Year 11 Methods Unit 1 — Graphs of Logarithmic Functions

Cambridge Methods 1&2, §13G  •  Wed 03 June 2026
📚 Also for this topic: 📄 Printable worksheet ✅ Solutions (answer key) ← Back: §13F Logs to solve exponentials

Today's lesson

$y=a^x$ and $y=\log_a(x)$ are inverse functions — they undo each other. Today we use that idea to graph logarithms. By the end of the lesson you should be able to:

Learning intentions

Part 1 — The inverse relationship

Key idea. Since $a^x=y \Leftrightarrow \log_a(y)=x$, swapping $x$ and $y$ in $y=a^x$ gives $x=a^y$ — i.e. $y=\log_a(x)$. So: $$ y=a^x \quad \text{and} \quad y=\log_a(x) \quad \text{are reflections of each other in } y=x. $$

📺 Walkthrough: $y=2^x$ and $y=\log_2(x)$ as reflections in $y=x$ — watch $(0,1)$ become $(1,0)$ and $(1,2)$ become $(2,1)$.

$y=a^x$  (basic, $a>1$)
Domain: $\mathbb{R}$
Range: $\mathbb{R}^+$ (i.e. $y>0$)
$y$-intercept: $(0,1)$
Asymptote: $y=0$ ($x$-axis)
Always increasing
$y=\log_a(x)$  ($a>1$)
Domain: $\mathbb{R}^+$ (i.e. $x>0$)
Range: $\mathbb{R}$
$x$-intercept: $(1,0)$
Asymptote: $x=0$ ($y$-axis)
Always increasing
EXAMPLE 1 — inverse of $y=2^x$
Find the inverse function of $y=2^x$.
  1. Swap $x$ and $y$: $x=2^y$.
  2. Rewrite in log form: $y=\log_2(x)$.
  3. So $f^{-1}(x)=\log_2(x)$, with domain $x>0$ and range $\mathbb{R}$.

Part 2 — Translations

$y=\log_a(x-h)+k$ is the basic log curve translated $h$ right and $k$ up. Three things to find for any sketch:

The "three things":
  1. Domain — the argument of the log must be $>0$. So $x-h>0\Rightarrow x>h$.
  2. Asymptote — vertical line $x=h$ (where the argument hits $0$).
  3. $x$-intercept — set $y=0$ and solve.
EXAMPLE 2 — basic $y=\log_2(x)$
Sketch $y=\log_2(x)$. Give the maximal domain, range, asymptote and any axis intercepts.
  1. Domain: $x>0$. Range: $\mathbb{R}$.
  2. Asymptote: $x=0$ (the $y$-axis).
  3. $x$-intercept: $\log_2(x)=0\Rightarrow x=1$, so $(1,0)$.
  4. Shape: increasing, passes through $(1,0)$ and $(2,1)$, very steep near $x=0^+$.
EXAMPLE 3 — translated right: $y=\log_2(x-2)$
Sketch $y=\log_2(x-2)$. State domain, asymptote, $x$-intercept.
  1. Domain: $x-2>0\Rightarrow x>2$.
  2. Asymptote: $x=2$.
  3. $x$-intercept: $\log_2(x-2)=0\Rightarrow x-2=1\Rightarrow x=3$, so $(3,0)$.
  4. Same shape as $y=\log_2(x)$, just shifted $2$ to the right.
EXAMPLE 4 — translated left: $y=\log_2(x+5)$
Sketch $y=\log_2(x+5)$. State domain, asymptote, $x$-intercept and $y$-intercept.
  1. Domain: $x+5>0\Rightarrow x>-5$.
  2. Asymptote: $x=-5$.
  3. $x$-intercept: $x+5=1\Rightarrow x=-4$.
  4. $y$-intercept: at $x=0$, $y=\log_2(5)\approx 2.32$.

Part 3 — Reflections

$y=\log_a(-x)$  — reflect in $y$-axis
Domain: $x<0$
Range: $\mathbb{R}$
Asymptote: $x=0$
$x$-intercept: $(-1,0)$
$y=-\log_a(x)$  — reflect in $x$-axis
Domain: $x>0$
Range: $\mathbb{R}$
Asymptote: $x=0$
$x$-intercept: $(1,0)$
Now decreasing
EXAMPLE 5 — show $\log_{1/2}(x)=-\log_2(x)$
Show that $\log_{1/2}(x)=-\log_2(x)$.
  1. Use change of base on $\log_{1/2}(x)$ with base $2$: $\log_{1/2}(x)=\dfrac{\log_2(x)}{\log_2(1/2)}=\dfrac{\log_2(x)}{-1}=-\log_2(x)$. ✓
  2. Graphical interpretation: $y=\log_a(x)$ for $0<a<1$ is the reflection of $y=\log_a(x)$ for $a>1$ in the $x$-axis. Same domain $x>0$, but now decreasing.

Part 4 — Putting it together: $y=\log_2(3x-2)$

EXAMPLE 6 — combined transformation
Sketch $y=\log_2(3x-2)$, labelling all key features.
  1. Domain. $3x-2>0\Rightarrow x>\tfrac{2}{3}$.
  2. Asymptote. $3x-2=0$ at $x=\tfrac{2}{3}$.
  3. $x$-intercept. $\log_2(3x-2)=0\Rightarrow 3x-2=1\Rightarrow x=1$, so $(1,0)$.
  4. A second helpful point. When $3x-2=2$, $y=1$. Solve $3x=4 \Rightarrow x=\tfrac{4}{3}$, so $(\tfrac{4}{3},1)$.
  5. Range: $\mathbb{R}$. Increasing, asymptote $x=\tfrac{2}{3}$ on the left.
EXAMPLE 7 — reflection + translation
Sketch $y=\log_{1/2}(x+1)-2$, labelling all key features.
  1. Rewrite using $\log_{1/2}(u)=-\log_2(u)$:   $y=-\log_2(x+1)-2$.
  2. Domain. $x+1>0\Rightarrow x>-1$.
  3. Asymptote. $x=-1$.
  4. $x$-intercept. $-\log_2(x+1)-2=0\Rightarrow \log_2(x+1)=-2\Rightarrow x+1=\tfrac{1}{4}\Rightarrow x=-\tfrac{3}{4}$, so $(-\tfrac{3}{4},0)$.
  5. $y$-intercept. At $x=0$, $y=-\log_2(1)-2=-2$.
  6. Now decreasing (because of the $-\log$). Range: $\mathbb{R}$.

Part 5 — Finding the inverse of a function

Recipe. To find $f^{-1}$:
  1. Write $y=f(x)$.
  2. Swap $x$ and $y$.
  3. Solve for $y$ — that's $f^{-1}(x)$.
  4. The domain of $f^{-1}$ equals the range of $f$, and vice versa.
EXAMPLE 8 — inverse of $f(x)=2^x-5$
Find the inverse of $f:\mathbb{R}\to\mathbb{R}$, $f(x)=2^x-5$. State the domain and range of $f^{-1}$.
  1. $y=2^x-5$. Swap: $x=2^y-5\Rightarrow 2^y=x+5\Rightarrow y=\log_2(x+5)$.
  2. Range of $f$: $2^x>0$ so $2^x-5>-5$, i.e. $(-5,\infty)$. So domain of $f^{-1}$ is $(-5,\infty)$.
  3. Domain of $f$ is $\mathbb{R}$, so range of $f^{-1}$ is $\mathbb{R}$.
  4. $\boxed{f^{-1}(x)=\log_2(x+5),\;\;x\in(-5,\infty)}$.
EXAMPLE 9 — inverse of $f(x)=2\times 5^x+2$
Find the inverse of $f:\mathbb{R}\to\mathbb{R}$, $f(x)=2\cdot 5^x+2$. State the domain and range of $f^{-1}$.
  1. $y=2\cdot 5^x+2$. Swap: $x=2\cdot 5^y+2\Rightarrow 5^y=\dfrac{x-2}{2}\Rightarrow y=\log_5\!\left(\dfrac{x-2}{2}\right)$.
  2. Range of $f$: $2\cdot 5^x>0$, so $f(x)>2$. Domain of $f^{-1}$ is $(2,\infty)$.
  3. $\boxed{f^{-1}(x)=\log_5\!\left(\dfrac{x-2}{2}\right),\;\;x\in(2,\infty)}$, range $\mathbb{R}$.
EXAMPLE 10 — inverse of $f(x)=\log_2(x+3)-2$
Find the inverse of $f:(-3,\infty)\to\mathbb{R}$, $f(x)=\log_2(x+3)-2$. State the domain and range of $f^{-1}$.
  1. $y=\log_2(x+3)-2$. Swap: $x=\log_2(y+3)-2\Rightarrow x+2=\log_2(y+3)\Rightarrow y+3=2^{x+2}\Rightarrow y=2^{x+2}-3$.
  2. Range of $f$: $\mathbb{R}$ (since the original log can take any real). So domain of $f^{-1}$ is $\mathbb{R}$.
  3. Domain of $f$ is $(-3,\infty)$, so range of $f^{-1}$ is $(-3,\infty)$.
  4. $\boxed{f^{-1}(x)=2^{x+2}-3,\;\;x\in\mathbb{R}}$, range $(-3,\infty)$.

Practice 5.1 — find the inverse and state its domain and range.

  1. $f(x)=3^x$, $f:\mathbb{R}\to\mathbb{R}$
  2. $f(x)=10^x+1$, $f:\mathbb{R}\to\mathbb{R}$
  3. $f(x)=\log_3(x)$, $f:\mathbb{R}^+\to\mathbb{R}$
a) $f^{-1}(x)=\log_3(x)$, domain $\mathbb{R}^+$, range $\mathbb{R}$.
b) Range of $f$: $f(x)>1$. $y=10^x+1\Rightarrow x=\log_{10}(y-1)$. So $f^{-1}(x)=\log_{10}(x-1)$, domain $(1,\infty)$, range $\mathbb{R}$.
c) $f^{-1}(x)=3^x$, domain $\mathbb{R}$, range $\mathbb{R}^+$ (inverse of part a).

Part 6 — Quick quiz (5 min)

Pick the correct answer for each, then click Mark.

Q1. The domain of $y=\log_2(x-2)$ is

Q2. The vertical asymptote of $y=\log_2(3x-2)$ is

Q3. The graph $y=\log_a(x)$ is the reflection of $y=a^x$ in

Q4. The inverse of $f(x)=2^x-5$ (with $f:\mathbb{R}\to\mathbb{R}$) is

Q5. $\log_{1/2}(x)$ is the same as

Working program — Cambridge Ex 13G

After the quiz, open Cambridge Chapter 13 and work through Exercise 13G:

QuestionFoundationStandard (set work)Advanced
Q1 — sketch $y=\log_a(x)$ and translations1 (a,c)1*1*
Q2 — reflections2 (a)2*2*
Q3 — combined transformations—3*3*
Q4 — inverse of an exponential4 (a)4*4*
Q5 — inverse of a log—5*5*

After 13G: applications of exponentials and logs — modelling questions, half-life, population growth.

Exit ticket — write in your book

Before you pack up, in your exercise book write one sentence each:
  1. What is the asymptote of $y=\log_2(x+5)$?
  2. If $f(x)=2^x$, what is $f^{-1}(x)$, and what is the domain of $f^{-1}$?
  3. How is the graph of $y=\log_{1/2}(x)$ related to $y=\log_2(x)$?

— Mr Wong