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Module 4 · L5 of 15 ~35 min ⚡ +95 XP available

Introduction to Logarithmic Functions

Logarithms answer the question “what power do I need?” Like a ruler that measures multiplicative steps instead of additive ones, logarithms turn products into sums and powers into products — and they are the key that unlocks every exponential equation.

Today's hook — If $2^x = 100$, you can't isolate $x$ by dividing or square-rooting. You need a new operation — one that undoes exponentiation. That operation is the logarithm, and by the end of this lesson you'll be able to evaluate it, find domains, and prove the inverse relationship.

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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

+5 XP warm-up

If $2^x = 100$, between which two whole numbers must $x$ lie? Lock in a guess before you read further — write the two numbers and your reasoning.

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The two moves

+5 XP to read

There is really only one move in this entire lesson — converting between logarithmic and exponential form. Lock $\log_a x = y \iff a^y = x$ into muscle memory and every evaluation becomes routine.

The logarithm and the exponent are two sides of the same coin: $\log_a x = y$ says “the exponent is $y$”, while $a^y = x$ says “raising $a$ to that exponent gives $x$”. They are logically identical.

$\log_a x = y \iff a^y = x$

Special values

$\log_a 1 = 0$ and $\log_a a = 1$ for any valid base $a$. These are guaranteed by $a^0=1$ and $a^1=a$.

Domain constraint

The argument of a logarithm must be strictly positive: $x > 0$. You cannot take $\log$ of zero or a negative.

Inverse identity

$\log_a(a^x) = x$ and $a^{\log_a x} = x$. Logarithms and exponentials perfectly undo each other.

What you'll master

Know

Key facts

The definition: $y = \log_a x$ means $a^y = x$
That $a > 0$, $a \neq 1$, and $x > 0$
Special cases: $\log_a 1 = 0$, $\log_a a = 1$

Understand

Concepts

Why logarithms and exponentials are inverse functions
Why the domain is restricted to $x > 0$
How common ($\log_{10}$) and natural ($\ln$) logs are special cases

Can do

Skills

Evaluate logarithmic expressions by converting to exponential form
Solve equations of the form $\log_a x = c$
Find the domain of $f(x) = \log_a(g(x))$

Key terms

LogarithmIf $a^y = x$ then $y = \log_a x$. The logarithm is the exponent needed to produce $x$ from base $a$.

Common logarithmLogarithm base 10: $\log_{10} x$, often written $\log x$.

Natural logarithmLogarithm base $e$: $\ln x = \log_e x$, where $e \approx 2.718$.

Inverse functionA function that reverses another; $f^{-1}(f(x)) = x$. Exponential and logarithmic functions are inverses of each other.

DomainFor $y = \log_a x$, the domain is $x > 0$; the argument must be positive.

RangeFor $y = \log_a x$, the range is all real numbers $(-\infty, \infty)$.

What is a logarithm?

core concept

A logarithm answers the question: "to what power must I raise the base to get this number?" The equation $\log_a x = y$ is completely equivalent to $a^y = x$. Switching between these two forms is the fundamental skill of this lesson.

$$\log_a x = y \iff a^y = x \qquad (a > 0,\ a \neq 1,\ x > 0)$$

Because $a^y > 0$ for all real $y$ when $a > 0$, the output of an exponential is always positive. This means when we invert to get $x = a^y$, the input $x$ must be positive — explaining why $\log_a 0$ and $\log_a(\text{negative})$ are undefined.

The inverse relationship in full. If $f(x) = a^x$ and $g(x) = \log_a x$, then $f(g(x)) = a^{\log_a x} = x$ and $g(f(x)) = \log_a(a^x) = x$. Both compositions give back $x$, confirming they are true inverses. Their graphs are reflections of each other across the line $y = x$.

Definition: $\log_a x = y \iff a^y = x$, where $a > 0$, $a \neq 1$, $x > 0$; Special values: $\log_a 1 = 0$ and $\log_a a = 1$ for any valid base

Pause — copy the logarithm definition ($\log_a x = y \iff a^y = x$) and the two special values ($\log_a 1 = 0$ and $\log_a a = 1$) into your book.

Did you get this? True or false: the expression $\log_3 9 = 2$ is equivalent to $3^2 = 9$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · EVALUATE

Evaluate $\log_2 8$ and $\log_3 \dfrac{1}{9}$.

$\log_2 8$: find $y$ such that $2^y = 8$. Since $8 = 2^3$, we have $\log_2 8 = 3$.

$2^3 = 8$, so the exponent needed is 3.

PROBLEM 2 · SOLVE

Solve $\log_5 x = 3$.

$\log_5 x = 3$ means $5^3 = x$

Convert from logarithmic form to exponential form.

PROBLEM 3 · DOMAIN

Find the domain of $f(x) = \log_2(x - 3)$.

For $\log_2(x - 3)$ to be defined, the argument must be positive: $x - 3 > 0$

The logarithm is only defined for positive inputs.

Quick check: Which value equals $\log_4 64$?

Common errors · the 3 traps that cost marks

Trap 01

Trying to evaluate $\log_a 0$ or $\log_a(\text{negative})$

$a^y$ is never zero or negative for real $y$ when $a > 0$. Therefore $\log_a x$ is only defined for $x > 0$. Writing $\log_2(-4)$ or $\log_3 0$ will always earn zero marks.

Trap 02

Confusing $\log_a(x + y)$ with $\log_a x + \log_a y$

$\log_a(x + y) \neq \log_a x + \log_a y$. The log of a sum is not the sum of logs. Only $\log_a(xy) = \log_a x + \log_a y$ is valid (log laws — next lesson).

Trap 03

Forgetting that $\log_a a^x = x$ for all real $x$

This identity is extremely useful for simplifying. $\log_2 2^5 = 5$ and $\ln e^3 = 3$. Students sometimes leave $\log_2 2^5$ unevaluated — always simplify using the inverse identity.

Fill in the blank: The domain of $f(x) = \ln(2x + 6)$ is $x >$ (enter as a number, e.g. -3).

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Evaluate $\log_2 16$ and $\log_{10} 1000$.

Evaluate $\log_3 \dfrac{1}{27}$.

Solve $\log_4 x = 2$.

Find the domain of $f(x) = \ln(2x + 1)$.

Simplify $\log_5 5^7$ and $e^{\ln 4}$.

Match each logarithm to its value. Drag or click to pair them.

$\log_2 32$
$\log_5 1$
$\log_3 \frac{1}{3}$
$\ln e^7$

$7$
$-1$
$0$
$5$

Revisit your thinking

Earlier you bracketed $x$ in $2^x = 100$. Since $2^6 = 64$ and $2^7 = 128$, the answer lives between 6 and 7. That value is exactly $\log_2 100 \approx 6.64$ — the logarithm answers the question “what power of the base gives this number?” Now that you have the definition and the conversion technique, you can place any such value precisely.

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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.

Short answer

ApplyBand 43 marks

Q1. Evaluate $\log_3 81$, $\log_2 \dfrac{1}{8}$, and $\ln e^4$. Show working for each. (3 marks)

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ApplyBand 43 marks

Q2. Solve $\log_2(x + 1) = 4$ and state the domain of $f(x) = \log_2(x + 1)$. (3 marks)

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AnalyseBand 53 marks

Q3. Show that $f(x) = 2^x$ and $g(x) = \log_2 x$ are inverse functions by verifying $f(g(x)) = x$ and $g(f(x)) = x$. (3 marks)

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Comprehensive answers (click to reveal)

Drill 1: $\log_2 16 = 4$ (since $2^4=16$) · $\log_{10} 1000 = 3$ (since $10^3=1000$)

Drill 2: $\log_3 \frac{1}{27} = -3$ (since $3^{-3} = \frac{1}{27}$)

Drill 3: $x = 4^2 = 16$

Drill 4: $2x+1>0 \Rightarrow x > -\frac{1}{2}$, domain $(-\frac{1}{2}, \infty)$

Drill 5: $\log_5 5^7 = 7$ · $e^{\ln 4} = 4$

Q1 (3 marks): $\log_3 81 = 4$ since $3^4=81$ [1] · $\log_2 \frac{1}{8} = -3$ since $2^{-3}=\frac{1}{8}$ [1] · $\ln e^4 = 4$ [1]

Q2 (3 marks): $x+1=2^4=16$, so $x=15$ [1.5] · Domain: $x+1>0 \Rightarrow x>-1$, i.e. $(-1,\infty)$ [1.5]

Q3 (3 marks): $f(g(x)) = 2^{\log_2 x} = x$ [1.5] · $g(f(x)) = \log_2(2^x) = x$ [1.5]

Boss battle · The Logarithm Lord

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

Enter the arena

Science Jump · platform challenge

Climb platforms by answering logarithm questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 4 Lesson 6 · Graphs of Logarithmic Functions →

Module overview · Maths Advanced