Logarithms — notes, question types & practice · Functions, Graphs & Logarithms

01

A log asks: what power?

A logarithm answers one question: this base, raised to what power, gives that number?

$2$ to the what gives $32$ ? Count the doublings: $2, 4, 8, 16, 32$ — five. So $\log_2 32 = 5$ .

The three players: $2$ is the base, $32$ is the argument, $5$ is the answer.

Read $\log_5 125$ as " $5$ to the what is $125$ ?" The answer is $3$ .

02

The definition is the master key

\log_a b = x \quad \Leftrightarrow \quad a^x = b

Any log question can be rewritten as a power question. Three small ones:

$\log_3 81$ : $3^4 = 81$ , so the answer is $4$ .
$\log_5 \dfrac{1}{25}$ : $5^{-2} = \dfrac{1}{25}$ , so the answer is $-2$ .
$\log_7 1$ : any power $0$ gives $1$ , so the answer is $0$ .

Conditions: the base is positive and not $1$ ; the argument must be positive. Logs of zero or negatives do not exist.

Rule: Stuck on any log — convert it to " $a^x = b$ " and match the powers.

03

Three laws, one idea

Logs turn multiplication into addition, division into subtraction, and powers into multiples.

$\log_a(mn) = \log_a m + \log_a n$ — a product becomes a sum.
$\log_a \dfrac{m}{n} = \log_a m - \log_a n$ — a quotient becomes a difference.
$\log_a m^p = p \log_a m$ — the power pops out front.

Worked: $\log_{10} 4 + \log_{10} 25 = \log_{10}(4 \times 25) = \log_{10} 100 = 2$ .

And $\log_{10} 8 = \log_{10} 2^3 = 3\log_{10} 2$ . With $\log_{10} 2 = 0.3010$ , this is $0.9030$ .

Watch: The laws work on products and quotients, never on sums. $\log(m + n)$ is not $\log m + \log n$ .

04

Change of base

To evaluate a log in an awkward base, divide two logs in a friendly base:

\log_a b = \frac{\log_{10} b}{\log_{10} a}

Worked: $\log_2 5 = \dfrac{\log 5}{\log 2} = \dfrac{0.6990}{0.3010} \approx 2.32$ .

The same idea links related bases. Since $4 = 2^2$ :

\log_4 x = \frac{1}{2}\log_2 x

So $\log_2 x + \log_4 x = \dfrac{3}{2}\log_2 x$ . If that equals $3$ , then $\log_2 x = 2$ and $x = 4$ .

One identity more: $a^{\log_a b} = b$ . Base and log cancel, leaving the argument: $5^{\log_5 9} = 9$ .

05

Log equations and the domain check

Same base on both sides: equate the powers.

$\log_2(x - 1) = 5$ means $x - 1 = 2^5 = 32$ , so $x = 33$ .

Products need one extra step. Solve $\log_{10} x + \log_{10}(x - 3) = 1$ :

Combine: $\log_{10}\big(x(x-3)\big) = 1$ .
So $x(x - 3) = 10$ , giving $x^2 - 3x - 10 = 0$ .
Roots: $x = 5$ and $x = -2$ .
Domain check: $x = -2$ makes $\log_{10}(-2)$ impossible. Reject it.

Answer: $x = 5$ only. This rejection is exactly where exam setters plant a trap option.

Watch: After solving a log equation, test every root in the original question. Arguments must stay positive.

06

Inequalities: the base rule

Base greater than $1$ : the log climbs, so the inequality sign stays.

$\log_2 x < 4$ gives $0 < x < 16$ — below by the power, and $x$ must stay positive.

Base between $0$ and $1$ : the log falls as $x$ grows, so the sign flips.

$\log_{1/2} x > 3$ : compare with the power $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$ . Bigger log now means smaller argument: $0 < x < \dfrac{1}{8}$ .

Rule: A small base ( $0 < a < 1$ ) reverses the inequality. Check with one number: $\log_{1/2} \dfrac{1}{16} = 4 > 3$ , and $\dfrac{1}{16} < \dfrac{1}{8}$ . Consistent.

07

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Evaluate a basic log

How to spot it:

A direct value like $\log_5 125$ or $\log_2 64$ is asked.

\log_a b = x \Leftrightarrow a^x = b

Method

Name the base and the argument.
Ask: base to the what gives the argument?
Write the argument as a power of the base.
Read off the exponent.

Why it works:

The definition converts every log into a power match.

Try this

Evaluate $\log_5 125$ .

Show solution

$125 = 5^3$ .
Exponent is $3$ .

Answer

3

Type 2common2 practice Q

Log of a fraction or a different base

How to spot it:

Arguments like $\dfrac{1}{32}$ or bases like $9$ with argument $27$ — the powers look mismatched.

a^{-p} = \frac{1}{a^p}, \quad (a^m)^n = a^{mn}

Method

Write base and argument as powers of one common number.
Use a negative power for fractions.
Equate the exponents.
Solve the small linear equation.

Why it works:

One common base turns any log into a simple exponent equation.

Try this

Evaluate $\log_2 \dfrac{1}{32}$ .

Show solution

$\dfrac{1}{32} = 2^{-5}$ .
Exponent is $-5$ .

Answer

-5

Type 3very common3 practice Q

Combine or expand with the laws

How to spot it:

A sum or difference of logs, or $\log$ of a big product, or a value like $\log 8$ built from $\log 2$ .

\log_a(mn) = \log_a m + \log_a n, \quad \log_a m^p = p\log_a m

Method

Spot products, quotients or powers inside logs.
Combine into one log, or split a big one.
Pop exponents out front.
Use given values or easy numbers like $\log 100 = 2$ .

Why it works:

The laws convert one messy log into known small ones.

Try this

Find $\log_{10} 4 + \log_{10} 25$ .

Show solution

Combine: $\log_{10}(4 \times 25) = \log_{10} 100$ .
$100 = 10^2$ .

Answer

2

Type 4common2 practice Q

Change of base

How to spot it:

Two different bases in one question, or logs given in base 10 while the question uses base 2 or 4.

\log_a b = \frac{\log b}{\log a}

Method

Write the awkward log as a quotient of base-10 logs.
Substitute the given values.
Divide.
For related bases, use $\log_{a^k} x = \frac{1}{k}\log_a x$ .

Why it works:

Change of base expresses every log through one common base, so given values apply.

Try this

Given $\log 2 = 0.3010$ and $\log 5 = 0.6990$ , find $\log_2 5$ .

Show solution

$\log_2 5 = \dfrac{\log 5}{\log 2}$ .
$= \dfrac{0.6990}{0.3010} \approx 2.32$ .

Answer

≈ 2.32

Type 5very common2 practice Q

Solve a log equation

How to spot it:

An equation with $\log$ terms equals a number; $x$ sits inside the log or its argument.

\log_a(f(x)) = c \Rightarrow f(x) = a^c

Method

Combine logs into a single log if there are several.
Convert to the power form.
Solve the resulting equation.
Reject roots that break the domain.

Why it works:

The definition converts the whole equation into ordinary algebra, but the domain still rules.

Try this

Solve $\log_2(x - 1) = 5$ .

Show solution

$x - 1 = 2^5 = 32$ .
$x = 33$ , and $33 - 1 > 0$ : valid.

Answer

x = 33

Type 6common2 practice Q

Log inequality with base care

How to spot it:

'Solve $\log_a x > c$ ' or '<' — check the base before answering.

0 < a < 1 \text{ flips the sign}

Method

Look at the base first.
Base $> 1$ : keep the direction; add $x > 0$ .
Base between 0 and 1: flip the direction.
Write the answer as one interval.

Why it works:

A base below 1 is a falling function, so bigger logs come from smaller arguments.

Try this

Solve $\log_{1/2} x > 3$ .

Show solution

Base $\dfrac{1}{2}$ is below 1, so flip.
$\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$ .
Answer: $0 < x < \dfrac{1}{8}$ .

Answer

0 < x < 1/8

Type 7occasional2 practice Q

Exponent-log identity

How to spot it:

An expression like $5^{\log_5 9}$ or $\log_5 5 + 3^{\log_3 4}$ mixing powers and logs.

a^{\log_a b} = b

Method

Match each power's base with the log's base below it.
Cancel the pair: the argument remains.
Add or multiply the leftovers.
State the final number.

Why it works:

Raising a base to its own log undoes the log, leaving the argument untouched.

Try this

Evaluate $5^{\log_5 9}$ .

Show solution

Base $5$ matches log base $5$ .
The pair cancels, leaving $9$ .

Answer

9

08

Formula sheet

Definition

\log_a b = x \Leftrightarrow a^x = b

Base $a > 0$, $a \ne 1$; argument $b > 0$.

Product / quotient laws

\log_a(mn) = \log_a m + \log_a n, \quad \log_a\frac{m}{n} = \log_a m - \log_a n

Multiplication becomes addition; division becomes subtraction.

Power law

\log_a m^p = p \log_a m

The exponent comes out front.

Change of base

\log_a b = \frac{\log b}{\log a}

Any common base works; base 10 is standard.

Special values

\log_a 1 = 0, \quad \log_a a = 1, \quad a^{\log_a b} = b

Log of 1 is 0; log of the base is 1; base and log cancel.

09

Shortcuts that save time

⚡ Translate to power language

Read every log aloud as 'base to the what gives this?'. Then match powers of the base. No formula needed for direct values.

Example

Evaluate $\log_2 64$ .

Show solution

$2$ to the what is $64$ ?
$64 = 2^6$ .

Answer

6

⚡ Fraction or root arguments: use negative and fractional powers

$1/8$ is $2^{-3}$ , $27$ is $3^3$ , $\sqrt{9}$ is $9^{1/2}$ . Write both sides with the same base and the answer falls out.

Example

Evaluate $\log_9 27$ .

Show solution

$9 = 3^2$ , $27 = 3^3$ .
$3^{2x} = 3^3$ , so $2x = 3$ .

Answer

$\dfrac{3}{2}$

⚡ Split products, pop powers

Seeing a product inside a log, split it. Seeing a power, bring it out front. Known values such as $\log 2 = 0.3010$ then finish the job.

Example

Given $\log 2 = 0.3010$ , find $\log 8$ .

Show solution

$8 = 2^3$ .
$\log 8 = 3 \times 0.3010$ .

Answer

0.9030

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Reading $\log_2 8$ as $8 \div 2$ — a log asks for the power: $2^3 = 8$ , so the answer is $3$ .

Mistake 02

Splitting $\log_a(m + n)$ into $\log_a m + \log_a n$ — the laws work on products and quotients, never sums.

Mistake 03

Forgetting the base condition in inequalities — with base $\dfrac{1}{2}$ the sign flips.

Mistake 04

Ignoring the domain: $\log(x - 2)$ needs $x > 2$ , so a root making the argument zero or negative must be rejected.

Mistake 05

Writing change of base upside down — $\log_a b = \dfrac{\log b}{\log a}$ , both logs of the same side stay together.

Mistake 06

Assuming $\log 100 = 2$ in every base — that is a base-10 fact; $\log_2 100$ is about $6.64$ .

11

Quick revision

Read this the night before the exam.

$\log_a b = x$ means $a^x = b$ .
$\log_a 1 = 0$ , $\log_a a = 1$ , $a^{\log_a b} = b$ .
Products add, quotients subtract, powers multiply: $\log m^p = p\log m$ .
$\log_a b = \dfrac{\log b}{\log a}$ .
Reject roots with non-positive arguments.
Base below 1 flips the inequality; always keep $x > 0$ .

12

Practice: 15 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.