b^x = y$
• Exponent rule for multiplying powers: $b^m \cdot b^n = b^{m+n}$
• Common Logarithms (Base 10): reading $\log x$ as $\log_{10} x$
• Properties of Logarithms: overview of the log rules

Discussion Starters

1. Why does multiplying two numbers add their logarithms? How is this the same as $b^m \cdot b^n = b^{m+n}$ ?
2. If $\log 2 \approx 0.301$ , what is $\log 20$ ? What about $\log 200$ ? What pattern do you see?
3. Your friend writes $\log(x + 5) = \log x + \log 5$ . How would you convince them it is wrong?
4. Before calculators, why would a table of logarithms make multiplying large numbers easier?

Common Misconceptions

Treating $\log$ like a number that distributes over addition: $\log(M + N) = \log M + \log N$

Remediation: Have students test it in the calculator sandbox with $\log(6 + 4)$ and $\log 6 + \log 4$ . Stress that $\log$ is a function, not a factor, and that the rule needs a product inside.

Thinking the log of a product is the product of the logs

Remediation: Compare $\log_2(4 \cdot 8) = 5$ with $\log_2 4 \cdot \log_2 8 = 6$ . Link back to exponents: $2^2 \cdot 2^3 = 2^5$ , the exponents are added, not multiplied.

Believing any two logs can be combined, even with different bases

Remediation: Revisit the proof: it needs one base $b$ for both powers. Show that $\log_2 8 + \log_4 16 = 3 + 2 = 5$ , but the product is $8 \cdot 16 = 128$ and $\log_2 128 = 7$ , $\log_4 128 = 3.5$ : neither base gives $5$ .

Differentiation Ideas

For Struggling Students:

• Start with base 2 and powers of 2 so every log is a whole number ( $\log_2 4 + \log_2 8 = \log_2 32$ )
• Write each number as a power first ( $4 = 2^2$ , $8 = 2^3$ ) and add the exponents
• Use the calculator sandbox to check every expansion numerically

For On-Level Students:

• Expand expressions like $\log_5(25xy)$ and condense sums like $\log_6 4 + \log_6 9$
• Approximate $\log 6$ , $\log 12$ and $\log 2000$ from $\log 2$ and $\log 3$
• Identify true and false log statements and correct the false ones

For Advanced Students:

• Prove the product rule from the definition of a logarithm in their own words
• Explain with the graph why $y = \log_2(ax)$ is a vertical shift of $y = \log_2 x$
• Condense expressions with variables and state the domain, like $\log x + \log(x - 4)$

Standards Alignment

HSF-LE.A.4 (CCSS.MATH.CONTENT.HSF.LE.A.4)
Express the solution to an exponential model as a logarithm and evaluate the logarithm using technology
HSF-BF.B.5 (CCSS.MATH.CONTENT.HSF.BF.B.5)
Understand the inverse relationship between exponents and logarithms and use it to solve problems
HSA-SSE.A.2 (CCSS.MATH.CONTENT.HSA.SSE.A.2)
Use the structure of an expression to identify ways to rewrite it

Lesson Resources

visualShifted Log Graph
Slide $a$ and see that $y = \log_2(a x)$ is the graph of $y = \log_2 x$ moved up by $\log_2 a$
activityCalculator Check
Students test true and false statements like $\log 20 = \log 4 + \log 5$ and $\log(6 + 4) = \log 6 + \log 4$
worksheetExpand and Condense
Mixed practice expanding products and condensing sums, plus estimates with $\log 2$ and $\log 3$

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

The product rule for logarithms says that the logarithm of a product is the sum of the logarithms of its factors:

\log_b(MN) = \log_b M + \log_b N

It holds for any base

b > 0

b \neq 1

, and any positive numbers

M

and

N

Why it works: a logarithm is an exponent. Write

M = b^m

and

N = b^n

, so

\log_b M = m

and

\log_b N = n

. When you multiply powers with the same base, you add the exponents:

MN = b^m \cdot b^n = b^{m+n}

So the exponent of

MN

m + n

, which means

\log_b(MN) = m + n = \log_b M + \log_b N

Check with numbers:

\log_2(4 \cdot 8) = \log_2 32 = 5

and

\log_2 4 + \log_2 8 = 2 + 3 = 5

Used from left to right, the rule expands a log of a product into a sum. Used from right to left, it condenses a sum of logs (same base) into one logarithm.

Worked Examples

Show that $\log_2(4 \cdot 8) = \log_2 4 + \log_2 8$ .

Evaluate the left side

$4 \cdot 8 = 32$ and $2^5 = 32$ → $\log_2 32 = 5$

Evaluate each log on the right side

$2^2 = 4$ and $2^3 = 8$ → $\log_2 4 = 2$ , $\log_2 8 = 3$

Add and compare

$2 + 3 = 5$ , the same as the left side → $5 = 5$

Answer: Both sides equal $5$ : multiplying $4$ by $8$ adds the exponents $2$ and $3$ .

Common Mistakes

Writing $\log(M + N) = \log M + \log N$

Why it's wrong: The rule is about a product inside the log, not a sum. Check: $\log(6 + 4) = \log 10 = 1$ , but $\log 6 + \log 4 \approx 0.778 + 0.602 = 1.380$ .

Correct: Only $\log(MN) = \log M + \log N$ . A log of a sum, like $\log(x + 3)$ , cannot be split.

Multiplying the logs: $\log_b(MN) = \log_b M \cdot \log_b N$

Why it's wrong: Multiplying the numbers adds the exponents, it does not multiply them. Check: $\log_2 4 \cdot \log_2 8 = 2 \cdot 3 = 6$ , but $\log_2 32 = 5$ .

Correct: A product inside becomes a sum outside: $\log_2(4 \cdot 8) = 2 + 3 = 5$ .

Combining logs with different bases: $\log_2 x + \log_3 y = \log_6(xy)$

Why it's wrong: The proof uses $b^m \cdot b^n = b^{m+n}$ , which only works when both powers have the same base $b$ .

Correct: Condense only logs with the same base. $\log_2 x + \log_3 y$ stays as it is (or use the Change of Base Formula first).

Using the rule with negative factors: $\log\big((-2)(-5)\big) = \log(-2) + \log(-5)$

Why it's wrong: $\log 10 = 1$ exists, but $\log(-2)$ and $\log(-5)$ do not: the argument of a logarithm must be positive.

Correct: The rule needs $M > 0$ and $N > 0$ . When you condense, keep the domain of the original logs (for example $x > 0$ ).

Why It Matters

The product rule turns multiplication into addition, which is the whole reason logarithms were invented:

Calculating by hand: before calculators, people multiplied large numbers by adding their logarithms from a table (and with slide rules, which add lengths).
Evaluating logs: if you know $\log 2 \approx 0.301$ and $\log 3 \approx 0.477$ , you get $\log 6 \approx 0.778$ without a calculator.
Sound and science: when a sound gets 100 times more intense, its decibel level goes up by a fixed amount, because $\log(100 \cdot I) = 2 + \log I$ .
Solving equations: in Solving Logarithmic Equations you will condense $\log_2 x + \log_2(x+2)$ into one logarithm before you can solve.
Computer science: combining independent choices multiplies the possibilities but adds the bits, since $\log_2(16 \cdot 64) = 4 + 6$ .

Real World Applications

Decibels and Sound Intensity

Sound level in decibels is $L = 10 \log\left(\frac{I}{I_0}\right)$. When the intensity $I$ is multiplied by a factor $k$, the product rule gives $\log(k \cdot \frac{I}{I_0}) = \log k + \log\frac{I}{I_0}$, so the level goes up by $10 \log k$.

Example:

A sound that becomes 10 times more intense goes up by $10 \log 10 = 10$ decibels, whatever level it started at.

1Try It Yourself

A street measures $60$ decibels. A construction machine makes the sound intensity $1000$ times greater.

What is the new sound level in decibels?

Step 1: Write the mathematical expression

Add $10 \log(1000)$ to the old level:

Slide Rules: Multiplying by Adding

A slide rule marks each number $x$ at a distance $\log x$ from the start. Sliding one scale along the other adds two distances, and by the product rule the sum of the distances is the log of the product.

Example:

To multiply $2 \cdot 3$ , add $\log 2 + \log 3 \approx 0.301 + 0.477 = 0.778$ , which is where $6$ is marked.

2Try It Yourself

On a slide rule, $4$ is at distance $\log 4 \approx 0.602$ and $5$ is at distance $\log 5 \approx 0.699$ . Placing them end to end gives a total distance.

Which product does the slide rule show?

Step 1: Write the mathematical expression

Find the number whose log is the total distance:

Bits of Information

In computer science, the number of bits needed to store one of $N$ equally likely options is $\log_2 N$. Combining two independent choices multiplies the options but adds the bits.

Example:

A choice from $8$ symbols needs $\log_2 8 = 3$ bits, a choice from $4$ needs $2$ bits, and both together have $32$ options: $\log_2 32 = 3 + 2 = 5$ bits.

3Try It Yourself

A game character is saved as one of $16$ colors and one of $64$ shapes.

How many bits are needed to store the color and shape together?

Step 1: Write the mathematical expression

Use $\log_2(16 \cdot 64) = \log_2 16 + \log_2 64$ :

Key Takeaways

1Product rule: $\log_b(MN) = \log_b M + \log_b N$ for $M > 0$ , $N > 0$ , $b > 0$ , $b \neq 1$
2It comes from the exponent rule $b^m \cdot b^n = b^{m+n}$ : multiplying numbers adds their exponents
3Expanding: split a log of a product into a sum, like $\log_3(9x) = 2 + \log_3 x$
4Condensing: combine a sum of logs with the same base into one log, like $\log_4 2 + \log_4 8 = \log_4 16$
5With $\log 2 \approx 0.301$ and $\log 3 \approx 0.477$ you can approximate logs of products such as $\log 6 \approx 0.778$
6There is no rule for $\log(M + N)$ : a log of a sum cannot be split

Frequently Asked Questions

Does the product rule work for more than two factors?

Yes. Apply it again and again:

\log_b(MNP) = \log_b M + \log_b N + \log_b P

. For example

\log 42 = \log 2 + \log 3 + \log 7

Does it work for the natural logarithm too?

Yes, it works for every valid base, including

e

\ln(MN) = \ln M + \ln N

. For example

\ln 2 + \ln 5 = \ln 10

Is there a rule for $\log(M + N)$ ?

No. A logarithm of a sum has no simpler form in general.

\log(6 + 4) = \log 10 = 1

, while

\log 6 + \log 4 \approx 1.380

How is this different from the quotient and power rules?

They are siblings of the same idea. The product rule turns multiplication into addition; the Quotient Rule for Logarithms turns division into subtraction; the Power Rule for Logarithms turns a power into multiplication by the exponent.

Glossary

Product rule: $\log_b(MN) = \log_b M + \log_b N$ : the log of a product is the sum of the logs
Expand: Rewrite one logarithm as a sum of simpler logarithms, like $\log(2x) = \log 2 + \log x$
Condense: Combine a sum of logarithms with the same base into a single logarithm
Argument: The number inside a logarithm, $M$ in $\log_b M$ ; it must be positive
Common logarithm: The logarithm with base 10, written $\log x$

Formula Card

Product rule

\log_b(MN) = \log_b M + \log_b N

For $M, N > 0$ and $b > 0$, $b \neq 1$

Where it comes from

b^m \cdot b^n = b^{m+n}

Multiplying powers adds the exponents

Three factors

\log_b(MNP) = \log_b M + \log_b N + \log_b P

Apply the rule repeatedly

Useful values

\log 2 \approx 0.301

\log 3 \approx 0.477

Common logarithms for estimating logs of products

Teacher Guide: Product Rule for Logarithms

Use this lesson with your class

For Teachers

For Struggling Students:

For On-Level Students:

For Advanced Students:

Lesson Content

Definition

Worked Examples

Common Mistakes

Why It Matters

Real World Applications

Decibels and Sound Intensity

Slide Rules: Multiplying by Adding

Bits of Information

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Product rule

Where it comes from

Three factors

Useful values

More in This Topic

Introduction to Logarithms

Definition of a Logarithm

Common Logarithms (Base 10)

Natural Logarithms (ln)