Graphs of Logarithmic Functions
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.
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.
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.
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 $y = x$
Concepts
- Why the vertical asymptote shifts when $h \neq 0$
- Why the domain changes for shifted logarithms
- How the inverse relationship manifests graphically
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
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 $y = x$.
Key features: vertical asymptote $x = 0$, $x$-intercept $(1, 0)$, passes through $(2, 1)$ and $(\frac{1}{2}, -1)$. Domain: $x > 0$; Range: $\mathbb{R}$.
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
Sketch $y = \ln x$, labelling all key features.
Sketch $y = \log_2(x - 1) + 1$ and state the domain.
A logarithmic curve has vertical asymptote $x = 2$, passes through $(3, 0)$ and $(5, 1)$. Find its equation.
Quick check: What is the domain of $y = \log_5(x + 3)$?
Common errors · the 3 traps that cost marks
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
State the vertical asymptote and domain of $y = \ln(x + 1)$.
Find the $x$-intercept of $y = \log_3(x - 2)$.
Find the $x$-intercept of $y = \log_2(x - 3) + 2$.
Describe the transformation from $y = \ln x$ to $y = \ln(2x)$. What is the asymptote?
Describe the transformation for $y = -\log_3 x$ and state its range.
Match each function to its vertical asymptote.
- $y = \log_2(x - 5)$
- $y = \ln(x + 4)$
- $y = \log_3 x$
- $y = \log_5(x - 1) + 2$
- $x = 1$
- $x = 0$
- $x = -4$
- $x = 5$
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Sketch $y = \log_2(x + 1) - 1$, labelling the vertical asymptote, $x$-intercept, and one other point. State the domain. (3 marks)
Q2. A logarithmic function has vertical asymptote $x = 1$ and passes through $(2, 0)$ and $(5, 2)$. Find its equation. (3 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)
Comprehensive answers (click to reveal)
Drill 1: Asymptote $x = -1$, domain $x > -1$
Drill 2: $\log_3(x-2)=0 \Rightarrow x-2=1 \Rightarrow x=3$. Intercept $(3,0)$.
Drill 3: $\log_2(x-3)+2=0 \Rightarrow \log_2(x-3)=-2 \Rightarrow x-3=2^{-2}=\frac{1}{4} \Rightarrow x=3.25$. Intercept $(3.25,0)$.
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]
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
Enter the arenaClimb platforms by answering logarithmic graph questions. Lighter alternative to the boss.
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