The logarithmic graph is the mirror image of its exponential counterpart, reflected across $y = x$. Where exponentials shoot upward, logarithms creep steadily, and knowing how to sketch, shift, and decode them is a core exam skill.
Today's hook, On the same axes, picture $y = 2^x$ and $y = \log_2 x$. What single straight line are they mirror images across? Commit to an answer before you read on, and by the end of this lesson you'll be able to sketch any logarithmic function from scratch and find its equation from a graph.
0/5QUESTS
1
You’re here
Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
On the same axes, picture $y = 2^x$ and $y = \log_2 x$. What straight line do they reflect across? Make a guess and jot down any features you expect the logarithmic graph to have.
auto-saved
2
You’re here
The two moves
Work through the core explanation before applying it.
02
The two moves
+5 XP to read
Sketching logarithmic functions comes down to two moves: start with the basic shape and key features, then apply transformations. Lock these into memory and every graph follows.
Basic shape:vertical asymptote at $x = 0$, passes through $(1, 0)$, and increases for $a > 1$. Transformations shift that shape predictably.
$y = \log_a(x-h)+k$: asymptote $x=h$, point $(1+h,\, k)$
Asymptote rule
$y = \log_a(x - h) + k$ has vertical asymptote $x = h$. A minus sign shifts right; domain becomes $x > h$.
Reflect to sketch
Sketch the exponential $y = a^x$ first, then reflect every point $(p, q)$ as $(q, p)$ to get the logarithmic curve across $y = x$.
x-intercept trick
Set $y = 0$ and solve $\log_a(\ldots) = 0$. Since $\log_a 1 = 0$, the argument equals 1, that pinpoints the intercept instantly.
3
You’re here
What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
03
What you'll master
Know
Key facts
Graph of $y = \log_a x$: asymptote $x = 0$, point $(1, 0)$, domain $x > 0$
Transformation rules for $y = \log_a(x-h) + k$
Logarithmic graphs are reflections of exponential graphs in $y = x$
Understand
Concepts
Why the vertical asymptote shifts when $h \neq 0$
Why the domain changes for shifted logarithms
How the inverse relationship manifests graphically
Can do
Skills
Sketch $y = \log_a(x-h) + k$ labelling asymptote, intercepts, and one other point
Find the equation of a log curve from its graph features
State domain and range of transformed logarithmic functions
04
Key terms
Vertical asymptoteA vertical line $x = c$ that the graph approaches but never crosses. For $y = \log_a x$, it is $x = 0$.
Inverse graphThe reflection of a graph across the line $y = x$. Logarithmic and exponential graphs are inverses.
Horizontal shift$y = \log_a(x - h)$ shifts the basic graph $h$ units to the right (left if $h < 0$) and moves the asymptote to $x = h$.
Vertical shift$y = \log_a x + k$ shifts the basic graph $k$ units up (down if $k < 0$); the asymptote stays at $x = 0$.
x-interceptWhere the graph crosses the x-axis. Find by setting $y = 0$ and solving: $\log_a(\text{expr}) = 0 \Rightarrow \text{expr} = 1$.
4
You’re here
Sketching logarithmic functions
Work through the core explanation before applying it.
05
Sketching logarithmic functions
core concept
The graph of $y = \log_a x$ (for $a > 1$) has these fixed features: a vertical asymptote at $x = 0$, an $x$-intercept at $(1, 0)$, and an ever-increasing curve that grows more and more slowly. It is the exact reflection of $y = a^x$ across the line $y = x$.
$$y = \log_a(x - h) + k: \quad \text{asymptote } x = h,\quad \text{domain } x > h, \quad \text{x-intercept at } x = a^{-k}+h$$
Sketching strategy. (1) Identify $h$ and $k$. (2) Draw the vertical asymptote $x = h$ as a dashed line. (3) Plot the x-intercept by solving $\log_a(x-h) + k = 0$. (4) Plot one more point (e.g. when $x - h = a$). (5) Draw a smooth increasing curve hugging the asymptote.
Basic graph: $y = \log_a x$: asymptote $x = 0$, $x$-intercept $(1,0)$, domain $x > 0$, range $\mathbb{R}$; $y = \log_a(x-h)+k$: asymptote $x = h$, domain $x > h$, shift right by $h$, up by $k$
Pause, copy the log graph features: basic $y = \log_a x$ (asymptote $x = 0$, $x$-int $(1,0)$, domain $x > 0$, range $\mathbb{R}$) and the shift rules for $y = \log_a(x-h) + k$ into your book.
Did you get this? True or false: the graph of $y = \log_3(x + 2)$ has a vertical asymptote at $x = -2$.
Worked examples · 3 in a row, reveal as you go
5
You’re here
Work examples end to end
Follow the reasoning through complete worked solutions.
The natural log is undefined at $x \leq 0$. Draw as a dashed line first.
2
$x$-intercept: $(1, 0)$ since $\ln 1 = 0$
$e^0 = 1$, so $\ln 1 = 0$ by the inverse relationship.
3
Second point: $(e, 1)$ since $\ln e = 1$
$\ln e = 1$ by definition of the natural log.
4
Draw an increasing curve from the bottom-left, approaching $x = 0$ without touching, passing through $(1, 0)$ and $(e, 1)$, rising slowly to the right.
$\ln x$ is the inverse of $e^x$, so it grows far slower than any linear function.
PROBLEM 2 · TRANSFORMED GRAPH
Sketch $y = \log_2(x - 1) + 1$ and state the domain.
1
Start with $y = \log_2 x$: shift right 1 unit and up 1 unit.
Identify the basic function $y = \log_2 x$ and the transformations $h = 1$, $k = 1$.
2
Vertical asymptote: $x = 1$ (dashed line)
Horizontal shift by $+1$ moves the asymptote from $x = 0$ to $x = 1$.
$\log_2 1 = 0$ for any base, this is always a reliable anchor point.
4
Domain: $x - 1 > 0 \Rightarrow x > 1$, i.e. $(1, \infty)$.
The argument must be positive; the asymptote is the left boundary of the domain.
PROBLEM 3 · FIND EQUATION FROM GRAPH
A logarithmic curve has vertical asymptote $x = 2$, passes through $(3, 0)$ and $(5, 1)$. Find its equation.
1
Asymptote $x = 2$ means the form is $y = \log_a(x - 2) + k$.
Horizontal shift of 2 units; $h = 2$.
2
At $(3, 0)$: $0 = \log_a(3-2) + k = \log_a 1 + k = 0 + k$, so $k = 0$.
$\log_a 1 = 0$ for any base, so $k$ vanishes immediately.
3
At $(5, 1)$: $1 = \log_a(5-2) = \log_a 3$, so $a^1 = 3$, giving $a = 3$.
Convert to exponential form: $\log_a 3 = 1$ means $a = 3$.
4
Equation: $y = \log_3(x - 2)$
Verify: asymptote $x = 2$ ✓, passes through $(3, 0)$ ✓ and $(5, 1)$ ✓.
Quick check: What is the domain of $y = \log_5(x + 3)$?
Common errors · the 3 traps that cost marks
6
You’re here
Dodge the traps, then apply
Meet the mistakes that cost marks, then do it yourself.
Trap 01
Drawing the curve crossing the vertical asymptote
The logarithmic graph approaches the vertical asymptote but never touches or crosses it. Curves that turn back and pass through the asymptote are always wrong, you lose marks for this in any sketch question.
Trap 02
Getting the sign wrong on $y = \log_a(x - h)$
$y = \log_a(x - 3)$ has asymptote $x = 3$, not $x = -3$. The minus sign shifts right. Students see a minus and write a negative asymptote, check by substituting $x = h + 1$ to confirm the argument is positive.
Trap 03
Forgetting to adjust the domain for horizontal shifts
$y = \log_a(x + 2)$ has domain $x > -2$, not $x > 0$. Always solve the inequality for the argument: $x + 2 > 0 \Rightarrow x > -2$. Stating $x > 0$ costs 1–2 marks in extended response.
Fill in the blank: The $x$-intercept of $y = \log_2(x - 3)$ is at $x =$ (enter a whole number).
Quick-fire practice · 5 problems
7
You’re here
Drill it, then lock it in
Run the quick drill and copy the summary into your book.
1
State the vertical asymptote and domain of $y = \ln(x + 1)$.
2
Find the $x$-intercept of $y = \log_3(x - 2)$.
3
Find the $x$-intercept of $y = \log_2(x - 3) + 2$.
4
Describe the transformation from $y = \ln x$ to $y = \ln(2x)$. What is the asymptote?
5
Describe the transformation for $y = -\log_3 x$ and state its range.
Match each function to its vertical asymptote.
$y = \log_2(x - 5)$
$y = \ln(x + 4)$
$y = \log_3 x$
$y = \log_5(x - 1) + 2$
$x = 1$
$x = 0$
$x = -4$
$x = 5$
12
Revisit your thinking
Earlier you predicted which line $y = 2^x$ and $y = \log_2 x$ are mirror images across. The answer is $y = x$, because logarithms and exponentials are inverse functions, every point $(a, b)$ on one becomes $(b, a)$ on the other. The vertical asymptote $x = 0$ of the log graph corresponds exactly to the horizontal asymptote $y = 0$ of the exponential graph, another manifestation of that reflection.
auto-saved
1
You’re here
Multiple choice
Answer the drill bank and rate your confidence.
01
Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
2
You’re here
Short answer
Write full responses, then check them against the model answers.
02
Short answer
ApplyBand 43 marks
Q1. Sketch $y = \log_2(x + 1) - 1$, labelling the vertical asymptote, $x$-intercept, and one other point. State the domain. (3 marks)
auto-saved
ApplyBand 43 marks
Q2. A logarithmic function has vertical asymptote $x = 1$ and passes through $(2, 0)$ and $(5, 2)$. Find its equation. (3 marks)
auto-saved
AnalyseBand 54 marks
Q3. The curve $y = e^{x-1}$ has inverse $y = \ln x + 1$. Verify that $(1, 1)$ lies on both curves and explain why the curves are reflections in $y = x$. (4 marks)
Drill 4: $\ln(2x)=\ln 2 + \ln x$, so it shifts $\ln x$ left by $\ln 2$ (horizontal compression). Asymptote still $x=0$.
Drill 5: Reflection in $x$-axis. Range is all reals (unchanged).
Q1 (3 marks): Asymptote $x = -1$ [0.5] · $x$-intercept: $\log_2(x+1)=1 \Rightarrow x+1=2 \Rightarrow x=1$, so $(1,0)$ [1] · Point at $x=0$: $y=\log_2 1 -1=-1$, point $(0,-1)$ [0.5] · Correct increasing shape [0.5] · Domain $x>-1$ [0.5]
Q2 (3 marks): Form $y=\log_a(x-1)+k$ [0.5] · At $(2,0)$: $k=0$ since $\log_a 1=0$ [0.5] · At $(5,2)$: $2=\log_a 4$, so $a^2=4$, $a=2$ [1.5] · $y=\log_2(x-1)$ [0.5]
Q3 (4 marks): At $x=1$: $y=e^0=1$ ✓ [1] · At $x=1$: $y=\ln 1+1=1$ ✓ [1] · $(1,1)$ lies on $y=x$, so it is invariant under reflection [1] · Exponential and logarithm with same base are inverses; their graphs are reflections in $y=x$ [1]
1
You’re here
Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
01
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
