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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.

For Teachers

Learning Objectives
  • State the product rule for logarithms and the conditions on the base and the arguments
  • Explain the rule using the exponent rule bm⋅bn=bm+nb^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≈0.301\log 2 \approx 0.301
Prerequisites
  • • Definition of a Logarithm: log⁡by=x\log_b y = x means bx=yb^x = y
  • • Exponent rule for multiplying powers: bm⋅bn=bm+nb^m \cdot b^n = b^{m+n}
  • • Common Logarithms (Base 10): reading log⁡x\log x as log⁡10x\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 bm⋅bn=bm+nb^m \cdot b^n = b^{m+n}?
  • 2. If log⁡2≈0.301\log 2 \approx 0.301, what is log⁡20\log 20? What about log⁡200\log 200? What pattern do you see?
  • 3. Your friend writes log⁡(x+5)=log⁡x+log⁡5\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⁡\log like a number that distributes over addition: log⁡(M+N)=log⁡M+log⁡N\log(M + N) = \log M + \log N

Remediation: Have students test it in the calculator sandbox with log⁡(6+4)\log(6 + 4) and log⁡6+log⁡4\log 6 + \log 4. Stress that log⁡\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⋅8)=5\log_2(4 \cdot 8) = 5 with log⁡24⋅log⁡28=6\log_2 4 \cdot \log_2 8 = 6. Link back to exponents: 22⋅23=252^2 \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 bb for both powers. Show that log⁡28+log⁡416=3+2=5\log_2 8 + \log_4 16 = 3 + 2 = 5, but the product is 8⋅16=1288 \cdot 16 = 128 and log⁡2128=7\log_2 128 = 7, log⁡4128=3.5\log_4 128 = 3.5: neither base gives 55.

Differentiation Ideas

For Struggling Students:

  • • Start with base 2 and powers of 2 so every log is a whole number (log⁡24+log⁡28=log⁡232\log_2 4 + \log_2 8 = \log_2 32)
  • • Write each number as a power first (4=224 = 2^2, 8=238 = 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)\log_5(25xy) and condense sums like log⁡64+log⁡69\log_6 4 + \log_6 9
  • • Approximate log⁡6\log 6, log⁡12\log 12 and log⁡2000\log 2000 from log⁡2\log 2 and log⁡3\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)y = \log_2(ax) is a vertical shift of y=log⁡2xy = \log_2 x
  • • Condense expressions with variables and state the domain, like log⁡x+log⁡(x−4)\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 aa and see that y=log⁡2(ax)y = \log_2(a x) is the graph of y=log⁡2xy = \log_2 x moved up by log⁡2a\log_2 a

  • activityCalculator Check

    Students test true and false statements like log⁡20=log⁡4+log⁡5\log 20 = \log 4 + \log 5 and log⁡(6+4)=log⁡6+log⁡4\log(6 + 4) = \log 6 + \log 4

  • worksheetExpand and Condense

    Mixed practice expanding products and condensing sums, plus estimates with log⁡2\log 2 and log⁡3\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⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N
It holds for any base b>0b > 0, b≠1b \neq 1, and any positive numbers MM and NN.
Why it works: a logarithm is an exponent. Write M=bmM = b^m and N=bnN = b^n, so log⁡bM=m\log_b M = m and log⁡bN=n\log_b N = n. When you multiply powers with the same base, you add the exponents:
MN=bm⋅bn=bm+nMN = b^m \cdot b^n = b^{m+n}
So the exponent of MNMN is m+nm + n, which means log⁡b(MN)=m+n=log⁡bM+log⁡bN\log_b(MN) = m + n = \log_b M + \log_b N.
Check with numbers: log⁡2(4⋅8)=log⁡232=5\log_2(4 \cdot 8) = \log_2 32 = 5 and log⁡24+log⁡28=2+3=5\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⋅8)=log⁡24+log⁡28\log_2(4 \cdot 8) = \log_2 4 + \log_2 8.
1

Evaluate the left side

4⋅8=324 \cdot 8 = 32 and 25=322^5 = 32 → log⁡232=5\log_2 32 = 5
2

Evaluate each log on the right side

22=42^2 = 4 and 23=82^3 = 8 → log⁡24=2\log_2 4 = 2, log⁡28=3\log_2 8 = 3
3

Add and compare

2+3=52 + 3 = 5, the same as the left side → 5=55 = 5

Common Mistakes

Writing log⁡(M+N)=log⁡M+log⁡N\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\log(6 + 4) = \log 10 = 1, but log⁡6+log⁡4≈0.778+0.602=1.380\log 6 + \log 4 \approx 0.778 + 0.602 = 1.380.

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

Multiplying the logs: log⁡b(MN)=log⁡bM⋅log⁡bN\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⁡24⋅log⁡28=2⋅3=6\log_2 4 \cdot \log_2 8 = 2 \cdot 3 = 6, but log⁡232=5\log_2 32 = 5.

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

Combining logs with different bases: log⁡2x+log⁡3y=log⁡6(xy)\log_2 x + \log_3 y = \log_6(xy)

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

Correct: Condense only logs with the same base. log⁡2x+log⁡3y\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⁡((−2)(−5))=log⁡(−2)+log⁡(−5)\log\big((-2)(-5)\big) = \log(-2) + \log(-5)

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

Correct: The rule needs M>0M > 0 and N>0N > 0. When you condense, keep the domain of the original logs (for example x>0x > 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≈0.301\log 2 \approx 0.301 and log⁡3≈0.477\log 3 \approx 0.477, you get log⁡6≈0.778\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⋅I)=2+log⁡I\log(100 \cdot I) = 2 + \log I.
  • Solving equations: in Solving Logarithmic Equations you will condense log⁡2x+log⁡2(x+2)\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⋅64)=4+6\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 10log⁡10=1010 \log 10 = 10 decibels, whatever level it started at.

1Try It Yourself

A street measures 6060 decibels. A construction machine makes the sound intensity 10001000 times greater.

What is the new sound level in decibels?

Step 1: Write the mathematical expression

Add 10log⁡(1000)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⋅32 \cdot 3, add log⁡2+log⁡3≈0.301+0.477=0.778\log 2 + \log 3 \approx 0.301 + 0.477 = 0.778, which is where 66 is marked.

2Try It Yourself

On a slide rule, 44 is at distance log⁡4≈0.602\log 4 \approx 0.602 and 55 is at distance log⁡5≈0.699\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 88 symbols needs log⁡28=3\log_2 8 = 3 bits, a choice from 44 needs 22 bits, and both together have 3232 options: log⁡232=3+2=5\log_2 32 = 3 + 2 = 5 bits.

3Try It Yourself

A game character is saved as one of 1616 colors and one of 6464 shapes.

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

Step 1: Write the mathematical expression

Use log⁡2(16⋅64)=log⁡216+log⁡264\log_2(16 \cdot 64) = \log_2 16 + \log_2 64:

Key Takeaways

  • 1Product rule: log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N for M>0M > 0, N>0N > 0, b>0b > 0, b≠1b \neq 1
  • 2It comes from the exponent rule bm⋅bn=bm+nb^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⁡3x\log_3(9x) = 2 + \log_3 x
  • 4Condensing: combine a sum of logs with the same base into one log, like log⁡42+log⁡48=log⁡416\log_4 2 + \log_4 8 = \log_4 16
  • 5With log⁡2≈0.301\log 2 \approx 0.301 and log⁡3≈0.477\log 3 \approx 0.477 you can approximate logs of products such as log⁡6≈0.778\log 6 \approx 0.778
  • 6There is no rule for log⁡(M+N)\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⁡bM+log⁡bN+log⁡bP\log_b(MNP) = \log_b M + \log_b N + \log_b P. For example log⁡42=log⁡2+log⁡3+log⁡7\log 42 = \log 2 + \log 3 + \log 7.

Does it work for the natural logarithm too?

Yes, it works for every valid base, including ee: ln⁡(MN)=ln⁡M+ln⁡N\ln(MN) = \ln M + \ln N. For example ln⁡2+ln⁡5=ln⁡10\ln 2 + \ln 5 = \ln 10.

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

No. A logarithm of a sum has no simpler form in general. log⁡(6+4)=log⁡10=1\log(6 + 4) = \log 10 = 1, while log⁡6+log⁡4≈1.380\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⁡bM+log⁡bN\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\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, MM in log⁡bM\log_b M; it must be positive
Common logarithm
The logarithm with base 10, written log⁡x\log x

Formula Card

Product rule

log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N

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

Where it comes from

bm⋅bn=bm+nb^m \cdot b^n = b^{m+n}

Multiplying powers adds the exponents

Three factors

log⁡b(MNP)=log⁡bM+log⁡bN+log⁡bP\log_b(MNP) = \log_b M + \log_b N + \log_b P

Apply the rule repeatedly

Useful values

log⁡2≈0.301\log 2 \approx 0.301, log⁡3≈0.477\log 3 \approx 0.477

Common logarithms for estimating logs of products

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