Introduces the inverse of the exponential and the tool for solving for an exponent. Establishes the equivalence between logarithmic and exponential form, evaluates logarithms by asking what power the base needs, derives the restricted domain from the exponential's restricted range, and names the common and natural logarithms.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 4 — Exponential and Logarithmic Functions
§4.3 Logarithmic Functions, pp. 501-512
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 501-512 — the pages these objectives are drawn from
Warm-up
Section 4.2 found an x-intercept it could only locate approximately. This is why.
Discussion prompt
Solve 2 to the power x equal to 8, then 2 to the power x equal to 10. What is different about the second?
Hint: The first is a guess you can check. Try guessing the second.
Answer:
The first is x equal to 3, because 2 cubed is 8. You can find it by trying small numbers, and it comes out exactly.
The second lies between 3 and 4, since 8 is too small and 16 is too big — but no whole number works and no obvious fraction does either. Guessing gets closer without ever arriving.
What is missing is an operation that undoes exponentiation, the way division undoes multiplication. That operation is the logarithm, and this section defines it. With it, the second answer is written down exactly rather than approximated.
Concept
The logarithm base b of y is the power to which b must be raised to give y. That single sentence is the definition, and every property in the chapter follows from it.
logarithm — For a positive base b other than 1, the logarithm base b of a positive number y is the exponent x for which b to the power x equals y. It is the inverse of the exponential function with that base.
\[ \log_b(y)=x \;\Longleftrightarrow\; b^x=y \]
The two forms say exactly the same thing, and being able to move between them instantly is the section's central skill. Almost every logarithm problem is easy in one of the two forms and awkward in the other, so the first move is usually to convert.
Figure (svg): The equivalence between logarithmic and exponential form, drawn with arrows showing which number moves where when the statement is rewritten
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 501-504
Section
Section 1
Concept
Every logarithmic equation has an exponential twin saying the same thing. Converting between them is the first move in most problems.
The reading-aloud habit is worth more than memorising which number goes where. Asked for the logarithm base 3 of 81, saying 'to what power must 3 be raised to give 81' turns the problem into arithmetic and removes any chance of putting the numbers in the wrong places.
Figure (svg): The equivalence between logarithmic and exponential form, drawn with arrows showing which number moves where when the statement is rewritten
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 501-506
Picture it
Three numbers, and each has a role in both forms.
Figure (svg): The equivalence between logarithmic and exponential form, drawn with arrows showing which number moves where when the statement is rewritten
The base is the one that never moves. The other two swap between being the exponent and being the result, which is the whole of the conversion.
Worked example
Identify the base, then place the other two.
\[ \text{Write } \log_4(64)=3 \text{ in exponential form.} \]
Identify the base
Why: The subscript.
\[ \text{base } 4 \]
Identify the exponent
Why: The logarithm's value.
\[ \text{exponent } 3 \]
Identify the result
Why: What the log was taken of.
\[ \text{result } 64 \]
Assemble
Why: Base to the exponent equals the result.
\[ 4 ^{3} = 64 \]
Figure (svg): The equivalence between logarithmic and exponential form, drawn with arrows showing which number moves where when the statement is rewritten
\[ 4^3=64 \]
Verify: read the original aloud
Why: 'To what power must 4 be raised to give 64?' The answer is 3, and 4 cubed is indeed 64. The two forms carry identical information, so the check is simply confirming the arithmetic the statement claims.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 502-503
Translation
The base stays put; the other two swap roles.
Match the pairs
Why: In every pair the base is unchanged and the exponent and the result exchange places. The last one is worth noting: the logarithm of 1 is 0 for every base, since any base to the power zero is 1.
Worked example
The same three roles, read the other way.
\[ \text{Write } 5^{-2}=\tfrac{1}{25} \text{ in logarithmic form.} \]
Identify the base
Why: It stays the base.
\[ \text{base } 5 \]
Identify the exponent
Why: It becomes the logarithm's value.
\[ \text{value } -2 \]
Identify the result
Why: It becomes what the log is taken of.
\[ \text{of } \frac{1}{25} \]
Assemble
Why: Log base 5 of one twenty-fifth.
\[ \log _{5}(\frac{1}{25}) = -2 \]
Figure (svg): The solution to Worked example convert to logarithmic form shown as a ladder of expressions, one row per legal move
\[ \log_5\Bigl(\tfrac{1}{25}\Bigr)=-2 \]
Verify: note that the value may be negative
Why: A logarithm's VALUE can be negative — it is an exponent, and exponents can be negative. What cannot be negative is the number you take the logarithm OF, since a positive base raised to any power stays positive. Keeping those two straight is the source of most domain errors in the section.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 503-506
Trap
\[ \log_2(8)=3 \;\Longrightarrow\; 8^3=2 \]
Convert to exponential form
Why: The three numbers 2, 8 and 3 are all placed into the exponential statement.
The base and the argument have been exchanged.
The base stays the base. The subscript 2 is the base in both forms, so the exponential statement is 2 cubed equals 8.
The wrong version claims 512 equals 2, which is visibly false — checking the arithmetic catches it immediately.
Read it as a question: 'to what power must 2 be raised to give 8?' The number being raised is the base and the number produced is the argument, and the sentence keeps them in the right roles.
Faded example
Rewrite the statement that the logarithm base 6 of 216 is 3.
Fill in the blanks
6^3} = 216
Why: The base 6 stays the base and the logarithm's value 3 becomes the exponent, giving 6 cubed equals 216. Checking: 6 times 6 is 36, times 6 again is 216. The conversion is mechanical once the roles are clear.
Prediction
Consider the logarithm base 5 of 1.
Predict first
What is it?
Correct: Zero, since 5 to the power 0 is 1.
Why: The question is what power 5 must be raised to in order to give 1, and any nonzero base to the power zero is 1. So the logarithm of 1 is zero for every base, which is why every logarithm graph passes through the point at input 1.
Socratic
The two forms carry identical information.
Discussion prompt
If the two forms say the same thing, why is converting useful?
Hint: Which form is easier to compute with?
Answer:
Because familiarity differs. Exponential statements are arithmetic you have done since Algebra 1, while logarithmic ones use notation introduced ten minutes ago.
So a logarithm that looks opaque usually becomes obvious in exponential form. 'The logarithm base 2 of one eighth' is unfamiliar; '2 to what power gives one eighth' is a question anyone can answer.
Later the direction reverses. Once the logarithm properties in §4.5 are available, some exponential problems are easier converted into logarithms — which is exactly how §4.6 solves for exponents. Fluency in both directions is the goal, and it starts with being able to convert without thinking.
Section
Section 2
Concept
To evaluate a logarithm, write the argument as a power of the base. The exponent that appears is the answer.
The third and fourth bullets cover most of the ones that look hard. The logarithm base 2 of one eighth is negative 3, and the logarithm base 9 of 3 is one half — both obvious once the argument is rewritten as a power of the base, and both opaque otherwise.
Figure (svg): The equivalence between logarithmic and exponential form, drawn with arrows showing which number moves where when the statement is rewritten
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 504-508
Picture it
Every evaluation is this one question, applied to particular numbers.
Figure (svg): The equivalence between logarithmic and exponential form, drawn with arrows showing which number moves where when the statement is rewritten
Writing the argument as a power of the base is the whole technique. Once it is in that form the answer is simply read off the exponent.
Worked example
A fraction means a negative exponent.
\[ \text{Evaluate } \log_3\Bigl(\tfrac{1}{81}\Bigr). \]
Express the argument as a power of 3
Why: Eighty one is 3 to the fourth.
\[ \frac{1}{81} = 1 / 3 ^{4} \]
Use the negative exponent rule
Why: A reciprocal is a negative power.
\[ = 3 ^{-4} \]
Read the exponent
Why: That is the logarithm's value.
\[ -4 \]
Check by converting
Why: Three to the negative 4 is one eighty-first.
Figure (svg): The solution to Worked example a negative value shown as a ladder of expressions, one row per legal move
\[ \log_3\Bigl(\tfrac{1}{81}\Bigr)=-4 \]
Verify: note which sign was allowed
Why: The VALUE came out negative, which is fine — it is an exponent. The ARGUMENT was one eighty-first, which is positive, as it must be. Confusing those two is what makes people think this problem is illegal, and keeping them apart resolves it.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 505-506
Faded example
Find the logarithm base 4 of one sixteenth.
Fill in the blanks
\tfrac-2___ = 4^___} \;\Longrightarrow\; \log_4\Bigl(\tfrac______\Bigr) = ___
Why: Sixteen is 4 squared, so one sixteenth is 4 to the negative 2, and the logarithm's value is negative 2. The value is negative because the argument is between zero and one, which is a reliable rule: arguments below 1 always give negative logarithms.
Worked example
A root means a fractional exponent.
\[ \text{Evaluate } \log_{25}(5). \]
Ask the question
Why: To what power must 25 be raised to give 5?
Recognise the relationship
Why: Five is the square root of 25.
\[ 5 = \sqrt{25} \]
Write the root as an exponent
Why: A square root is the power one half.
\[ 25 ^{\frac{1}{2}} = 5 \]
Read the exponent
Why: That is the answer.
\[ \frac{1}{2} \]
Figure (svg): The solution to Worked example a fractional value shown as a ladder of expressions, one row per legal move
\[ \log_{25}(5)=\tfrac{1}{2} \]
Verify: check with the other direction
Why: Twenty five to the power one half is the square root of 25, which is 5 — correct. Note that the logarithm base 5 of 25 is 2, the reciprocal. Swapping the base and the argument inverts the value, which is a pattern worth recognising.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 506-508
Error analysis
A student evaluates a logarithm.
Annotate
On: \( \log_2(8) \overset{?}{=} \frac{8}{2} = 4 \)
Convert to exponential form and check. A proposed logarithm value can always be tested in one step by raising the base to it and comparing with the argument.
Sorting
The argument's position relative to 1 decides, for a base above 1.
Sort into buckets
Sort each logarithm by the sign of its value.
Prediction
The logarithm base 9 of 3 is one half.
Predict first
What is the logarithm base 3 of 9?
Correct: 2, the reciprocal.
Why: Swapping the base and the argument inverts the value. Nine to the power one half is 3, and 3 squared is 9 — the same relationship read from the other end. This reciprocal pattern holds whenever both logarithms are defined, and it is a quick way to check one against the other.
Explain it to yourself
A logarithm can be negative, but only in one of its two slots.
Discussion prompt
Explain which number in a logarithm may be negative and which may not.
Hint: Which one is the exponent and which is the result?
Answer:
The value of a logarithm may be negative, because it is an exponent and exponents can be negative. A negative exponent produces a reciprocal, which is a perfectly ordinary positive number.
The argument may not be negative, because it is the result of raising a positive base to a power, and that result is always positive. There is no exponent that turns a positive base into a negative number.
So 'the logarithm of a negative number' is undefined while 'a negative logarithm' is entirely ordinary. The restriction is on what goes in, not on what comes out — which is exactly the exponential's restriction seen from the other side.
Section
Section 3
Concept
Because the logarithm is the exponential's inverse, its graph is the exponential's reflected in the line y equals x — so every feature is the exponential's with the roles swapped.
The slow growth is worth noticing. Because the exponential climbs so steeply, its reflection climbs very gradually: the logarithm base 10 of a million is only 6. That is precisely what makes logarithms useful for compressing enormous ranges, as decibels and the Richter scale do.
Figure (svg): An exponential and its logarithm drawn as reflections in the line y equals x, with corresponding points marked showing their swapped coordinates
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 508-511
Picture it
The exponential and its logarithm are mirror images in the diagonal.
Figure (svg): An exponential and its logarithm drawn as reflections in the line y equals x, with corresponding points marked showing their swapped coordinates
Each marked point on one curve has its coordinates swapped on the other. That single fact generates every entry in the comparison of their features.
Worked example
Read them off the exponential and swap.
\[ \text{Give the domain, range and asymptote of } f(x)=\log_2(x). \]
Recall the exponential's range
Why: Positive numbers only.
\[ \text{exponential range } (0, \infty) \]
Swap for the logarithm's domain
Why: The inverse's domain is the original's range.
\[ \text{domain } x > 0 \]
Recall the exponential's domain
Why: Every real number.
Swap for the range and asymptote
Why: The horizontal asymptote reflects to a vertical one.
Figure (svg): A card contrasting the exponential and the logarithm, showing that the domain and range are exchanged and that the logarithm's domain is only the positive numbers
\[ \text{domain } (0,\infty), \; \text{range } (-\infty,\infty), \; \text{asymptote } x=0 \]
Verify: test near the asymptote
Why: The logarithm base 2 of one thousandth is about negative 10, and of one millionth about negative 20. The outputs plunge without bound as the input approaches zero, which is what a vertical asymptote looks like — and it explains why zero itself has no logarithm.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 509-510
Discrimination
Three kinds of expression restrict a domain, and they differ at the endpoint.
Sort into buckets
Sort each by whether zero is allowed.
Worked example
The argument must be positive, whatever it is.
\[ \text{Find the domain of } f(x)=\log_5(x-3). \]
State the requirement
Why: The argument must be positive.
\[ x - 3 > 0 \]
Solve
Why: Add 3 to both sides.
\[ x > 3 \]
Note the asymptote's position
Why: It moves with the shift.
\[ \text{asymptote } x = 3 \]
State the domain
Why: Strictly greater, not at or above.
\[ (3, \infty) \]
Figure (svg): The solution to Worked example find a restricted domain shown as a ladder of expressions, one row per legal move
\[ \text{domain } (3,\infty), \; \text{asymptote } x=3 \]
Verify: check the boundary is excluded
Why: At x equal to 3 the argument is zero, and zero has no logarithm — so 3 is excluded and the bracket is round. This is a genuinely new kind of domain restriction: §1.2 listed denominators and even roots, and logarithms are the third, arriving exactly here.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 510-511
Trap
\[ \log_2(x-4): \quad \text{domain } x\ge 4 \]
Require the argument to be nonnegative
Why: The condition is written with a non-strict inequality, by analogy with a square root.
The input 4 is included in the domain.
Zero has no logarithm. No power of 2 gives zero, since a positive base raised to any power stays positive, so the argument must be strictly positive.
The domain is x strictly greater than 4, with a round bracket, and there is a vertical asymptote at 4.
A logarithm's argument is strictly positive, unlike an even root's radicand which may be zero. The two restrictions look alike and differ exactly at the endpoint.
Faded example
For the natural logarithm of the quantity 2x plus 6.
Fill in the blanks
2x + 6 > 0 \;\Longrightarrow\; x > -3, \text___ x = ___
Why: The argument must be strictly positive, giving 2x greater than negative 6 and so x greater than negative 3. The vertical asymptote sits at the boundary, where the argument would be zero. The inequality is strict, so the bracket is round.
Prediction
Consider the common logarithm, base 10.
Predict first
What is the logarithm base 10 of one million?
Correct: 6, since a million is ten to the sixth.
Why: A million is 10 multiplied by itself six times, so the logarithm is 6. This is what makes logarithms so useful for compressing ranges: an input spanning six orders of magnitude produces an output spanning six units, which is why scales like decibels and the Richter scale are logarithmic.
Matching
The reflection swaps everything.
Match the pairs
Why: Every pairing is the same swap: reflecting in the diagonal exchanges the two axes, so domains become ranges, horizontal asymptotes become vertical ones, and every point's coordinates are reversed. Knowing the exponential's features is therefore enough to know the logarithm's.
Section
Section 4
Concept
Base ten and base e are used so often that they have abbreviations which leave the base unwritten. Everything else must have its base stated explicitly.
The two abbreviations exist because the two bases dominate different fields. Base ten matches the decimal system and so suits measurement scales; base e arises from continuous growth and so suits anything modelled by a rate of change, which is most of science.
Figure (svg): A card naming the two special logarithm bases and their notations: the common logarithm base ten and the natural logarithm base e
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 508-512
Picture it
Neither writes its base, and they are different bases.
Figure (svg): A card naming the two special logarithm bases and their notations: the common logarithm base ten and the natural logarithm base e
Reading log as base ten and ln as base e is a convention that has to be internalised, since nothing in the notation announces it.
Worked example
The base is ten, unwritten.
\[ \text{Evaluate } \log(0.001). \]
Identify the base
Why: No base written means ten.
\[ \text{base } 10 \]
Express the argument as a power of ten
Why: One thousandth.
\[ 10 ^{-3} \]
Read the exponent
Why: That is the value.
\[ -3 \]
Check by converting
Why: Ten to the negative 3 is one thousandth.
Figure (svg): A card naming the two special logarithm bases and their notations: the common logarithm base ten and the natural logarithm base e
\[ \log(0.001)=-3 \]
Verify: notice the pattern for powers of ten
Why: The common logarithm of a power of ten is simply its exponent, so it counts decimal places: 1000 gives 3, 0.01 gives negative 2. That is exactly why base ten is the convenient base for measurement scales, where quantities span many orders of magnitude.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 509-510
Sorting
Two abbreviations, two different bases.
Sort into buckets
Sort each notation.
Worked example
The base is e, unwritten.
\[ \text{Evaluate } \ln(e^5) \text{ and } \ln(1). \]
Identify the base
Why: ln means base e.
Ask the question for the first
Why: To what power must e be raised to give e to the fifth?
\[ 5 \]
Ask it for the second
Why: To what power must e be raised to give 1?
\[ 0 \]
Note the general pattern
Why: The logarithm undoes the exponential.
\[ \ln(e ^{x}) = x \]
Figure (svg): The solution to Worked example evaluate a natural logarithm shown as a ladder of expressions, one row per legal move
\[ \ln(e^5)=5, \qquad \ln(1)=0 \]
Verify: state the general rule that emerged
Why: The logarithm base b of b to the power x is always x, because the two operations are inverses and composing them returns the input. That is §1.7's composition condition, and it is the most-used simplification in the whole chapter.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 510-512
Error analysis
A student evaluates a common logarithm.
Annotate
On: \( \log(e^2) \overset{?}{=} 2 \)
The two abbreviations look similar and mean different bases. Checking which one is written before applying an inverse simplification prevents this entirely.
Faded example
Simplify the natural logarithm of e to the power 7.
Fill in the blanks
\ln(e^7) = 7, \qquad \text3 e^___ = ___
Why: The two operations are inverses, so composing them in either order returns the input. That is §1.7's two-sided condition, and it is why both simplifications work — one undoes the exponential and the other undoes the logarithm.
Prediction
A quantity is modelled by continuous growth.
Predict first
Which logarithm is most likely to appear in the analysis?
Correct: The natural logarithm, base e.
Why: Continuous growth produces base e, as §4.1's compounding limit showed, so the logarithm that undoes it is the natural one. Base ten dominates where quantities are described in orders of magnitude instead. The choice makes no mathematical difference, since §4.5's change-of-base formula converts freely, but it makes the formulas much tidier.
Real world
Decibels, the Richter scale and pH are all logarithmic.
Discussion prompt
What problem does a logarithmic scale solve?
Hint: How wide a range do these quantities span?
Answer:
These quantities span enormous ranges — sound intensities differ by factors of trillions, and earthquake energies by similar amounts. A linear scale would need numbers with a dozen digits and would compress everything interesting into an invisible sliver.
A logarithm turns multiplication into addition and factors into differences, so a range spanning twelve orders of magnitude becomes a scale from 0 to 12. Each step up is a fixed multiple rather than a fixed amount.
That is why a magnitude 7 earthquake is not slightly worse than a magnitude 6 but around thirty times more energetic. The scale is logarithmic and the intuition it invites is linear, which is a reliable source of public misunderstanding.
Section
Section 5
Concept
The exponential and the logarithm with the same base undo each other, so composing them in either order gives back what you started with.
\[ \log_b(b^x)=x, \qquad b^{\log_b(x)}=x \]
The asymmetry in the domains is worth noticing and is exactly §1.7's point about restricted domains. The two compositions have different domains because the two functions do, and the identity holds wherever both sides are defined.
Figure (svg): An exponential and its logarithm drawn as reflections in the line y equals x, with corresponding points marked showing their swapped coordinates
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 506-511
Picture it
The reflection in the diagonal is the geometric statement of the composition identity.
Figure (svg): An exponential and its logarithm drawn as reflections in the line y equals x, with corresponding points marked showing their swapped coordinates
Going out along one curve and back along the other returns to the starting input, which is what the two identities say in symbols.
Worked example
Match the bases and the composition collapses.
\[ \text{Simplify } \log_7(7^{12}) \text{ and } 10^{\log(45)}. \]
Check the bases match in the first
Why: Both are 7.
Apply the identity
Why: The logarithm undoes the exponential.
\[ 12 \]
Check the second
Why: The log has no base written, so base ten.
Apply the identity
Why: The exponential undoes the logarithm.
\[ 45 \]
Figure (svg): The solution to Worked example simplify compositions shown as a ladder of expressions, one row per legal move
\[ \log_7(7^{12})=12, \qquad 10^{\log(45)}=45 \]
Verify: check the bases really had to match
Why: The logarithm base 2 of 7 to the twelfth does not simplify, because the bases differ. The identities require the same base on both operations, and checking that before applying them is what prevents a very tempting wrong simplification.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 507-508
Sorting
The identity requires matching bases.
Sort into buckets
Sort each expression.
Worked example
This is the tool the warm-up was missing.
\[ \text{Solve } 2^x=10. \]
Convert to logarithmic form
Why: The exponent is what we want.
\[ x = \log _{2}(10) \]
Note that this IS the answer
Why: Exact, though not a familiar number.
Estimate it
Why: Between 3 and 4, since 8 and 16 straddle 10.
\[ \text{about } 3.32 \]
Check
Why: Two to the power 3.32 is about 10.
Figure (svg): The solution to Worked example use the inverse to solve shown as a ladder of expressions, one row per legal move
\[ x=\log_2(10)\approx 3.32 \]
Verify: compare with the warm-up
Why: The warm-up could only say the answer was between 3 and 4. Now it is written down exactly, as a logarithm, and evaluated to as many places as wanted. That is what the section set out to provide, and §4.5's change-of-base formula will show how a calculator produces the decimal.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 508-511
Trap
\[ \log_3(5^7) \overset{?}{=} 7 \]
Apply the inverse identity
Why: A logarithm of a power is simplified by reading off the exponent.
The answer is given as 7.
The bases do not match. The identity requires the logarithm's base and the power's base to be the same, and here they are 3 and 5.
This expression does simplify, but by §4.5's power property rather than by cancellation — it becomes 7 times the logarithm base 3 of 5, which is about 11.1, not 7.
Check the bases before cancelling. The identity is about a function and its own inverse, and 5 to the seventh is not something a base-3 logarithm undoes.
Faded example
Solve the equation setting 5 to the power x equal to 40.
Fill in the blanks
x = \log_5}(40)
Why: Converting to logarithmic form isolates the exponent immediately: the base stays 5, the result 40 becomes the argument, and the exponent becomes the logarithm's value. That expression is the exact answer, and it evaluates to about 2.29.
Prediction
Consider raising a base to the logarithm of x with the same base.
Predict first
For which x does that composition equal x?
Correct: Only for positive x.
Why: The inner logarithm requires a positive argument, so the composition is undefined for zero and negatives. The other composition, taking the logarithm of a power, works for every real input because an exponential accepts anything. The two identities have different domains, which is §1.7's point about restricted inverses appearing concretely.
Explain it
The warm-up posed a problem this section solved.
Discussion prompt
Explain to a classmate what problem logarithms solve, and why nothing earlier in the course could solve it.
Hint: Where is the unknown, and what operations do you know that undo what?
Answer:
The problem is an unknown in the exponent. Every technique before this chapter isolates a variable by undoing operations around it — subtracting, dividing, taking roots — but none of those undoes exponentiation.
A root undoes a power when the variable is in the base: the cube root undoes cubing. But when the variable is in the exponent, roots are the wrong tool, and there was simply no operation available.
The logarithm is that missing operation. It brings the exponent down where it can be worked with, which is why every equation with the unknown in an exponent is solved by taking a logarithm of both sides — the technique §4.6 develops.
Comparison
Fill the blanks from memory. Every row is the same swap seen from a different angle.
Comparison matrix
| exponential | logarithm | |
|---|---|---|
| domain | all real numbers | positive numbers only |
| range | positive numbers only | all real numbers |
| asymptote | horizontal, at y = 0 | vertical, at x = 0 |
| passes through | (0, 1) | (1, 0) |
| growth | very fast | very slow |
Every row follows from the reflection in the diagonal, so none of them needs separate memorising once the inverse relationship is understood.
Pattern
Five steps, and the first one solves most problems by itself.
Step 4's check is worth making explicit. The inverse identities are so convenient that they get applied to expressions where the bases differ, which produces a confidently wrong answer.
OpenStax Algebra and Trigonometry 2e, §6.3 Logarithmic Functions §6.3
Check
The base stays the base.
Check your understanding
Which exponential statement is equivalent to the logarithm base 3 of 81 being 4?
Answer: A
Why: The base stays the base, the logarithm's value becomes the exponent, and the argument becomes the result. So 3 to the fourth equals 81, which checks arithmetically.
Check
Strictly positive.
Check your understanding
What is the domain of the logarithm base 2 of the quantity x minus 5?
Answer: A
Why: The argument must be strictly positive, so x minus 5 must exceed zero, giving x greater than 5. There is a vertical asymptote at 5, and 5 itself is excluded because zero has no logarithm.
Check
Check the bases match.
Check your understanding
What is the value of e raised to the power of the natural logarithm of 12?
Answer: A
Why: The natural logarithm has base e and so does the exponential, so the two are inverses and composing them returns the input. The answer is 12 exactly, with no approximation involved.
Real world
Logarithms turn multiplicative questions into additive ones, which is why they are everywhere.
Discussion prompt
Why is 'how long until my investment doubles' a logarithm question, and 'how much will it be worth in ten years' an exponential one?
Hint: In each, where is the unknown?
Answer:
In the second question the time is known and the amount is not, so it is an evaluation: substitute the time into the exponential model and compute.
In the first the amount is known — double the original — and the time is not. But the time sits in the exponent, so isolating it requires an operation that brings an exponent down.
That is exactly what a logarithm does, which is why doubling-time and half-life questions are logarithm questions and value-after-n-years questions are exponential ones. The distinction is §1.1's evaluating-against-solving, applied to a family where solving needs a new tool.
Commit first
State your confidence along with your answer.
Predict first
What is the value of the logarithm base 8 of 2?
Correct: One third, since 8 to the power one third is 2.
Why: The question is what power 8 must be raised to in order to give 2, and 2 is the cube root of 8, which is the power one third. The answer 3 is the logarithm base 2 of 8, with the base and argument swapped — and note the two values are reciprocals, which is the pattern for any such swap.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why you can never take the logarithm of a negative number.
Hint: What question is a logarithm asking, and can it ever be answered for a negative?
Answer:
A logarithm asks to what power the base must be raised to give the argument. So the question 'the logarithm of negative 4' means 'what power of the base gives negative 4'.
A positive base raised to any power stays positive. Positive exponents give large positives, negative exponents give small positives, and zero gives 1. Nothing in that list is negative.
So there is no answer, and the expression is undefined. A good explanation adds the contrast: the logarithm's value may be negative — that is just a negative exponent — but its argument may not be. Those two slots have completely different rules and are easy to confuse.
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 first underlies all the others, and fluency in it makes the rest routine. The fourth is the most-used simplification in the chapter and the one most often applied where it does not hold.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the logarithmic-to-exponential equivalence with all three numbers labelled by their roles. Beneath it, sketch an exponential and its logarithm reflected in the line y equals x, marking one pair of corresponding points with their swapped coordinates and both asymptotes. Then list, side by side, the domain and range of each, and note which one is restricted and why.
If your two asymptotes are perpendicular to each other and your two domains are each other's ranges, the reflection is doing the work rather than memory.
Recap
Five things, and the first is what the whole chapter runs on.
| if you remember one thing | it should be this |
|---|---|
| about the definition | a logarithm is an exponent, and reading it aloud says which |
| about the domain | the argument must be strictly positive, never zero |
| about the value | it may be negative; that is just a negative exponent |
| about compositions | they cancel only when the bases match |
Section 4.4 graphs these functions and applies Chapter 1's transformations, where the vertical asymptote moves with a horizontal shift — the mirror image of what happened with exponentials.
OpenStax, Precalculus, §4.3 Logarithmic Functions §4.3, pp. 501-512 — everything on these slides traces back here
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