The product, quotient and power properties of logarithms, expanding a single logarithm into several and condensing several into one, the change-of-base formula for evaluating any logarithm on a calculator, and applying the properties to a decibel model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 7 — Exponential and Logarithmic Functions
Apply Properties of Logarithms
Objectives
Five outcomes. Three properties, and they all come from the exponent rules you already have.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 506-513 — the lesson these objectives are drawn from
Warm-up
Lesson 5.1 gave you the exponent rules, and Lesson 7.4 made a logarithm an exponent.
Discussion prompt
Two cubed times 2 to the fifth is 2 to the eighth. If a logarithm is an exponent, what should happen to logarithms when the numbers they describe are multiplied?
Hint: Exponents add when powers multiply.
Answer:
\[ 2^3 \cdot 2^5 = 2^{3+5} = 2^8 \]
The exponents added. Since a logarithm is the exponent, the logarithm of a product must be the sum of the logarithms.
\[ \log_2(8 \cdot 32) = \log_2 8 + \log_2 32 = 3 + 5 = 8 \]
Every property in this lesson is one of Lesson 5.1's exponent rules, translated. Nothing new is being asserted; the same facts are being read from the other side.
Concept
The logarithm of a product is a sum, the logarithm of a quotient is a difference, and the logarithm of a power is a multiple. Each property replaces an operation with the easier one below it.
product property of logarithms — For positive b, m and n with b not 1, the logarithm of m times n equals the logarithm of m plus the logarithm of n.
\[ \log_b mn = \log_b m + \log_b n \]
This is why logarithm tables ran navigation and astronomy for three hundred years: multiplying seven-digit numbers by hand is brutal, and adding them is not.
Figure (svg): The three properties of logarithms, each turning one operation into a simpler one
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 507-507
Section
Section 1
Concept
For positive b, m and n with b not 1: the logarithm of a product is the sum of the logarithms, the logarithm of a quotient is their difference, and the logarithm of a power is the exponent times the logarithm.
\[ \log_b mn = \log_b m + \log_b n, \quad \log_b\tfrac{m}{n} = \log_b m - \log_b n, \quad \log_b m^n = n\log_b m \]
The two warnings printed beside them matter. The logarithm of a quotient is a difference of logarithms, not a quotient of them, and the logarithm of a product is a sum, not a product.
Figure (svg): The three properties of logarithms, each turning one operation into a simpler one
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 507-507 — Properties of Logarithms
Picture it
Multiplication drops to addition, division to subtraction, and an exponent to a factor.
Figure (svg): The three properties of logarithms, each turning one operation into a simpler one
The bottom bar is the pair of mistakes the book flags directly. Both come from reading the property as if the logarithm distributed over the operation.
Worked example
Example 1, all three parts.
\[ \text{Given } \log_4 3 \approx 0.792 \text{ and } \log_4 7 \approx 1.404, \text{ evaluate } \log_4\tfrac{3}{7}, \; \log_4 21, \; \log_4 49. \]
First: use the quotient property
Why: The logarithm of 3 over 7 is the difference of the two given values.
\[ 0.792 - 1.404 = -0.612 \]
Second: rewrite 21 as a product
Why: Twenty-one is 3 times 7, both of which are given.
\[ \log _{4}(3 \cdot 7) \]
Second: use the product property
Why: The logarithm of the product is the sum.
\[ 0.792 + 1.404 = 2.196 \]
Third: rewrite 49 as a power and use the power property
Why: Forty-nine is 7 squared, so the logarithm is twice the given value.
\[ 2(1.404) = 2.808 \]
Figure (svg): Three logarithms evaluated from two given values using the three properties
\[ -0.612, \quad 2.196, \quad 2.808 \]
Verify: check the third against the second
Why: Forty-nine is larger than 21, and 2.808 is larger than 2.196 — consistent. And since 3 is below 4 while 7 is above it, the first logarithm should be below 1 and the second above it, which the given values confirm. Sign and size checks like these catch a property applied in the wrong direction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 507-507
Matching
Each lowers one operation.
Match the pairs
Why: The last row is the power property and the fact that the logarithm of the base is 1, combined: n times 1 is n. Lesson 7.4's inverse property and this lesson's power property agree, as they must.
Worked example
Guided Practice 1 to 4.
\[ \text{Given } \log_6 5 \approx 0.898 \text{ and } \log_6 8 \approx 1.161, \text{ evaluate } \log_6\tfrac{5}{8}, \; \log_6 40, \; \log_6 64, \; \log_6 125. \]
First: quotient property
Why: The difference of the two values.
\[ 0.898 - 1.161 = -0.263 \]
Second: 40 is 5 times 8
Why: So the logarithm is the sum.
\[ 0.898 + 1.161 = 2.059 \]
Third: 64 is 8 squared
Why: So the logarithm is twice the second value.
\[ 2(1.161) = 2.322 \]
Fourth: 125 is 5 cubed
Why: So the logarithm is three times the first value.
\[ 3(0.898) = 2.694 \]
Figure (svg): The solution to Worked example four more from given values shown as a ladder of expressions, one row per algebraic move
\[ -0.263, \; 2.059, \; 2.322, \; 2.694 \]
Verify: check the first sign
Why: Five is smaller than 8, so five eighths is below 1, and the logarithm of a number below 1 to a base above 1 must be negative — which negative 0.263 is. The other three arguments all exceed 6, so all three logarithms must exceed 1, and they do.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 507-507
Trap
\[ \log_b\tfrac{m}{n} \]
Split it as a quotient of logarithms
Why: The fraction bar is carried through the logarithm sign.
\[ = \tfrac{\log_b m}{\log_b n} \quad \text{(wrong)} \]
Test it: the logarithm base 2 of 8 over 2 is the logarithm of 4, which is 2. But 3 divided by 1 is 3. The two disagree.
\[ \log_b\tfrac{m}{n} = \log_b m - \log_b n \]
Lower the operation instead of copying it
Why: Division becomes subtraction, not division.
\[ \log_2\tfrac{8}{2} = \log_2 8 - \log_2 2 = 3 - 1 = 2 \quad \checkmark \]
The same warning applies to products: the logarithm of a product is a sum, not a product. The book prints both cautions beside the properties for exactly this reason.
Fill the middle
Guided Practice 2.
Fill in the blanks
\log_6 40 = \log_6(5 \cdot 8) = 0.898 + 1.161
Why: Forty had to be rewritten as 5 times 8 before either given value could be used. Spotting that rewriting is the real work; applying the property afterwards is one line of arithmetic.
Two truths and a lie
All three are about the properties of logarithms.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The logarithm of a quotient is a DIFFERENCE of logarithms, not a quotient of them. Testing base 2 with m equal to 8 and n equal to 2 gives 2 on the correct side and 3 on the false one. The book prints this warning beside the properties because it is the most common error in the lesson.
Prediction
Commit before reasoning.
Predict first
Why is the logarithm of a product the sum of the logarithms?
Correct: Because exponents add when powers with the same base are multiplied.
\[ m = b^s, \; n = b^t \;\Rightarrow\; mn = b^{s+t} \;\Rightarrow\; \log_b mn = s+t \]
Why: Write m as b to the s and n as b to the t. Then mn is b to the s plus t, so its logarithm is s plus t — which is the logarithm of m plus the logarithm of n. Every property in this lesson is one of Lesson 5.1's exponent rules read through the definition of a logarithm, which is why none of them needs to be memorised separately from the rules you already have.
Section
Section 2
Concept
To expand, apply the properties from the outside in: split a quotient first, then a product, then bring any exponents down as coefficients. Assume every variable is positive.
\[ \log_6\frac{5x^3}{y} = \log_6 5 + 3\log_6 x - \log_6 y \]
The order matters only for tidiness, not correctness, but working outermost operation first keeps every line short and makes an error easy to spot.
Figure (svg): One logarithm expanded into three simple ones, one property applied per line
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 508-508 — Expand a logarithmic expression
Picture it
Example 2, expanded in three steps.
Figure (svg): One logarithm expanded into three simple ones, one property applied per line
The fraction went first, then the product, then the exponent. Three lines, three properties, and no step doing two things at once.
Worked example
Example 2.
\[ \text{Expand } \log_6\frac{5x^3}{y}. \]
Split the quotient
Why: The outermost operation is the division, so use the quotient property first.
\[ \log _{6}(5 x ^{3}) - \log _{6} y \]
Split the product
Why: Five times x cubed is a product, so use the product property.
\[ \log _{6} 5 + \log _{6}(x ^{3}) - \log _{6} y \]
Bring the exponent down
Why: The power property turns the 3 into a coefficient.
\[ \log _{6} 5 + 3 \log _{6} x - \log _{6} y \]
Stop
Why: Nothing left has a product, quotient or power inside a logarithm.
Figure (svg): One logarithm expanded into three simple ones, one property applied per line
\[ \log_6 5 + 3\log_6 x - \log_6 y \]
Verify: check at a value
Why: Take x equal to 6 and y equal to 6. The original is the logarithm base 6 of 5 times 216 over 6, which is the logarithm of 180. The expansion gives the logarithm of 5, plus 3, minus 1 — that is the logarithm of 5 plus 2. Since 180 is 5 times 36, its logarithm is the logarithm of 5 plus 2. They agree.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 508-508
Sorting
Work from the outermost operation inward.
Sort into buckets
Sort each expression by the property you would use first.
Choosing the outermost operation first is a habit, not a requirement — but it keeps every line to one property and makes a slip easy to locate.
Worked example
Guided Practice 5 and lesson exercises 15 and 22.
\[ \text{Expand } \log 3x^4, \; \log_3 4x, \text{ and } \ln 4x^2y. \]
First: split the product
Why: Three times x to the fourth is a product.
\[ \log 3 + \log(x ^{4}) \]
First: bring the exponent down
Why: The power property gives a coefficient of 4.
\[ \log 3 + 4 \log x \]
Second: split the product
Why: There is no exponent to lower here.
\[ \log _{3} 4 + \log _{3} x \]
Third: split twice, then lower
Why: Three factors give three logarithms, and the square on x becomes a 2.
\[ \ln 4 + 2 \ln x + \ln y \]
Figure (svg): The solution to Worked example two more expansions shown as a ladder of expressions, one row per algebraic move
\[ \log 3 + 4\log x; \quad \log_3 4 + \log_3 x; \quad \ln 4 + 2\ln x + \ln y \]
Verify: check that only x carried the exponent in the third
Why: In 4x squared y the square attaches to x alone, so only that term gets a coefficient of 2. Had the whole product been squared, every term would have. Reading which factor an exponent belongs to is the step that decides the answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 508-508
Error analysis
A student expands a logarithm of a product, from lesson exercise 31.
Annotate
On: \( \log_2 5x = (\log_2 5)(\log_2 x) \)
Every property lowers the operation by one level. A logarithm of a product is a sum, and a logarithm of a quotient is a difference.
Fill the middle
Guided Practice 5.
Fill in the blanks
\log 3x^4 = \log 3 + 4\log x
Why: The exponent 4 becomes a coefficient in front of the logarithm of x. Only x is raised to the fourth here — the 3 is not — which is why the first term keeps no coefficient.
Matching
Watch which factor carries the exponent.
Match the pairs
Why: The last two used a second fact after expanding: the logarithm base 6 of 36 is 2, and the logarithm base 2 of 2 is 1. Simplifying any numerical logarithm you recognise is always worth doing once the expansion is finished.
Prediction
Commit before reasoning.
Predict first
Expand ln of 8 x cubed. What happens to the 8?
Correct: It stays as ln 8, since only x is cubed.
\[ \ln 8x^3 = \ln 8 + 3\ln x, \text{ not } 3\ln 8 + \ln x \]
Why: In 8 x cubed the exponent attaches to x alone, so the correct expansion is the natural logarithm of 8 plus 3 times the natural logarithm of x. Writing 3 ln 8 plus ln x, which is lesson exercise 32's printed error, gets both terms wrong at once. Deciding what an exponent is attached to is a Lesson 5.1 reading skill, not a new one.
Section
Section 3
Concept
To condense, run the properties backwards: clear every coefficient with the power property first, then join sums into products and differences into quotients. The result is a single logarithm.
\[ \log 3 + 3\log 4 - \log 6 = \log\frac{3 \cdot 4^3}{6} = \log 32 \]
Coefficients must go first. A number in front of a logarithm blocks the product and quotient rules, which only join logarithms that stand alone.
Figure (svg): Three separate logarithms condensed into one, one property applied per line
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 508-508 — Condense a logarithmic expression
Picture it
Example 3, condensed in three steps.
Figure (svg): Three separate logarithms condensed into one, one property applied per line
The 3 went back up as an exponent before anything could be joined. Then the sum became a product and the difference became a quotient.
Worked example
Example 3, a multiple-choice item.
\[ \text{Condense } \log 3 + 3\log 4 - \log 6. \]
Clear the coefficient
Why: The power property sends the 3 back up as an exponent on 4.
\[ \log 3 + \log(4 ^{3}) - \log 6 \]
Join the sum
Why: The product property turns the sum into one logarithm of a product.
\[ \log(3 \cdot 64) = \log 192 \]
Join the difference
Why: The quotient property turns the subtraction into a division.
\[ \log(\frac{192}{6}) \]
Simplify
Why: One hundred ninety-two divided by 6 is 32.
\[ \log 32 \]
Figure (svg): Three separate logarithms condensed into one, one property applied per line
\[ \log 32 \]
Verify: check numerically
Why: The original is 0.4771 plus 3 times 0.6021 minus 0.7782, which is 0.4771 plus 1.8062 minus 0.7782, or 1.5051. And the common logarithm of 32 is 1.5051. They match, which confirms both the property work and the arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 508-508
Ranking
Condensing a sum and difference with coefficients.
Put in order
Why: Step one cannot be skipped or delayed: a coefficient in front of a logarithm blocks the joining rules entirely. The middle two may be done in either order, but doing sums first keeps the numerator assembled before the division appears.
Worked example
Guided Practice 6 and lesson exercise 43.
\[ \text{Condense } \ln 4 + 3\ln 3 - \ln 12, \text{ and } 3\log_4 6. \]
First: clear the coefficient
Why: Three times the natural logarithm of 3 becomes the natural logarithm of 27.
\[ \ln 4 + \ln 27 - \ln 12 \]
First: join the sum, then the difference
Why: Four times 27 is 108, divided by 12.
\[ \ln(\frac{108}{12}) \]
First: simplify
Why: One hundred eight divided by 12 is 9.
\[ \ln 9 \]
Second: one property is enough
Why: The power property alone turns the coefficient into an exponent, and 6 cubed is 216.
\[ \log _{4} 216 \]
Figure (svg): The solution to Worked example two more condensations shown as a ladder of expressions, one row per algebraic move
\[ \ln 9 \quad \text{and} \quad \log_4 216 \]
Verify: check the first numerically
Why: The original is 1.3863 plus 3 times 1.0986 minus 2.4849, which is 1.3863 plus 3.2958 minus 2.4849, or 2.1972. And the natural logarithm of 9 is 2.1972. They agree, and 9 is indeed 108 over 12.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 508-508
Trap
\[ 2\log x + \log 11 \]
Join the sum straight away
Why: The product property is applied while a coefficient is still in front.
\[ = 2\log(11x) \quad \text{(wrong)} \]
At x equal to 10 the correct value is 2 plus 1.041, or 3.041. The wrong version gives 2 times 2.041, which is 4.083.
\[ 2\log x + \log 11 = \log x^2 + \log 11 \]
Send the coefficient up first, then join
Why: The product property joins logarithms that stand alone, with no coefficient attached.
\[ = \log 11x^2 \]
Order matters in one direction only: condensing needs the power property first, while expanding needs it last. Both times it is the coefficient that has to be dealt with at the right moment.
Fill the middle
Example 3, first step.
Fill in the blanks
3\log 4 = \log 4^3} = \log 64
Why: The coefficient 3 becomes the exponent, and 4 cubed is 64. This is the power property read right to left, which is the only direction that is useful when condensing.
Matching
Coefficients first.
Match the pairs
Why: The first needed no power property at all, since neither term carried a coefficient. The other three all did, and in each case sending the coefficient up was what made the joining possible.
Comparison
Fill the blanks. Same properties, opposite directions.
Comparison matrix
| Question | Expanding | Condensing |
|---|---|---|
| Starts with | one logarithm | several logarithms |
| Power property used | last, to lower exponents | first, to clear coefficients |
| Products become | sums | products again |
| Typical use | preparing to solve | finishing an answer |
Knowing which direction a question wants decides where the power property goes, and that single choice is what most condensing errors turn on.
Section
Section 4
Concept
For positive a, b and c with b and c not 1, the logarithm of a to base c equals the logarithm of a to base b divided by the logarithm of c to base b. Choosing base 10 or base e makes any logarithm computable.
\[ \log_c a = \frac{\log_b a}{\log_b c} = \frac{\log a}{\log c} = \frac{\ln a}{\ln c} \]
A calculator has only two logarithm keys, common and natural. This formula is what lets those two keys evaluate a logarithm to any base at all.
Figure (svg): The change-of-base formula and the same logarithm evaluated two ways
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 508-509 — Change-of-Base Formula
Picture it
Example 4: the logarithm base 3 of 8, by common logarithms and by natural ones.
Figure (svg): The change-of-base formula and the same logarithm evaluated two ways
Both give 1.893. The new base cancels between the numerator and the denominator, so it can be anything convenient.
Worked example
Example 4, both routes.
\[ \text{Evaluate } \log_3 8 \text{ using common logarithms and using natural logarithms.} \]
Set up with common logarithms
Why: The argument goes on top and the old base underneath.
\[ \log 8 / \log 3 \]
Compute
Why: Point nine oh three one divided by 0.4771.
\[ \text{about } 1.893 \]
Set up with natural logarithms
Why: The same arrangement, with ln in place of log.
\[ \ln 8 / \ln 3 \]
Compute
Why: Two point oh seven nine four divided by 1.0986.
\[ \text{about } 1.893 \]
Figure (svg): The change-of-base formula and the same logarithm evaluated two ways
\[ \log_3 8 \approx 1.893 \]
Verify: bracket the answer
Why: Three to the first is 3 and 3 squared is 9, and 8 lies between them, so the logarithm must lie between 1 and 2 — and 1.893 does, close to 2 because 8 is close to 9. Bracketing before computing catches an inverted fraction, which is by far the most common slip here.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 509-509
Fill the middle
Guided Practice 7.
Fill in the blanks
\log_5 8 = \frac5___}} \approx 1.292
Why: The old base, 5, goes in the denominator and the argument, 8, on top. Getting these the wrong way round produces the logarithm base 8 of 5 instead, which is the most common error in this idea.
Worked example
Guided Practice 7 to 10.
\[ \text{Evaluate } \log_5 8, \; \log_8 14, \; \log_{26} 9, \; \log_{12} 30. \]
First: natural logarithms
Why: Two point oh seven nine four over 1.6094.
\[ \text{about } 1.292 \]
Second
Why: Two point six three nine one over 2.0794.
\[ \text{about } 1.269 \]
Third
Why: Two point one nine seven two over 3.2581.
\[ \text{about } 0.674 \]
Fourth
Why: Three point four oh one two over 2.4849.
\[ \text{about } 1.369 \]
Figure (svg): The solution to Worked example four evaluations shown as a ladder of expressions, one row per algebraic move
\[ 1.292, \; 1.269, \; 0.674, \; 1.369 \]
Verify: check the third against the rest
Why: Nine is smaller than its base 26, so that logarithm must be below 1 — and it is the only one of the four that is. The other three all have arguments larger than their bases, so all three exceed 1. Comparing the argument with the base predicts the answer's size before any division.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 509-509
Error analysis
A student evaluates a logarithm with base 3 and argument 8.
Annotate
On: \( \log_3 8 = \frac{\log 3}{\log 8} \approx 0.528 \)
The answer the student found is the logarithm base 8 of 3, a genuine quantity but not the one asked for. A single bracketing check rules it out immediately.
Sorting
Compare the argument with the base.
Sort into buckets
Sort each logarithm before computing it.
This single comparison predicts whether an answer should be above or below 1, which is enough to catch an inverted change-of-base fraction every time.
Prediction
Commit before reasoning.
Predict first
Example 4 evaluated the same logarithm with base 10 and with base e and got 1.893 both times. Coincidence?
Correct: No — the new base cancels, so any base gives the same value.
\[ \frac{\log_b a}{\log_b c} = \frac{k\log_b c}{\log_b c} = k, \text{ for every allowed } b \]
Why: Write a as c to the k, so that k is the answer. Then the logarithm of a to any base b is k times the logarithm of c to base b, and dividing by that second logarithm returns k regardless of what b was. Base 10 and base e are used only because those are the keys a calculator has; base 2 or base 7 would give 1.893 as well.
Matching
Predict the size first, then match.
Match the pairs
Why: The third is the only value below 1, and it is the only one whose argument is smaller than its base. Sorting by that comparison narrows four choices to one before any arithmetic is done.
Section
Section 5
Concept
Loudness in decibels is ten times the common logarithm of a sound's intensity divided by a reference intensity. Because the logarithm turns a product into a sum, multiplying the intensity adds a fixed amount to the loudness, no matter what the original intensity was.
\[ L(I) = 10\log\frac{I}{I_0}, \quad I_0 = 10^{-12} \]
That fixed amount is the whole reason the decibel scale is useful: the difference between two sounds depends only on their ratio, never on how loud either one is.
Figure (svg): Repeated doublings of sound intensity plotted against the rise in loudness
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 509-509 — Use properties of logarithms in real life
Picture it
Example 5: repeated doublings against the decibel gain.
Figure (svg): Repeated doublings of sound intensity plotted against the rise in loudness
Eight doublings multiply the intensity by 256 and raise the loudness by only 24 decibels. The graph is a straight line because each doubling adds the same 3.01.
Worked example
Example 5.
\[ \text{With } L(I) = 10\log\tfrac{I}{I_0}, \text{ find the increase in loudness when the intensity doubles.} \]
Write the increase
Why: The new loudness minus the old one, with 2I in place of I.
\[ L(2 I) - L(I) \]
Substitute and distribute
Why: Ten times the difference of the two logarithms.
\[ 10(\log(2 I / I _{0}) - \log(I / I _{0})) \]
Use the product property
Why: Two I over I nought splits into the logarithm of 2 plus the logarithm of I over I nought.
\[ 10(\log 2 + \log(I / I _{0}) - \log(I / I _{0})) \]
Simplify
Why: The two matching terms cancel, leaving ten times the logarithm of 2.
\[ 10 \log 2 = 3.01 \]
Figure (svg): Repeated doublings of sound intensity plotted against the rise in loudness
\[ 10\log 2 \approx 3.01 \text{ dB} \]
Verify: notice what cancelled
Why: The term carrying I disappeared entirely, so the answer holds for a whisper and for a jet engine alike. That independence is what the product property delivered, and it is the reason a decibel is a ratio unit rather than an absolute one.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 509-509
Prediction
Commit before reasoning.
Predict first
Each doubling adds about 3 decibels. What does multiplying the intensity by 1024 add?
Correct: About 30 decibels, since 1024 is 2 to the tenth.
\[ 10\log 2^{10} = 10 \cdot 10\log 2 \approx 30.1 \]
Why: Ten doublings, each worth 3.01, give 30.1 decibels. Equivalently, ten times the common logarithm of 1024 is 30.1 directly. A thousandfold jump in physical intensity registers as a thirty-point move on the scale, which is exactly the compression that makes the decibel scale readable — and it is the power property in disguise.
Worked example
Guided Practice 11.
\[ \text{Find the increase in loudness when the intensity triples instead.} \]
Repeat the setup with 3I
Why: The new loudness minus the old, with 3I in place of I.
\[ L(3 I) - L(I) \]
Use the product property
Why: Three I over I nought splits into the logarithm of 3 plus the logarithm of I over I nought.
\[ 10(\log 3 + \log(I / I _{0}) - \log(I / I _{0})) \]
Cancel and simplify
Why: The matching terms cancel again, leaving ten times the logarithm of 3.
\[ 10 \log 3 \]
Compute
Why: The common logarithm of 3 is about 0.4771.
\[ \text{about } 4.77\text{ dB} \]
Figure (svg): The solution to Worked example tripling the intensity shown as a ladder of expressions, one row per algebraic move
\[ 10\log 3 \approx 4.77 \text{ dB} \]
Verify: check it against doubling
Why: Tripling gains 4.77 dB and doubling gains 3.01. Doubling twice, which multiplies by 4, should gain 6.02 — and tripling falls between them, as it must, since 3 lies between 2 and 4. The gains add exactly as the factors multiply.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 509-509
Trap
\[ L(2I) - L(I) \text{ depends on } I \]
Assume a loud sound gains more from doubling
Why: Bigger inputs are expected to give bigger changes.
\[ \text{gain} = \text{something involving } I \quad \text{(wrong)} \]
Doubling a 30 decibel sound and doubling a 90 decibel sound both add the same 3 decibels.
\[ L(2I) - L(I) = 10\log 2 \approx 3.01 \]
Let the product property cancel the original intensity
Why: The logarithm of a product separates the factor of 2 from everything else.
\[ 10\log\tfrac{2I}{I_0} - 10\log\tfrac{I}{I_0} = 10\log 2 \]
Only the RATIO of the two intensities survives. This is the same flattening seen in the tornado model of Lesson 7.4, expressed as a unit rather than as a curve.
Fill the middle
Example 5.
Fill in the blanks
L(2I) - L(I) = 10\log 2 \approx 3.01
Why: Everything except the factor of 2 cancelled, so only that factor survives inside the logarithm. Whatever the intensity is multiplied by is exactly what appears there.
Sorting
Each application below leans on one property.
Sort into buckets
Sort each situation.
Every one of these is a property doing a job in an application rather than in an exercise, which is what the properties were invented for.
Ranking
Smallest gain first.
Put in order
Why: The gains are 3.01, 4.77, 10, 20 and 30 decibels. Notice the last three: multiplying the factor by ten each time adds a flat 10 decibels each time, because the common logarithm of a power of ten is just its exponent. That is the whole design of the scale.
Comparison
Fill the blanks. Each lowers the operation by one level.
Comparison matrix
| Property | Statement | What it does |
|---|---|---|
| Product | log_b(mn) = log_b m + log_b n | multiplication becomes addition |
| Quotient | log_b(m/n) = log_b m - log_b n | division becomes subtraction |
| Power | log_b(m^n) = n log_b m | an exponent becomes a factor |
| Change of base | log_c a = (log a)/(log c) | any base becomes a calculator base |
The first three are Lesson 5.1's exponent rules read backwards; the fourth is a consequence of them, not an independent fact.
Pattern
One routine for expanding, one for condensing, one for evaluating.
The logarithm of a quotient is a difference of logarithms, never a quotient of them. That single caution accounts for most of the errors in this lesson.
OpenStax Algebra and Trigonometry 2e, §6.5 Logarithmic Properties §6.5
Check
Expanding. Watch which factor carries the exponent.
Check your understanding
Expand log_6(5x^3/y).
Answer: A
Why: Split the quotient, then the product, then bring the 3 down from x alone.
Check
Condensing. Coefficients first.
Check your understanding
Condense log 3 + 3 log 4 - log 6.
Answer: A
Why: The 3 goes up first, giving 3 times 64 over 6, which is 32.
Check
Change of base. Which one goes on top?
Check your understanding
Which expression equals log_3 8?
Answer: A
Why: The argument goes on top and the old base underneath, giving about 1.893.
Real world
The pH of a solution is the negative of the common logarithm of its hydrogen ion concentration in moles per litre.
Discussion prompt
Vinegar has a hydrogen ion concentration about a thousand times that of pure water, whose pH is 7. Find vinegar's pH, and explain which property makes the calculation one line.
Hint: A thousand is ten cubed.
Answer:
\[ \text{pH} = -\log[H^+], \quad [H^+]_{\text{vinegar}} = 1000 \cdot [H^+]_{\text{water}} \]
\[ -\log(1000 \cdot [H^+]) = -\log 1000 - \log[H^+] = -3 + 7 = 4 \]
Vinegar has a pH of about 4. The product property did the whole job: multiplying the concentration by a thousand subtracts exactly 3 from the pH, whatever the starting concentration was.
This is the same structure as the decibel calculation. In both scales a multiplication of the physical quantity becomes an addition on the reported number, which is why chemists and acousticians can compare enormous ranges on a scale of single digits. Every logarithmic scale in science — pH, decibels, Richter, stellar magnitude — is built on that one property.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the logarithm base 2 of 8 over 2 equal to the logarithm base 2 of 8 divided by the logarithm base 2 of 2?
Correct: No — the quotient property gives a difference, not a quotient.
\[ \log_2\tfrac{8}{2} = 2, \quad \tfrac{\log_2 8}{\log_2 2} = 3 \]
Why: The left side is the logarithm of 4, which is 2. The right side is 3 divided by 1, which is 3. They are not equal, and no choice of base or argument makes the false version work. The book prints this warning directly beside the properties: the logarithm of m over n is the DIFFERENCE of the logarithms, never their quotient. The genuine quotient of two logarithms shows up in the change-of-base formula instead, where it means something entirely different.
Explain it
They know that exponents add when powers multiply, and have just met logarithms.
Discussion prompt
In four sentences or fewer, explain why the logarithm of a product is the sum of the logarithms.
Hint: Start from the exponent rule they already trust.
Answer:
A logarithm is an exponent. If m is the base raised to some exponent and n is the base raised to another, then multiplying them adds those exponents — that is the rule you already know.
So the exponent belonging to the product is the sum of the two exponents. Since the logarithm reports that exponent, the logarithm of the product is the sum of the two logarithms. Nothing new has been claimed; the old rule has been read from the other side.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For evaluating from given values, rewrite the argument in terms of the given numbers before touching a property. For expanding, work outermost operation first. For condensing, clear every coefficient before joining anything. For change of base, put the argument on top and bracket the answer between two powers of the base. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a one-page reference for the properties. Top left: write all three properties, and beside each write the exponent rule from Lesson 5.1 it comes from. Top right: write the two false versions the book warns about, and next to each give a numerical test with base 2 that shows it failing. Middle left: expand a logarithm of your own invention containing a fraction, a product and an exponent, one property per line, labelling each line. Middle right: condense a three-term expression with at least one coefficient, again one property per line, and state which property had to come first and why. Bottom left: write the change-of-base formula and use it twice on the same logarithm, once with common logarithms and once with natural, showing the answers agree. Bottom right: work the doubling calculation for decibels and write one sentence explaining what cancelled and why that makes the decibel a ratio unit.
If your condensing line joined two logarithms while a coefficient was still in front of one of them, that is the step to redo — the joining rules only apply to logarithms standing alone.
Recap
Five things, and all three properties come from exponent rules you already had.
| If you see | Then |
|---|---|
| log_b(mn) | log_b m + log_b n |
| log_b(m/n) | log_b m - log_b n, never a quotient |
| log_b(m^n) | n log_b m |
| A coefficient, and you are condensing | Send it up as an exponent first |
| An unusual base | Divide log of the argument by log of the base |
| A factor multiplying the input of a log model | It adds a constant to the output |
Lesson 7.6 uses these properties to solve exponential and logarithmic equations, where condensing is what turns a many-term equation into one a logarithm can undo.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.5 Apply Properties of Logarithms §7.5, pp. 506-513 — everything on these slides traces back here
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