Graphs of logarithmic functions (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“The logarithmic graph is the reflection of the exponential graph in the line y = x, with a vertical asymptote and a characteristic slow growth.”

8 min readUpdated 2 Jun 2026

Quick answer

The graph of $y=\log_a x$ (with $a>1$ ) passes through $(1,0)$ , rises slowly, and has a vertical asymptote at $x=0$ . Its domain is $x>0$ and its range is all real numbers. It is the reflection of $y=a^x$ in the line $y=x$ , because logarithm and exponential are inverse functions.

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What this dot point is asking
Key features of $y=\log_a x$ ( $a>1$ )
Reflection of the exponential
Transformations
Vertical stretches and reflections
Logarithmic scales in context
Common errors
Why it matters
Exam-style questions

What this dot point is asking

The logarithmic function is the inverse of the exponential function, so its graph is the exponential graph reflected in the line $y=x$ . Knowing the basic shape lets you sketch any transformed log curve. Because reflection in $y=x$ swaps the $x$ and $y$ roles, every feature of the exponential has a mirror feature on the log: the exponential's $y$ -intercept becomes the log's $x$ -intercept, and the exponential's horizontal asymptote becomes the log's vertical asymptote.

Key features of $y=\log_a x$ ( $a>1$ )

Features of the basic log graph

Domain: $x>0$ (you cannot take the log of zero or a negative number).
Range: all real numbers, $y\in\mathbb{R}$ .
Vertical asymptote: the y-axis, $x=0$ ; as $x\to 0^+$ , $y\to -\infty$ .
x-intercept: $(1,0)$ , since $\log_a 1=0$ . There is no y-intercept.
Shape: increasing and concave down - it rises forever but ever more slowly.

The point $(a,1)$ also lies on the curve, since $\log_a a=1$ - a handy second point when sketching.

Reflection of the exponential

The exponential $y=a^x$ has a horizontal asymptote ( $y=0$ ), domain all reals, and range $y>0$ . Reflecting in $y=x$ swaps these:

Feature	$y=a^x$	$y=\log_a x$
Domain	all $x$	$x>0$
Range	$y>0$	all $y$
Asymptote	$y=0$ (horizontal)	$x=0$ (vertical)
Key point	$(0,1)$	$(1,0)$

This is exactly the inverse relationship: every point $(p,q)$ on $y=a^x$ becomes $(q,p)$ on $y=\log_a x$ .

Transformations

A transformed log function combines shifts and stretches. For $y=\log_a(x-h)+k$ :

$h$ shifts the graph (and its asymptote) horizontally to $x=h$ .
$k$ shifts the graph vertically by $k$ .

Sketching a transformed log curve

Describe the key features of $y=\ln(x-2)+1$ .

Step 1: Start from the parent curve

The parent function is $y=\ln x$ , which has a vertical asymptote at $x=0$ , passes through $(1,0)$ (since $\ln 1=0$ ), and passes through $(e,1)$ . All transformations will move this baseline shape.

Step 2: Find the asymptote and domain

The vertical asymptote of $y=\ln x$ is where the argument equals zero, namely $x=0$ . In the transformed function $y=\ln(x-2)+1$ , the argument is $(x-2)$ , which equals zero when $x=2$ . So the asymptote shifts to $x=2$ , and the domain becomes $x>2$ because the log is only defined for a positive argument.

Step 3: Track the transformations on a known point

The transformation $x \to x-2$ shifts every point two units to the right, and adding $1$ shifts every point one unit up. Applying both to the key point $(1,0)$ on the parent curve:

(1,0) \to (1+2,\ 0+1) = (3,1).

So $(3,1)$ lies on the transformed curve. This gives a concrete point to mark when sketching.

Step 4: Find the $x$ -intercept

The $x$ -intercept is where $y=0$ . Set the function equal to zero and solve:

\ln(x-2)+1=0 \implies \ln(x-2)=-1 \implies x-2=e^{-1} \implies x=2+e^{-1}\approx 2.37.

We removed the logarithm by rewriting $\ln(x-2)=-1$ in exponential form as $x-2=e^{-1}$ .

Final answer: The curve has a vertical asymptote at $x=2$ , domain $x>2$ , crosses the $x$ -axis at $x=2+e^{-1}\approx 2.37$ , and passes through $(3,1)$ .

Vertical stretches and reflections

A coefficient in front of the log, as in $y=c\log_a x$ , stretches the graph vertically by factor $c$ ; if $c$ is negative the graph also reflects in the $x$ -axis, so it falls instead of rising while keeping the same asymptote and $x$ -intercept at $(1,0)$ . A coefficient inside, as in $y=\log_a(bx)$ , looks like a horizontal stretch but, by the product law, equals $\log_a x+\log_a b$ , which is just a vertical shift by $\log_a b$ . Recognising that an inside multiplier becomes an additive constant is a favourite SACE testing point because it catches students who assume every inside change is a horizontal one.

Logarithmic scales in context

The slow growth of the log graph is exactly why logarithmic scales are used to display quantities that span many orders of magnitude. The Richter scale, the decibel scale and pH are all logarithmic: a one-unit increase corresponds to a tenfold change in the underlying quantity. On such a scale, equal spacings represent equal ratios rather than equal differences, which is the graphical signature of the log function and a context SACE often frames modelling questions around.

Common errors

Forgetting the domain restriction or the asymptote shift

The log function is only defined for a positive argument, so $y=\ln(x-2)$ requires $x-2>0$ , i.e. $x>2$ - the domain shifts with the graph. A common slip is leaving the asymptote at $x=0$ after a horizontal shift; the asymptote moves to wherever the argument equals zero.

Find the asymptote from the argument

Set the argument of the log equal to zero to locate the vertical asymptote. For $y=\log_a(x-h)+k$ , solving $x-h=0$ gives the asymptote $x=h$ and tells you the domain $x>h$ in one step.

Why it matters

Recognising the log graph's shape, domain and asymptote is examined directly and underpins your interpretation of logarithmic models (such as decibel, pH and Richter scales). It also reinforces the inverse relationship with exponentials that drives the equation-solving in this topic.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style2 marks

Calculator-free. Consider the graphs of

f(x) = \ln(5x)

and

g(x) = \ln x

. Point

A

is at

(2, f(2))

and point

B

is at

(2, g(2))

. Find the exact distance between

A

and

B

Show worked answer →

$A$ and $B$ share the same $x$ -coordinate, so the distance is the vertical gap $|f(2) - g(2)|$ .

$f(2) = \ln(5 \times 2) = \ln 10$ and $g(2) = \ln 2$ . By the quotient law,

\ln 10 - \ln 2 = \ln\!\left(\frac{10}{2}\right) = \ln 5.

So the exact distance is $\ln 5$ units. Marks: one for both heights, one for combining to the exact value $\ln 5$ . Leaving it exact (not $1.61$ ) is required.

SACE 2022-style2 marks

Calculator-assumed. Comment on the relationship between the graphs of

g(x) = \ln x

and

h(x) = \ln(kx)

, where

k > 0

, and state the effect on the asymptote.

Show worked answer →

Use the product law: $h(x) = \ln(kx) = \ln k + \ln x = g(x) + \ln k$ .

So $h$ is $g$ translated vertically by $\ln k$ (upwards if $k > 1$ , downwards if $0 < k < 1$ ). The vertical asymptote stays at $x = 0$ , because adding a constant does not move a vertical asymptote.

Marks: one for identifying the vertical translation by $\ln k$ , one for noting the asymptote is unchanged. The key insight is that a horizontal scale factor inside a log becomes an additive constant, a vertical shift, not a horizontal stretch.

SACE 2021-style3 marks

Calculator-assumed. The curve

y = \log_2(x + 4)

is given. State its domain, the equation of its vertical asymptote, and its

x

-intercept.

Show worked answer →

The argument must be positive: $x + 4 > 0$ , so the domain is $x > -4$ .

The vertical asymptote is where the argument is zero: $x + 4 = 0$ , giving $x = -4$ .

The $x$ -intercept is where $y = 0$ : $\log_2(x + 4) = 0$ means $x + 4 = 2^0 = 1$ , so $x = -3$ , giving the point $(-3, 0)$ .

Marks: one for the domain $x > -4$ , one for the asymptote $x = -4$ , one for the $x$ -intercept $(-3, 0)$ .

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board

What this dot point is asking

Key features of y=log⁡axy=\log_a xy=loga​x (a>1a>1a>1)

Reflection of the exponential

Transformations

Step 1: Start from the parent curve

Step 2: Find the asymptote and domain

Step 3: Track the transformations on a known point

Step 4: Find the xxx-intercept

Vertical stretches and reflections

Logarithmic scales in context

Common errors

Why it matters

Exam-style questions

Practise this

Sources & how we know this

Key features of $y=\log_a x$ ( $a>1$ )

Step 4: Find the $x$ -intercept