Logarithmic functions and their graphs (QCE Math Methods Unit 2)

Syllabus dot point

“Sketch and analyse logarithmic functions of the form $y = \log_b(x - h) + k$ , state their domain, range and asymptote, relate them to exponential functions as inverses, and solve logarithmic equations”

7 min readUpdated 7 Oct 2026

Quick answer

The function $y=\log_b x$ is the inverse of $y=b^x$ , so its graph is the reflection of the exponential graph in $y=x$ . For $b>1$ it passes through $(1,0)$ and $(b,1)$ , has the vertical asymptote $x=0$ , domain $x>0$ and range all real numbers. For $y=\log_b(x-h)+k$ the asymptote moves to $x=h$ and the domain becomes $x>h$ .

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What this dot point is asking
The basic graph
Transformations
Inverse relationship
Solving logarithmic equations
Practice questions

What this dot point is asking

In the 2025 QCAA Mathematical Methods syllabus, Unit 2 Topic 2 is titled Logarithms and logarithmic functions. The laws of logarithms are covered on the logarithms and log laws page. This page covers the functions and their graphs, and how to solve logarithmic equations. The official syllabus is at qcaa.qld.edu.au.

The basic graph

For $b>1$ , the graph of $y=\log_b x$ :

passes through $(1,0)$ , because $b^0=1$
passes through $(b,1)$ , because $b^1=b$
has the vertical asymptote $x=0$ and increases slowly as $x$ grows
has domain $x>0$ and range $\mathbb{R}$

It is the reflection of $y=b^x$ in the line $y=x$ . The exponential has the horizontal asymptote $y=0$ and the logarithm has the vertical asymptote $x=0$ .

Transformations

For $y=\log_b(x-h)+k$ :

the vertical asymptote is $x=h$ and the domain is $x>h$
the point $(1,0)$ moves to $(h+1,k)$
the point $(b,1)$ moves to $(h+b,k+1)$

Sketch

y=\log_2(x-3)+1

The vertical asymptote is $x=3$ and the domain is $x>3$ .

The key points come from shifting $(1,0)$ and $(2,1)$ by $3$ right and $1$ up: $(4,1)$ and $(5,2)$ . Check: $\log_2(5-3)+1=\log_2 2+1=2$ .

The $x$ -intercept needs $\log_2(x-3)=-1$ , so $x-3=2^{-1}=\tfrac12$ and $x=3.5$ .

There is no $y$ -intercept, because $x=0$ is outside the domain. The range is all real numbers.

Inverse relationship

To find the inverse of $y=3^x+1$ , swap and solve: $x=3^y+1$ , so $3^y=x-1$ and $y=\log_3(x-1)$ . The inverse has the domain $x>1$ , which is the range of the original function. The asymptote $y=1$ of the exponential becomes the asymptote $x=1$ of the logarithm.

Solving logarithmic equations

Use the laws of logarithms to combine terms, convert to exponential form, solve, and then check each answer against the domain.

Solve

\log_3 x+\log_3(x-2)=1

Combine with the product law: $\log_3\big(x(x-2)\big)=1$ .

Convert: $x(x-2)=3^1=3$ , so $x^2-2x-3=0$ and $(x-3)(x+1)=0$ .

The candidates are $x=3$ and $x=-1$ . The original equation needs $x>2$ , because both $x$ and $x-2$ must be positive. Reject $x=-1$ .

The solution is $x=3$ . Check: $\log_3 3+\log_3 1=1+0=1$ .

Common trap

Always check logarithmic solutions in the original equation. Combining logs can introduce an extraneous answer, such as $x=-1$ above. Also, do not write $\log(x+y)$ as $\log x+\log y$ , because the laws apply to products and quotients and not to sums.

Practice questions

1. State the domain, range and asymptote of $y=\log_2(x+4)$ .

Model answer: The domain is $x>-4$ , the range is all real numbers and the vertical asymptote is $x=-4$ .

2. Find the inverse of $y=3^x+1$ and state its domain.

Model answer: Swapping gives $x=3^y+1$ , so $y=\log_3(x-1)$ . The domain is $x>1$ .

3. Solve $\log_3 x+\log_3(x-2)=1$ and explain why one candidate is rejected.

Model answer: The equation reduces to $x^2-2x-3=0$ , so $x=3$ or $x=-1$ . The value $x=-1$ is rejected because the logarithm of a negative number is undefined, and only $x=3$ satisfies the equation.

4. A graph of $y=\log_2(x-3)+1$ passes through $(5,2)$ . Verify this.

Model answer: Substituting $x=5$ gives $\log_2 2+1=1+1=2$ , so the point $(5,2)$ lies on the graph.

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA