Establishes the three logarithm properties as exponent rules read through the definition: products become sums, quotients become differences, and powers become multipliers. Expands and condenses logarithmic expressions, warns against the false property for sums, and supplies the change-of-base formula that makes any logarithm computable from the two a calculator provides.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 4 — Exponential and Logarithmic Functions
§4.5 Logarithmic Properties, pp. 538-552
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 538-552 — the pages these objectives are drawn from
Warm-up
Every property in this section is an exponent rule you already know, wearing new notation.
Discussion prompt
What is 2 to the third, times 2 to the fifth? What happened to the exponents?
Hint: Write both out as repeated multiplication if it helps.
Answer:
It is 2 to the eighth. The exponents added, because multiplying three twos by five twos gives eight twos altogether.
Now recall that a logarithm is an exponent. So if exponents add when the numbers multiply, then logarithms should add when the arguments multiply.
That is the product property, and it is not a new fact at all — it is the exponent rule you have had since Algebra 1, translated through the definition of a logarithm. All three properties in this section come from exponent rules the same way.
Concept
Because a logarithm is an exponent, every rule about combining exponents becomes a rule about combining logarithms. Multiplying arguments adds logarithms, dividing subtracts them, and raising to a power multiplies.
\[ \log_b(MN)=\log_b M+\log_b N, \quad \log_b\tfrac{M}{N}=\log_b M-\log_b N, \quad \log_b(M^p)=p\log_b M \]
This makes logarithms a computational tool rather than only an inverse. They convert multiplication into addition, which is why they were invented in the seventeenth century and used for three hundred years to multiply large numbers by hand.
Figure (svg): The three logarithm properties listed with the exponent rule each one comes from, showing that products become sums, quotients become differences, and powers become multipliers
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 538-543
Section
Section 1
Concept
The three properties convert multiplicative structure inside a logarithm into additive structure outside it. Each one comes directly from the corresponding exponent rule.
The power property is the one that matters most for what follows. Bringing an exponent down in front is exactly what is needed to solve for an unknown exponent, which is the whole technique of §4.6 — so of the three, it is the one to have at your fingertips.
Figure (svg): The three logarithm properties listed with the exponent rule each one comes from, showing that products become sums, quotients become differences, and powers become multipliers
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 538-544
Picture it
Each property is paired with the rule it comes from.
Figure (svg): The three logarithm properties listed with the exponent rule each one comes from, showing that products become sums, quotients become differences, and powers become multipliers
Reading the right-hand column first makes the left-hand one unsurprising. Nothing here is a new fact about logarithms; it is an old fact about exponents in new notation.
Worked example
Four lines, using only the definition and one exponent rule.
\[ \text{Show that } \log_b(MN)=\log_b M+\log_b N. \]
Name the two logarithms
Why: Write each argument as a power.
\[ M = b ^{m}, N = b ^{n} \]
Multiply
Why: Using the exponent rule.
\[ MN = b ^{m + n} \]
Read it as a logarithm
Why: The exponent is the logarithm.
\[ \log _{b}(MN) = m + n \]
Substitute back
Why: m and n were the two logarithms.
\[ = \log _{b} M + \log _{b} N \]
Figure (svg): A derivation of the product property, showing two numbers written as powers of the base, multiplied, and the exponents added
\[ \log_b(MN)=\log_b M+\log_b N \]
Verify: check on numbers
Why: Take base 10, M equal to 4 and N equal to 25. The left side is the logarithm of 100, which is 2. The right side is about 0.602 plus 1.398, which is also 2. The derivation and the arithmetic agree, and the numerical check is worth doing whenever a property is in doubt.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 539-540
Matching
Each one is a translation of the other.
Match the pairs
Why: Every logarithm property is the corresponding exponent rule read through the definition. The last pairing is worth noting: the logarithm of 1 being zero is just the statement that any base to the power zero is 1, which makes it a property rather than a special case.
Worked example
Work outward from the largest structure.
\[ \text{Expand } \log\Bigl(\frac{x^3y}{z^2}\Bigr). \]
Apply the quotient property first
Why: The outermost structure is a division.
\[ \log(x ^{3} y) - \log(z ^{2}) \]
Apply the product property
Why: The numerator is a product.
\[ \log(x ^{3}) + \log(y) - \log(z ^{2}) \]
Apply the power property twice
Why: Bring the exponents down.
\[ 3 \log x + \log y - 2 \log z \]
Check the signs
Why: Everything from the denominator is subtracted.
\[ \text{the } 2 \log z\text{ is negative} \]
Figure (svg): The solution to Worked example apply all three at once shown as a ladder of expressions, one row per legal move
\[ 3\log x+\log y-2\log z \]
Verify: test with numbers
Why: Take x, y and z all equal to 10 in base 10. The original is the logarithm of 1000 over 100, which is the logarithm of 10, giving 1. The expansion gives 3 plus 1 minus 2, which is also 1. Testing with convenient values is the reliable check on a multi-step expansion.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 541-544
Trap
\[ \log_2(x)+\log_3(y) \overset{?}{=} \log_6(xy) \]
Apply the product property to the two logarithms
Why: Two logarithms are being added, so they are combined into one.
The bases are multiplied along with the arguments.
The property requires the same base throughout. These have bases 2 and 3, so they cannot be combined at all.
The bases certainly do not multiply — nothing in the derivation does anything of the kind, since it uses one base from start to finish.
Check the bases match before combining anything. Mismatched bases can be reconciled with the change-of-base formula later in this section, but not by any of the three properties.
Faded example
Expand the logarithm of x squared times y.
Fill in the blanks
\log(x^2y) = \log(x^2) + \log y = 2\log x + \log y
Why: The product property splits the two factors and the power property brings the exponent 2 down in front. Note the exponent multiplies only the logarithm of x, not the whole expression — a distinction that matters as soon as there is more than one term.
Sorting
Look at the structure inside the logarithm.
Sort into buckets
Sort each expression by the property that expands it first.
Socratic
Logarithms were invented before calculators, for a practical reason.
Discussion prompt
Why would a seventeenth-century astronomer have found these properties worth a lifetime's work to tabulate?
Hint: Which is easier by hand: multiplying two six-digit numbers, or adding two six-digit numbers?
Answer:
Multiplying large numbers by hand is slow and error-prone; adding them is neither. The product property converts one into the other, so a multiplication becomes a table lookup, an addition, and a lookup back.
That is exactly what logarithm tables were for, and they cut the labour of astronomical calculation by so much that Laplace said they doubled the working life of an astronomer.
The same idea survives in the slide rule, which is a physical logarithm table: sliding two scales adds their logarithms, which multiplies the numbers. Every engineer used one until the 1970s, and it works entirely on this section's product property.
Section
Section 2
Concept
To expand a complicated logarithm, apply the properties in order of the expression's structure: the outermost operation first, then work inward.
The order matters for keeping the signs right. Applying the power property to a numerator before splitting off the denominator is a common route to losing a minus sign, because it is easy to forget which terms came from below the line.
Figure (svg): The three logarithm properties listed with the exponent rule each one comes from, showing that products become sums, quotients become differences, and powers become multipliers
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 544-548
Picture it
Expanding is applying these repeatedly until nothing is left to split.
Figure (svg): The three logarithm properties listed with the exponent rule each one comes from, showing that products become sums, quotients become differences, and powers become multipliers
A fully expanded expression has a single variable inside each logarithm and every exponent brought out in front.
Worked example
A root is a fractional power.
\[ \text{Expand } \ln\Bigl(\sqrt{\frac{x}{y}}\Bigr). \]
Rewrite the root as a power
Why: A square root is the power one half.
\[ \ln((\frac{x}{y}) ^{\frac{1}{2}}) \]
Apply the power property
Why: Bring the one half down.
\[ (\frac{1}{2}) \ln(\frac{x}{y}) \]
Apply the quotient property
Why: Inside the remaining logarithm.
\[ (\frac{1}{2}) (\ln x - \ln y) \]
Distribute
Why: The one half multiplies both terms.
\[ (\frac{1}{2}) \ln x - (\frac{1}{2}) \ln y \]
Figure (svg): The solution to Worked example expand with a root shown as a ladder of expressions, one row per legal move
\[ \tfrac{1}{2}\ln x-\tfrac{1}{2}\ln y \]
Verify: check the distribution
Why: The one half must multiply BOTH terms, not just the first. Forgetting to distribute is the standard slip here, and it produces an expression that disagrees with the original at almost every input. Testing with x equal to e squared and y equal to e confirms: the original gives one half and the expansion gives 1 minus one half, which is also one half.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 545-546
Faded example
Expand the logarithm of 3 over x cubed.
Fill in the blanks
\log 3 - \log(x^3) = \log 3 - 3\log x
Why: The quotient property gives a minus before the denominator's logarithm, and the power property brings the exponent 3 down. The result is log 3 minus 3 log x. Both the sign and the coefficient come from separate properties, and either can be dropped independently.
Worked example
Every term from a denominator is subtracted.
\[ \text{Expand } \log\Bigl(\frac{7}{x^2z}\Bigr). \]
Apply the quotient property
Why: The outermost structure.
\[ \log 7 - \log(x ^{2} z) \]
Note what the minus applies to
Why: The whole of the denominator's logarithm.
Expand the denominator's product
Why: Inside the brackets.
\[ \log 7 - (\log(x ^{2}) + \log z) \]
Distribute the minus and apply the power property
Why: Both denominator terms become negative.
\[ \log 7 - 2 \log x - \log z \]
Figure (svg): The solution to Worked example expand with everything below the line shown as a ladder of expressions, one row per legal move
\[ \log 7-2\log x-\log z \]
Verify: check both denominator terms are negative
Why: Both x and z were in the denominator, so both of their logarithms must be subtracted. Writing the bracket explicitly at step 3 and then distributing is what guarantees it; splitting the denominator without the bracket is how the second minus sign gets lost.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 546-548
Error analysis
A student expands a logarithm with two factors below the line.
Annotate
On: \( \log\Bigl(\frac{a}{bc}\Bigr) = \log a - \log b + \log c \)
Write the denominator's logarithm in a bracket before expanding it, then distribute the minus. That one bracket prevents the whole class of sign errors here.
Ranking
For an expression that is a quotient of products of powers.
Put in order
Why: Working outward from the largest structure means the quotient first, then the products within each part, then distributing the minus that the quotient introduced, and finally bringing down the exponents. Doing the powers first is not wrong but makes the sign tracking harder.
Sorting
Everything from the denominator is subtracted.
Sort into buckets
For the logarithm of a b over c d, sort each factor.
Explain it to yourself
Writing the denominator's logarithm in a bracket prevents a whole class of errors.
Discussion prompt
Explain what goes wrong without the bracket, using an example.
Hint: What does the minus sign apply to?
Answer:
The quotient property puts a minus in front of the entire logarithm of the denominator, not in front of its first factor. Without a bracket that scope is invisible.
So expanding the denominator's product without a bracket produces a minus on the first term and nothing on the rest, which is wrong for every factor after the first.
Writing it as minus a bracketed sum and then distributing makes the scope explicit and gets every sign right automatically. It costs one bracket, and it is the same discipline as distributing a minus in ordinary algebra — which is exactly what it is.
Section
Section 3
Concept
Condensing collects a sum or difference of logarithms into a single one. Coefficients go back up as exponents, sums become products, and differences become quotients.
Doing the coefficients first is not optional. The product and quotient properties combine bare logarithms, and a logarithm with a coefficient in front is not in that form — so the coefficient has to be absorbed before anything can be combined.
Figure (svg): The three logarithm properties listed with the exponent rule each one comes from, showing that products become sums, quotients become differences, and powers become multipliers
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 548-550
Picture it
Condensing is expanding run backwards.
Figure (svg): The three logarithm properties listed with the exponent rule each one comes from, showing that products become sums, quotients become differences, and powers become multipliers
Reading each row from right to left gives the condensing rule. Nothing new is needed, only the willingness to use the properties in the other direction.
Worked example
Coefficients up first, then combine.
\[ \text{Condense } 2\log x+\log y-3\log z. \]
Move the coefficients up
Why: Using the power property backwards.
\[ \log(x ^{2}) + \log y - \log(z ^{3}) \]
Combine the added terms
Why: Product property backwards.
\[ \log(x ^{2} y) - \log(z ^{3}) \]
Combine the subtracted term
Why: Quotient property backwards.
\[ \log(x ^{2} y / z ^{3}) \]
Check the structure
Why: One logarithm, correct placement.
Figure (svg): The solution to Worked example condense a sum and a difference shown as a ladder of expressions, one row per legal move
\[ \log\Bigl(\frac{x^2y}{z^3}\Bigr) \]
Verify: expand it back
Why: Expanding gives 2 log x plus log y minus 3 log z — the original. Expanding back is the reliable check on any condensation, and it takes three steps. Note that the subtracted term ended up in the denominator, which is where subtraction always sends things.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 548-549
Faded example
Condense 4 log x minus log y.
Fill in the blanks
\log(x^4}) - \log y = \log\Bigl(\fracy}}}___}\Bigr)
Why: The coefficient 4 goes up as an exponent, and the subtraction sends y into the denominator. The result is the logarithm of x to the fourth over y. Expanding it back recovers the original, which is the check worth performing.
Worked example
The assembled argument may collapse.
\[ \text{Condense } \log(8)+\log(x)-\log(2). \]
Combine the added terms
Why: Product property backwards.
\[ \log(8 x) \]
Combine the subtracted term
Why: Quotient property backwards.
\[ \log(8 x / 2) \]
Simplify the argument
Why: Eight over two is four.
\[ \log(4 x) \]
Check nothing further simplifies
Why: Four x has no more structure.
Figure (svg): The solution to Worked example condense and simplify shown as a ladder of expressions, one row per legal move
\[ \log(4x) \]
Verify: test numerically
Why: Take x equal to 25 in base 10. The original is about 0.903 plus 1.398 minus 0.301, which is 2. The answer is the logarithm of 100, which is also 2. The simplification of the argument is worth doing — leaving it as 8x over 2 is correct but not finished.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 549-550
Trap
\[ 3\log x+\log y = \log(3xy) \]
Combine the two logarithms into one
Why: The product property is applied to the sum.
The coefficient 3 is carried into the argument as a factor.
The coefficient becomes an exponent, not a factor. The power property says 3 log x is the logarithm of x cubed.
So the correct condensation is the logarithm of x cubed times y, not of 3xy.
Absorb every coefficient before combining anything. The product property applies to bare logarithms, and a coefficient in front means the expression is not yet in that form.
Ranking
The order that avoids errors.
Put in order
Why: Coefficients must be absorbed first, because the other two properties combine bare logarithms only. Then the added terms form a product and the subtracted ones a denominator, and simplifying the argument is a final tidy that is easy to forget.
Prediction
An expression contains a logarithm being subtracted.
Predict first
Where does its argument end up after condensing?
Correct: In the denominator.
Why: The quotient property says a difference of logarithms is the logarithm of a quotient, so the subtracted argument becomes the denominator. Putting a minus sign inside the argument instead would produce a negative number, which usually has no logarithm at all.
Counterexample
A classmate condenses a coefficient by multiplying it into the argument.
Discussion prompt
Show with numbers that 2 log 3 is not the logarithm of 6.
Hint: Compute both in base 10.
Answer:
Two times the logarithm of 3 is about 0.954. The logarithm of 6 is about 0.778. They are not equal.
What 2 log 3 actually equals is the logarithm of 3 squared, which is the logarithm of 9 — about 0.954, matching. The coefficient becomes an exponent, not a factor.
The confusion comes from the coefficient sitting where a multiplier normally would. Testing on small numbers settles it in ten seconds, and it is worth doing whenever a property is applied in an unfamiliar direction.
Section
Section 4
Concept
The properties convert products, quotients and powers. A sum inside a logarithm cannot be split at all, and the expression that looks as though it should work is simply false.
The connection with §1.4 is worth making explicit. That section warned that a function name is not a multiplier and does not distribute over its argument. This is the same warning for a specific function, and it is the error's most common habitat.
Figure (svg): A warning card contrasting the true product property with the false claim that a logarithm distributes over a sum, with a numerical counterexample
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 542-546
Picture it
The two expressions look alike and behave completely differently.
Figure (svg): A warning card contrasting the true product property with the false claim that a logarithm distributes over a sum, with a numerical counterexample
One numerical test separates them. The true property agrees exactly; the false one is out by an amount that varies with the inputs, so there is no fixing it.
Worked example
One numerical test is enough.
\[ \text{Test whether } \log(M+N)=\log M+\log N. \]
Choose convenient values
Why: Base 10, with round logarithms.
\[ M = 10, N = 10 \]
Compute the left side
Why: The logarithm of 20.
\[ \text{about } 1.301 \]
Compute the right side
Why: One plus one.
\[ 2 \]
Compare
Why: They differ substantially.
Figure (svg): A warning card contrasting the true product property with the false claim that a logarithm distributes over a sum, with a numerical counterexample
\[ \log(20)\ne \log 10+\log 10 \]
Verify: check the true property on the same numbers
Why: The product property with the same values gives the logarithm of 100 on the left, which is 2, and 1 plus 1 on the right, which is also 2 — they agree. So the product property survives the test the sum version fails, which is the clearest way to see that the two are genuinely different claims.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 543-544
Sorting
Only products, quotients and powers can.
Sort into buckets
Sort each expression.
Worked example
Factor it first, if it factors.
\[ \text{Expand } \log(x^2+5x) \text{ as far as possible.} \]
Note that a sum cannot be split
Why: No property applies to it directly.
Look for a factorisation
Why: Both terms share an x.
\[ x(x + 5) \]
Now it is a product
Why: The product property applies.
\[ \log x + \log(x + 5) \]
Check whether more can be done
Why: The second argument is a sum with no factors.
Figure (svg): The solution to Worked example when a sum can be handled shown as a ladder of expressions, one row per legal move
\[ \log x+\log(x+5) \]
Verify: note what was and was not achieved
Why: Factoring turned the sum into a product, which the property could then split. But the remaining sum inside the second logarithm cannot be split further, and no amount of algebra will change that. Factoring is the only route from a sum to a form the properties handle.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 545-546
Error analysis
A student expands a logarithm of a sum.
Annotate
On: \( \ln(x+y) = \ln x + \ln y \)
There is no property for the logarithm of a sum. If the sum factors, factoring first produces a product the properties can handle; if it does not, the expression stays as it is.
Prediction
Someone splits the logarithm of a sum as though it were a product.
Predict first
How wrong is the result?
Correct: Wrong by an amount that varies with the inputs.
Why: The two expressions are unrelated functions, so their difference changes as the inputs change. There is no correction factor and no restricted range where it happens to work — which is why the error cannot be patched and the expression simply must be left alone.
Faded example
Expand the logarithm of x squared minus 4 as far as possible.
Fill in the blanks
\log((x-2)(x+2)) = \log(x-2) + \log(x+___)
Why: The difference of squares factors into two linear brackets, which is a product, and the product property then splits it. Without factoring, the original is a difference and no property applies at all. Factoring is the only bridge from additive structure to multiplicative.
Analogy
The same mistake appears in several sections.
Match the pairs
Why: All three are one error: treating a function as though it distributed over addition. §1.4 warned about it in general, and it recurs for every unfamiliar function. The reliable defence is the same in every case — test on small numbers, and the claim collapses immediately.
Section
Section 5
Concept
A logarithm in one base can be written as a ratio of logarithms in any other base. Since calculators provide base ten and base e, that makes every logarithm computable.
\[ \log_b(M)=\frac{\log_c(M)}{\log_c(b)} \]
That last observation is worth noticing. Since one base's logarithm is a constant multiple of another's, every logarithm function is a vertical stretch of every other — which is why they all have the same domain, the same asymptote and the same crossing point, and differ only in steepness.
Figure (svg): The change of base formula shown with a worked instance, converting a base 2 logarithm into a ratio of natural logarithms so a calculator can evaluate it
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 550-552
Picture it
The argument on top, the original base underneath, in whatever new base is convenient.
Figure (svg): The change of base formula shown with a worked instance, converting a base 2 logarithm into a ratio of natural logarithms so a calculator can evaluate it
The check at the bottom is worth doing: raise the original base to the computed value and confirm it gives the argument back.
Worked example
Convert to natural logarithms and divide.
\[ \text{Evaluate } \log_2(10) \text{ to four decimal places.} \]
Apply the formula
Why: Argument on top, base underneath.
\[ \ln(10) / \ln(2) \]
Evaluate the numerator
Why: The natural logarithm of ten.
\[ \text{about } 2.3026 \]
Evaluate the denominator
Why: The natural logarithm of two.
\[ \text{about } 0.6931 \]
Divide
Why: Numerator over denominator.
\[ \text{about } 3.3219 \]
Figure (svg): The change of base formula shown with a worked instance, converting a base 2 logarithm into a ratio of natural logarithms so a calculator can evaluate it
\[ \log_2(10)\approx 3.3219 \]
Verify: check by exponentiating
Why: Two raised to the power 3.3219 is about 10.000, so the value is right. This is also the answer §4.3's worked example left as an exact expression — the change-of-base formula is what turns that exact form into a decimal.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 550-551
Faded example
Write the logarithm base 7 of 50 as a ratio of common logarithms.
Fill in the blanks
\log_7(50) = \frac50})}7})}
Why: The argument 50 goes on top and the original base 7 underneath. Evaluating gives about 1.699 over 0.845, which is about 2.01 — and 7 squared is 49, close to 50, so the answer is sensible.
Worked example
Any two logarithms differ by a constant factor.
\[ \text{Show that } \log_2(x) \text{ is a constant multiple of } \ln(x). \]
Apply the change-of-base formula
Why: Converting to natural logarithms.
\[ \ln(x) / \ln(2) \]
Identify the constant
Why: The denominator does not depend on x.
\[ 1 / \ln(2) \]
Rewrite
Why: As a multiple.
\[ \text{about } 1.4427 \times \ln(x) \]
Interpret graphically
Why: A constant multiple is a vertical stretch.
Figure (svg): The solution to Worked example the formula's other use shown as a ladder of expressions, one row per legal move
\[ \log_2(x)=\frac{\ln x}{\ln 2}\approx 1.4427\ln x \]
Verify: check the graphical consequence
Why: Every logarithm is a vertical stretch of every other, so they all share the same domain, the same vertical asymptote and the same crossing at input 1 — which is exactly what §4.4's comparison of two bases found by inspection. The formula explains why that had to be so.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 551-552
Trap
\[ \log_2(10) = \frac{\ln 2}{\ln 10} \approx 0.301 \]
Apply the change-of-base formula
Why: The two logarithms are formed and divided.
The base has been put in the numerator and the argument underneath.
The argument goes on top. The correct ratio is the logarithm of 10 over the logarithm of 2, giving about 3.32.
A sanity check catches it: 2 to the power 0.301 is about 1.23, not 10. The answer should be between 3 and 4, since 2 cubed is 8 and 2 to the fourth is 16.
Estimate the answer before computing it. Knowing roughly what a logarithm should be makes an inverted ratio obvious, since the two answers are reciprocals and usually far apart.
Prediction
The change-of-base formula is applied with base ten, and then again with base e.
Predict first
Do the two give the same answer?
Correct: Yes: the choice of new base does not matter.
Why: The formula holds for any new base, and the ratio comes out the same because both numerator and denominator scale by the same factor when the base changes. So base ten and base e give identical results, and the choice is purely a matter of which the calculator provides.
Estimation
Consider the logarithm base 3 of 100.
Predict first
Roughly what is it?
Correct: Between 4 and 5.
Why: Three to the fourth is 81 and three to the fifth is 243, so the answer lies between 4 and 5, closer to 4. The exact value is about 4.19. Estimating this way before using the formula catches an inverted ratio immediately, since the inverted answer would be about 0.24.
Socratic
The change-of-base formula can be derived in three lines.
Discussion prompt
Derive it, starting from the definition.
Hint: Let the logarithm you want be x, convert to exponential form, and take logarithms in the new base.
Answer:
Let x be the logarithm base b of M. Converting to exponential form gives b to the power x equals M.
Now take the logarithm base c of both sides. The left becomes x times the logarithm base c of b, by the power property; the right is the logarithm base c of M.
Dividing gives x, which is the formula. So it rests on the power property and nothing else, which is a good illustration of how much that one property does — it is also the property §4.6 will use to solve every exponential equation.
Comparison
Fill the blanks from memory. Each converts multiplicative structure into additive.
Comparison matrix
| inside the logarithm | becomes | from the exponent rule | |
|---|---|---|---|
| product | a product | a sum | exponents add |
| quotient | a quotient | a difference | exponents subtract |
| power | a power | a multiplier out front | exponents multiply |
| sum | a sum | nothing: no property exists | there is no such rule |
The last row is the one to remember as firmly as the other three. Its absence is not an oversight — there simply is no relationship between the logarithm of a sum and the logarithms of its parts.
Pattern
Two procedures, each the other run backwards.
Step 3's ordering is the one people skip. The product and quotient properties apply to bare logarithms, so a coefficient must be absorbed before those properties are available at all.
OpenStax Algebra and Trigonometry 2e, §6.5 Logarithmic Properties §6.5
Check
Coefficients become exponents.
Check your understanding
What does 3 log x equal?
Answer: A
Why: The power property says the logarithm of a power is the exponent times the logarithm, so running it backwards turns a coefficient into an exponent. Testing with x equal to 10 in base 10 confirms: 3 times 1 is 3, and the logarithm of 1000 is also 3.
Check
No property for sums.
Check your understanding
Which expansion is valid?
Answer: A
Why: The product property converts a product inside the logarithm into a sum outside it. The other three are all false, and the two involving sums are false because no property applies to a sum at all.
Check
Argument on top.
Check your understanding
Which expression equals the logarithm base 5 of 20?
Answer: A
Why: The change-of-base formula puts the argument on top and the original base underneath. Evaluating gives about 3.00 over 1.61, which is about 1.86 — and 5 squared is 25, so a value just under 2 is sensible.
Real world
The slide rule was a physical implementation of the product property, and it built the twentieth century.
Discussion prompt
How does a slide rule multiply, and which property is it using?
Hint: What does sliding one ruler along another do to the lengths?
Answer:
A slide rule has scales marked logarithmically, so the distance from the start to a number is proportional to that number's logarithm.
Sliding one scale along the other adds two distances, which adds the two logarithms. By the product property, that is the logarithm of the product — so reading off the total distance gives the product itself.
Every engineer used one until the mid-1970s, including the ones who designed the Apollo missions. A mechanical adder becomes a multiplier, purely because of this section's first property — which is a striking demonstration that converting multiplication into addition is worth doing.
Commit first
State your confidence along with your answer.
Predict first
What does the logarithm of x squared over y equal, expanded?
Correct: 2 log x minus log y.
Why: The quotient property splits the fraction and the power property brings the exponent 2 down onto the numerator's term only. The last option distributes the 2 over both terms, which is wrong because the exponent applied only to x — that would be the expansion of the logarithm of x over y, all squared.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why the logarithm of a product is a sum, without just stating the rule.
Hint: Write both numbers as powers of the base.
Answer:
Write each number as a power of the base: M is b to the m and N is b to the n. Then m and n are exactly their logarithms, by the definition.
Multiplying gives b to the m plus n, because that is what multiplying powers of the same base does. So the product's logarithm is m plus n.
And m plus n is the sum of the two logarithms. So the property is the exponent rule you already knew, stated in the other notation — which is worth emphasising, because it means there is nothing new to memorise, only a translation to become fluent in.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second and third are where the routine errors live, and both are fixed by one discipline each — brackets for expanding, coefficients first for condensing. The power property within the first is the one §4.6 will assume you can apply without thinking.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the three properties in one column and the exponent rule each comes from in the next, so the pairing is visible. Underneath, write the false property for a sum with a numerical counterexample beside it. Then take one complicated logarithm of a quotient of powers, expand it fully, and condense it back to check.
If your expansion and condensation return each other exactly, and your counterexample shows two clearly different numbers, both halves of the section are on the page.
Recap
Five things, and the power property is the one the next section depends on.
| if you remember one thing | it should be this |
|---|---|
| about the properties | each one is an exponent rule read through the definition |
| about expanding | bracket the denominator's logarithm before distributing the minus |
| about condensing | coefficients become exponents before anything is combined |
| about sums | there is no property, and testing on numbers proves it |
Section 4.6 uses these properties to solve equations, where the power property is what brings an unknown exponent down to where it can be isolated.
OpenStax, Precalculus, §4.5 Logarithmic Properties §4.5, pp. 538-552 — everything on these slides traces back here
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