Transforms the logarithm parent, where the vertical asymptote and the domain boundary are the same line and therefore move together. Establishes that a horizontal shift moves both, that a reflection in the vertical axis moves the domain to the negative inputs, and that the domain can always be found by requiring the argument to be strictly positive.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 4 — Exponential and Logarithmic Functions
§4.4 Graphs of Logarithmic Functions, pp. 513-537
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 513-537 — the pages these objectives are drawn from
Warm-up
Section 4.3 derived its features from the exponential. Now they need to be usable at a glance.
Discussion prompt
Sketch the logarithm base 2 of x. Where does it cross an axis, and what happens as the input approaches zero?
Hint: What is the logarithm of 1, and what is the logarithm of a very small positive number?
Answer:
It crosses at the point where the input is 1, since the logarithm of 1 is zero for every base. It never crosses the vertical axis at all, because zero is not in its domain.
As the input approaches zero the outputs plunge without bound: the logarithm base 2 of one millionth is about negative 20. The graph runs down alongside the vertical axis.
That vertical line is the asymptote, and it is also the edge of the domain. Those two facts being the same line is what makes this section's transformations behave differently from §4.2's.
Concept
A logarithm's vertical asymptote sits exactly at the boundary of its domain. So any transformation that moves one moves the other, and finding either gives the other free.
\[ f(x)=a\log_b(x-h)+k \;\Longrightarrow\; \text{asymptote } x=h, \; \text{domain } x>h \]
This coupling has no precedent in the earlier chapters. A rational function's vertical asymptote also excluded a point, but it excluded a single point from an otherwise unrestricted domain; here the asymptote is the endpoint of a ray, so moving it slides the entire domain.
Figure (svg): The parent logarithm graph with its features labelled: the vertical asymptote along the vertical axis, the point at input one, and the domain marked as only the positive inputs
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 513-519
Section
Section 1
Concept
The logarithm parent has a vertical asymptote at zero, passes through the point at input one, and accepts only positive inputs.
The absence of a y-intercept is worth stating explicitly, because every function since Chapter 1 has had one. It is a direct consequence of the domain: the vertical axis sits outside it, so there is nowhere for the graph to cross.
Figure (svg): The parent logarithm graph with its features labelled: the vertical asymptote along the vertical axis, the point at input one, and the domain marked as only the positive inputs
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 513-520
Picture it
The shaded region is the domain, and its edge is the asymptote.
Figure (svg): The parent logarithm graph with its features labelled: the vertical asymptote along the vertical axis, the point at input one, and the domain marked as only the positive inputs
The graph occupies only the shaded strip, running down alongside the asymptote and rising slowly to the right. The point at input 1 is where every logarithm of every base crosses.
Worked example
Three features, read off without computing.
\[ \text{State the domain, range and asymptote of } f(x)=\log_3(x). \]
State the domain
Why: Arguments must be positive.
\[ x > 0 \]
State the range
Why: A logarithm's value is unrestricted.
Name the asymptote
Why: At the edge of the domain.
\[ x = 0 \]
Find the intercept
Why: The logarithm of 1 is zero.
\[ (1, 0) \]
Figure (svg): The parent logarithm graph with its features labelled: the vertical asymptote along the vertical axis, the point at input one, and the domain marked as only the positive inputs
\[ \text{domain } (0,\infty), \; \text{range } (-\infty,\infty), \; \text{asymptote } x=0 \]
Verify: test very close to zero
Why: The logarithm base 3 of one thousandth is about negative 6.3, and of one millionth about negative 12.6. The outputs fall without bound as the input approaches zero, confirming the vertical asymptote — and confirming that no output is ever attained at zero itself.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 514-516
Sorting
Compare against the exponential parent from §4.2.
Sort into buckets
Sort each statement.
Worked example
The base changes the steepness and nothing else.
\[ \text{Compare } \log_2(x) \text{ with } \log_{10}(x). \]
Compare at input 1
Why: Both give zero.
Compare at input 100
Why: About 6.6 against exactly 2.
\[ \text{base } 2\text{ is higher} \]
Interpret
Why: A smaller base rises faster.
Check the shared features
Why: Same domain, range and asymptote.
Figure (svg): The solution to Worked example compare two bases shown as a ladder of expressions, one row per legal move
\[ \text{both through } (1,0), \text{ asymptote } x=0; \; \log_2 \text{ steeper} \]
Verify: explain the direction of the effect
Why: A smaller base means the corresponding exponential grows more slowly, so its reflection rises more quickly. The base 2 logarithm counts doublings and the base 10 one counts decimal places, and there are far more doublings than decimal places in any given number — which is why base 2 gives the larger value.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 516-520
Trap
\[ f(x)=\log_2(x): \quad y\text{-intercept} = \log_2(0) \]
Substitute zero to find the y-intercept
Why: That is how a y-intercept is found for every other function.
The intercept is written as the logarithm of zero.
Zero is not in the domain, so there is no y-intercept at all. The logarithm of zero is undefined, not some number waiting to be computed.
The graph runs down alongside the vertical axis without touching it, so it never crosses. There is nothing there.
Check whether zero is in the domain before looking for a y-intercept. For the logarithm parent it never is, and for a shifted one it is only when the shift moved the asymptote past zero.
Prediction
Several logarithms with different bases are graphed together.
Predict first
Where do they all meet?
Correct: At the point where the input is 1.
Why: The logarithm of 1 is zero for every base, since any base to the power zero is 1. So every logarithm passes through that point and they fan out from it. The origin is not on any of them, since zero is outside the domain.
Faded example
For the logarithm base 6 of x.
Fill in the blanks
\text0 x > 0; \quad \text___; \quad \text___ x = ___
Why: Both are zero for the parent, and they are the same boundary seen two ways — the asymptote is the line, and the domain is everything to the right of it. A horizontal shift later will move both together, which is why keeping them linked is the right habit.
Explain it to yourself
The logarithm rises very slowly compared with anything else in the course.
Discussion prompt
Explain why, using the exponential it inverts.
Hint: How far along the input axis do you have to go for the output to increase by one?
Answer:
The exponential climbs steeply: one step along the input multiplies the output by the base. Reflecting a steep curve in the diagonal produces a shallow one.
Concretely, for the output of a base-10 logarithm to increase by 1, the input has to be multiplied by 10. Going from an output of 6 to an output of 7 means going from a million to ten million.
So the logarithm compresses multiplicative distances into additive ones, which is exactly what makes it useful for scales spanning many orders of magnitude — and it is why a logarithm graph looks almost flat once you are any distance to the right.
Section
Section 2
Concept
A horizontal shift slides the asymptote, and because the asymptote is the domain's boundary, the domain slides with it.
The last bullet is the reliable method and it is worth using rather than tracking the shift by eye. Setting the argument positive and solving handles every case at once — shifts, stretches inside, and reflections — without any need to reason about which way something moved.
Figure (svg): A logarithm shifted horizontally, showing the vertical asymptote and the domain boundary moving together, drawn against the parent
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 520-527
Picture it
The dashed curve is the parent and the solid one has shifted right 3.
Figure (svg): A logarithm shifted horizontally, showing the vertical asymptote and the domain boundary moving together, drawn against the parent
The asymptote moved from zero to 3 and the domain moved with it. There is no way to move one without the other, because they are the same boundary.
Worked example
Set the argument positive; everything follows.
\[ \text{Describe } f(x)=\log_2(x-3). \]
Set the argument positive
Why: This is the domain condition.
\[ x - 3 > 0 \]
Solve
Why: Add 3.
\[ x > 3 \]
Place the asymptote
Why: At the domain's edge.
\[ x = 3 \]
Find the x-intercept
Why: Where the argument is 1.
\[ x = 4 \]
Figure (svg): A logarithm shifted horizontally, showing the vertical asymptote and the domain boundary moving together, drawn against the parent
\[ \text{domain } (3,\infty), \; \text{asymptote } x=3, \; (4,0) \]
Verify: check the intercept
Why: At x equal to 4 the argument is 1 and the logarithm of 1 is zero, so the graph crosses there. Note the x-intercept moved right 3 along with everything else, which is what a horizontal shift does. The output is unrestricted, so the range is unchanged.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 521-523
Faded example
For the logarithm of the quantity x plus 7.
Fill in the blanks
x + 7 > 0 \;\Longrightarrow\; x > -7, \text___ x = ___
Why: Setting the argument positive gives x greater than negative 7, and the asymptote sits at that boundary. The plus inside shifted the graph left, which is §1.5's reversal — and solving the inequality produces the sign automatically without needing to remember the rule.
Worked example
The contrast with the previous example is the point.
\[ \text{Describe } f(x)=\log_2(x)+4. \]
Check the argument
Why: It is just x, unchanged.
\[ x > 0 \]
Place the asymptote
Why: Unmoved.
\[ x = 0 \]
State the range
Why: Still every real number.
Find the x-intercept
Why: Set the whole rule to zero.
\[ \log _{2}(x) = -4,\text{ so } x = \frac{1}{16} \]
Figure (svg): The solution to Worked example a vertical shift changes nothing horizontal shown as a ladder of expressions, one row per legal move
\[ \text{domain } (0,\infty), \; \text{asymptote } x=0, \; (\tfrac{1}{16},0) \]
Verify: notice which intercept moved
Why: The vertical shift left the domain and asymptote alone but moved the x-intercept, from 1 to one sixteenth. In the previous example the horizontal shift moved everything. The two shifts affect disjoint sets of features, which is the cleanest way to keep them apart.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 524-527
Error analysis
A student finds the asymptote of a shifted logarithm.
Annotate
On: \( f(x)=\log(x+5): \quad \text{asymptote } x=5 \)
Set the argument strictly greater than zero and solve. That method carries the sign automatically and never requires remembering which way an inside change goes.
Sorting
For a logarithm, the horizontal transformations are the ones that matter.
Sort into buckets
Sort each transformation.
Prediction
A logarithm is shifted up 3, with no horizontal shift.
Predict first
What happens to its x-intercept?
Correct: It moves, since the whole graph rose.
Why: The crossing point was where the output was zero, and raising the graph means a different input now produces zero. Since a logarithm's range is all real numbers, the crossing still exists — it has just moved left, to where the logarithm equals negative 3.
Socratic
For an exponential the asymptote and the domain were independent.
Discussion prompt
Explain why a logarithm's asymptote and domain must move together, and why an exponential's did not.
Hint: Where does each asymptote sit relative to the domain?
Answer:
A logarithm's asymptote sits at the edge of its domain — it is the boundary line itself, the place where the argument would be zero. So anything that moves the boundary moves both descriptions of it.
An exponential's asymptote is horizontal, describing the outputs, while its domain is every real number and has no boundary at all. The two are about different axes, so they cannot interact.
This is the reflection at work again. The exponential's asymptote is a range feature and the logarithm's is a domain feature, because swapping the axes swapped which set the asymptote bounds. Everything about this section is §4.2 read through that mirror.
Section
Section 3
Concept
A coefficient outside stretches the graph vertically. A negative outside flips it; a negative inside moves the entire domain to the negative inputs.
The inside reflection is the striking case. The argument must still be positive, so negating it means the input itself must be negative — and the whole graph relocates to the left of the vertical axis. No earlier family did anything like this, because none had a one-sided domain to flip.
Figure (svg): Two reflections of the logarithm parent: one over the horizontal axis flipping the curve, and one over the vertical axis moving the whole graph to the negative inputs
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 527-533
Picture it
Outside on the left, inside on the right, and only one of them moves the domain.
Figure (svg): Two reflections of the logarithm parent: one over the horizontal axis flipping the curve, and one over the vertical axis moving the whole graph to the negative inputs
The outside reflection flips the curve within the same domain. The inside one relocates the entire graph to the negative inputs, which is a change of a different kind.
Worked example
The argument must still be positive, which forces the input negative.
\[ \text{Find the domain of } f(x)=\log_2(-x). \]
Set the argument positive
Why: As always.
\[ -x > 0 \]
Solve
Why: Divide by negative 1 and reverse.
\[ x < 0 \]
Interpret
Why: Only negative inputs are legal.
Place the asymptote
Why: Still at the boundary.
\[ x = 0 \]
Figure (svg): Two reflections of the logarithm parent: one over the horizontal axis flipping the curve, and one over the vertical axis moving the whole graph to the negative inputs
\[ \text{domain } (-\infty,0), \; \text{asymptote } x=0 \]
Verify: test one input from each side
Why: At x equal to negative 4 the argument is 4, giving 2 — legal. At x equal to 4 the argument is negative 4, which has no logarithm. So the graph really does live entirely to the left of the axis, which is the reflection of the parent about that axis.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 528-530
Faded example
For the natural logarithm of the quantity 6 minus x.
Fill in the blanks
6 - x > 0 \;\Longrightarrow\; x < 6, \text___ x = ___
Why: Solving the inequality gives x below 6, so the domain runs leftward from the asymptote at 6 rather than rightward. The negative coefficient on x inside reflects the graph, so it approaches the asymptote from the left instead of the right.
Worked example
Neither touches the domain.
\[ \text{Describe } f(x)=-3\log(x). \]
Check the argument
Why: It is just x.
\[ \text{domain } x > 0 \]
Read the coefficient's size
Why: Three.
\[ \text{stretch by } 3 \]
Read its sign
Why: Negative, acting on the output.
State the range
Why: A logarithm's range is unrestricted either way.
Figure (svg): The solution to Worked example a stretch and an outside reflection shown as a ladder of expressions, one row per legal move
\[ \text{domain } (0,\infty), \; \text{range } (-\infty,\infty) \]
Verify: note what did not change
Why: The range is every real number both before and after, since reflecting and stretching a set that is already everything leaves it everything. That contrasts with §4.2's exponentials, where a reflection genuinely changed the range from the positives to the negatives — because there the range was bounded and here it is not.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 531-533
Trap
\[ f(x)=\log(4-x): \quad \text{domain } x>0 \]
Recall that a logarithm needs positive inputs
Why: The domain is written as the positive numbers, as for the parent.
The domain is given as the positive inputs.
The ARGUMENT must be positive, not the input. Setting 4 minus x greater than zero gives x less than 4.
So the domain is everything below 4, which includes plenty of negative inputs and excludes plenty of positive ones — nearly the opposite of what was claimed.
Always solve the inequality on the argument. The parent's domain being the positives is a special case of that rule, not a rule in itself, and it stops being right the moment the argument is anything other than x.
Sorting
Only changes to the argument can.
Sort into buckets
Sort each transformation of a logarithm.
Prediction
A logarithm is reflected over the horizontal axis.
Predict first
What happens to its range?
Correct: Nothing: it was already every real number.
Why: Reflecting a set that is already all of the real numbers leaves it unchanged. This is the opposite of §4.2's exponentials, whose range was bounded and did flip under the same reflection. The difference is whether the range was one-sided to begin with.
Counterexample
A classmate says a logarithm always has the positive numbers as its domain.
Discussion prompt
Give two counterexamples with quite different domains.
Hint: Change the argument.
Answer:
The logarithm of negative x has the negatives as its domain, since the argument is positive only when the input is not.
The logarithm of 4 minus x has everything below 4, which mixes negatives and positives. Neither has the positives as its domain, and neither is exotic.
The correct rule is that the argument must be positive, and the parent's domain is what that gives when the argument happens to be x itself. Solving the inequality is the general method, and it produces all of these cases without any special reasoning.
Section
Section 4
Concept
A logarithm crosses the horizontal axis where its output is zero, which is where its argument equals one. A y-intercept exists only if zero is in the domain.
The second bullet is where the work is. With a vertical shift the crossing is no longer where the argument is 1, so the equation has to be solved by converting to exponential form — which is the technique §4.6 develops properly.
Figure (svg): A table showing which transformations move a logarithm's vertical asymptote and which affect its domain, contrasted with the exponential case
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 523-531
Picture it
The horizontal transformations control the domain, the asymptote, and therefore the intercepts.
Figure (svg): A table showing which transformations move a logarithm's vertical asymptote and which affect its domain, contrasted with the exponential case
Whether a y-intercept exists is entirely a question about the domain, which is entirely a question about the horizontal transformations.
Worked example
Check whether the y-intercept can exist at all.
\[ \text{Find the intercepts of } f(x)=\log_2(x+8). \]
Find the domain
Why: Argument positive.
\[ x > -8 \]
Check whether zero is in it
Why: Zero exceeds negative 8.
Find the y-intercept
Why: Substitute zero.
\[ \log _{2}(8) = 3 \]
Find the x-intercept
Why: Argument equal to 1.
\[ x = -7 \]
Figure (svg): The solution to Worked example find both intercepts shown as a ladder of expressions, one row per legal move
\[ (0,\,3) \text{ and } (-7,\,0) \]
Verify: confirm both lie in the domain
Why: Both 0 and negative 7 exceed negative 8, so both are legal inputs. The left shift of 8 brought the vertical axis inside the domain, which is why a y-intercept exists here and does not for the parent. Checking the domain first is what makes the y-intercept question answerable without guessing.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 524-526
Sorting
It has one exactly when zero is in the domain.
Sort into buckets
Sort each function.
Worked example
Set the whole rule to zero and convert.
\[ \text{Find the x-intercept of } f(x)=\log_3(x)-2. \]
Set the rule to zero
Why: That is what an x-intercept means.
\[ \log _{3}(x) - 2 = 0 \]
Isolate the logarithm
Why: Add 2.
\[ \log _{3}(x) = 2 \]
Convert to exponential form
Why: Base to the value.
\[ x = 3 ^{2} \]
Evaluate
Why: Nine.
\[ x = 9 \]
Figure (svg): The solution to Worked example a vertical shift moves the crossing shown as a ladder of expressions, one row per legal move
\[ (9,\,0) \]
Verify: check by substitution
Why: At x equal to 9 the logarithm base 3 of 9 is 2, minus 2 gives zero — correct. The parent crossed at 1 and the vertical shift moved the crossing to 9, which is a large move for a shift of 2, because a logarithm rises so slowly that a small vertical change corresponds to a large horizontal one.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 527-531
Error analysis
A student looks for the y-intercept of a right-shifted logarithm.
Annotate
On: \( f(x)=\log(x-2): \quad y\text{-intercept} = \log(-2) \)
Check whether zero lies in the domain before looking for a y-intercept. For a logarithm shifted right, it never does.
Faded example
For the logarithm base 5 of x, minus 3.
Fill in the blanks
\log_5(x) = 3 \;\Longrightarrow\; x = 5^3} = 125
Why: Setting the rule to zero and isolating gives the logarithm equal to 3, which converts to x equal to 5 cubed, or 125. The vertical shift of 3 moved the crossing from 1 all the way out to 125, which shows how slowly a logarithm climbs.
Prediction
A logarithm parent is shifted right 6, with no vertical shift.
Predict first
Where does it cross the horizontal axis?
Correct: At input 7, where the argument is 1.
Why: The crossing happens where the argument equals 1, since the logarithm of 1 is zero. Here the argument is x minus 6, which is 1 at x equal to 7. Input 6 is the asymptote, where the argument is zero and the function is undefined.
Step zero
You are asked for the intercepts of a transformed logarithm.
Discussion prompt
What do you check first, and why does it save work?
Hint: One of the two intercepts may not exist.
Answer:
Find the domain first, by setting the argument strictly greater than zero. It takes one line.
Then check whether zero is in it. If not, there is no y-intercept and the substitution need not be attempted at all — which saves reaching an undefined expression and having to interpret it.
The domain also gives the asymptote for free, since they share a boundary, and it tells you which side of the asymptote the graph lives on. One inequality answers three questions, which is why it is worth doing before anything else.
Section
Section 5
Concept
A fully transformed logarithm has four parameters. Reading the argument first gives the domain and the asymptote, and everything else follows.
\[ f(x)=a\log_b(x-h)+k \]
Reading h first is the efficient order for the same reason k was first for exponentials: it is the parameter that unlocks the most. The domain, the asymptote, and whether a y-intercept can exist all depend on it, and none of them can be settled without it.
Figure (svg): A table showing which transformations move a logarithm's vertical asymptote and which affect its domain, contrasted with the exponential case
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 531-537
Picture it
Only the horizontal shift touches the domain and the asymptote.
Figure (svg): A table showing which transformations move a logarithm's vertical asymptote and which affect its domain, contrasted with the exponential case
Comparing with §4.2's table shows the mirror clearly: there the vertical shift was the special one, and here the horizontal shift is.
Worked example
Argument first, then the rest.
\[ \text{Describe } f(x)=-2\log_3(x+1)+5 \text{ completely.} \]
Set the argument positive
Why: This gives the domain.
\[ x > -1 \]
Place the asymptote
Why: At the boundary.
\[ x = -1 \]
Read the coefficient
Why: Two, negative.
\[ \text{stretch } 2,\text{ reflected} \]
Read the vertical shift
Why: Plus 5.
\[ \text{up } 5 \]
Figure (svg): The solution to Worked example read all four parameters shown as a ladder of expressions, one row per legal move
\[ \text{domain } (-1,\infty), \; \text{asymptote } x=-1, \; \text{range all reals} \]
Verify: find the y-intercept as a check
Why: Zero exceeds negative 1, so it is in the domain. Substituting gives negative 2 times the logarithm base 3 of 1, plus 5, which is 0 plus 5, giving 5. The left shift brought the vertical axis into the domain, so a y-intercept exists — which the domain check predicted before any computation.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 532-534
Comparison
Fill the blanks from memory. The two families mirror each other exactly.
Comparison matrix
| exponential | logarithm | |
|---|---|---|
| asymptote is | horizontal | vertical |
| it bounds the | range | domain |
| moved by | a vertical shift | a horizontal shift |
| the unrestricted set | the domain | the range |
Every row is the same swap. Knowing one family's arrangement gives the other's by reflecting, which is faster and more reliable than learning two tables.
Worked example
Asymptote, then one easy point, then the direction.
\[ \text{Sketch } f(x)=\log_2(x-1)+3. \]
Draw the asymptote
Why: Where the argument vanishes.
\[ \text{vertical line } x = 1 \]
Find the easy point
Why: Where the argument is 1.
\[ (2, 3) \]
Determine the direction
Why: Positive coefficient, base above 1.
Note the behaviour near the asymptote
Why: Plunges as the input approaches 1.
\[ \text{falls alongside } x = 1 \]
Figure (svg): The solution to Worked example sketch from the parameters shown as a ladder of expressions, one row per legal move
\[ \text{asymptote } x=1, \text{ through } (2,3) \]
Verify: check the x-intercept exists
Why: The range is all reals, so the curve must reach zero somewhere. Setting the rule to zero gives the logarithm equal to negative 3, so the argument is one eighth and x is nine eighths — just to the right of the asymptote, which matches the sketch. Finding the input that makes the argument 1 is the fastest exact point on any transformed logarithm.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 535-537
Trap
\[ f(x)=\log(x-2)+5: \quad \text{range } y>5 \]
Apply the exponential reasoning about ranges
Why: The vertical shift is taken to bound the range from below, as it would for an exponential.
The range is given as everything above 5.
A logarithm's range is every real number, and shifting it vertically leaves it every real number.
The confusion is with §4.2's exponentials, whose range genuinely was bounded by their asymptote. A logarithm's asymptote is vertical and bounds the domain instead.
Check which axis the asymptote is on. A horizontal asymptote bounds the range; a vertical one bounds the domain. The two families have opposite arrangements, which is exactly what the reflection produces.
Faded example
For the function 4 times the logarithm base 2 of the quantity x minus 5, minus 1.
Fill in the blanks
\text5 x > all reals, \quad \text___ x = ___, \quad \text______
Why: The argument x minus 5 is positive when x exceeds 5, which gives both the domain and the asymptote. The range is every real number regardless of the coefficient 4 and the vertical shift, since neither can bound an already unbounded set.
Prediction
A logarithm with a negative coefficient has its asymptote at x equal to 2.
Predict first
What happens as the input approaches 2 from the right?
Correct: The outputs rise without bound.
Why: The parent plunges downward near its asymptote, and the negative coefficient reflects that, so the transformed graph shoots upward instead. The outputs do not approach any value — that is what an unbounded range near a vertical asymptote means, and it contrasts with an exponential, whose outputs do approach its horizontal asymptote.
Explain it
This section and §4.2 are the same content reflected.
Discussion prompt
Explain to a classmate how knowing the exponential's transformations gives the logarithm's for free.
Hint: What does reflecting in the diagonal do to horizontal and vertical?
Answer:
Reflecting in the line y equals x exchanges the two axes. So everything horizontal becomes vertical and everything vertical becomes horizontal.
For an exponential the asymptote is horizontal and bounds the range, and the vertical shift moves it. Reflecting all three words gives the logarithm: the asymptote is vertical, it bounds the domain, and the horizontal shift moves it.
So there is one set of facts, read two ways. That is worth more than memorising two tables, because the reflection also explains why the two families' unrestricted sets are opposite — the exponential's domain and the logarithm's range are the same set seen from either side.
Comparison
Fill the blanks from memory. Compare with §4.2's table, which is this one reflected.
Comparison matrix
| moves the asymptote | changes the domain | changes the range | |
|---|---|---|---|
| horizontal shift | yes | yes, it moves with it | no |
| vertical shift | no | no | no |
| vertical stretch | no | no | no |
| reflection over the y axis | no | yes, it flips sides | no |
The last column is entirely 'no', because a logarithm's range is already every real number and nothing can enlarge or restrict it.
Pattern
Five steps, and the first answers three questions at once.
Step 1 is the workhorse. Solving the inequality handles shifts, reflections and inside stretches all at once, and it carries every sign correctly without any rule needing to be recalled.
OpenStax Algebra and Trigonometry 2e, §6.4 Graphs of Logarithmic Functions §6.4
Check
Set the argument positive.
Check your understanding
What is the domain of the logarithm of the quantity x plus 9?
Answer: A
Why: The argument x plus 9 must exceed zero, giving x greater than negative 9. The vertical asymptote sits at negative 9, and the plus inside shifted the graph left, which the inequality produces automatically.
Check
The asymptote is vertical here.
Check your understanding
What is the range of the function 3 times the logarithm base 2 of x, plus 7?
Answer: A
Why: A logarithm's range is every real number, and stretching or shifting vertically cannot bound an already unbounded set. The asymptote here is vertical and bounds the domain, not the range.
Check
Zero must be in the domain.
Check your understanding
Which function has a y-intercept?
Answer: A
Why: At input zero the argument is 3, which is positive, so zero is in the domain and the graph crosses the vertical axis at the logarithm of 3. The left shift brought the vertical axis inside the domain.
Real world
Logarithmic axes are used constantly in science and finance, and they are this section's graph.
Discussion prompt
Why does plotting data on a logarithmic axis turn exponential growth into a straight line?
Hint: What does the axis do to the numbers before plotting them?
Answer:
A logarithmic axis plots the logarithm of each value rather than the value. So exponential growth, which multiplies by a fixed factor each step, becomes a fixed addition each step.
A fixed addition per step is exactly a constant rate of change, which §2.1 says produces a straight line. So the curve straightens out, and its slope is the growth rate.
This is why log plots are the standard tool for spotting exponential behaviour: straight on a log axis means exponential, and the eye is far better at judging straightness than at judging exponential curvature. It also compresses enormous ranges onto one page, which is the same benefit §4.3 noted for decibels and magnitudes.
Commit first
State your confidence along with your answer.
Predict first
A logarithm has a vertical asymptote at x equal to 4. What is its domain likely to be?
Correct: Either everything above 4 or everything below it, depending on the argument.
Why: The asymptote marks the boundary, but which side is legal depends on the argument's sign. The logarithm of x minus 4 gives everything above; the logarithm of 4 minus x gives everything below. Unlike a rational function's asymptote, which excludes a single point from an otherwise full domain, a logarithm's asymptote is the endpoint of a ray.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why a logarithm's asymptote and domain always move together, unlike anything in Chapter 3.
Hint: Where does the asymptote sit relative to the domain?
Answer:
A logarithm's asymptote is at the input where its argument would be zero — the exact point where the function stops being defined. So the asymptote IS the domain's boundary rather than a line the graph merely avoids.
Contrast a rational function: its vertical asymptote removes a single point from a domain that continues on both sides. Moving it changes which point is missing but the domain is still nearly everything.
For a logarithm, moving the asymptote slides the whole ray. That is why one inequality — the argument being positive — answers both questions at once, and why finding either one gives the other with no extra work.
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 is the method that handles every case and is worth making automatic. The fourth is worth the time even though it is not directly examinable, because understanding the mirror halves what has to be remembered about both sections.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Sketch the logarithm parent with its vertical asymptote, its point at input one, and its domain shaded. Beside it sketch a transformed version with a horizontal shift and a reflection, marking the new asymptote and the new domain. Underneath, write the inequality you solved to get the domain, and note which transformations left it alone.
If the asymptote in your second sketch sits exactly at the boundary of the shaded domain, you have drawn the coupling that makes this section different from every earlier one.
Recap
Five things, and the second one does most of the work.
| if you remember one thing | it should be this |
|---|---|
| about the domain | set the argument positive and solve; that handles every case |
| about the asymptote | it is the domain's boundary, so they always move together |
| about the range | always every real number, whatever the transformations |
| about the mirror | everything is Section 4.2's with horizontal and vertical swapped |
Section 4.5 turns from graphs to algebra, establishing the properties that let a logarithm of a product be split apart — which is what makes logarithms a computational tool rather than just an inverse.
OpenStax, Precalculus, §4.4 Graphs of Logarithmic Functions §4.4, pp. 513-537 — everything on these slides traces back here
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