The laws of logarithms (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“The logarithm laws turn products into sums, quotients into differences, and powers into multipliers, mirroring the index laws.”

8 min readUpdated 2 Jun 2026

Quick answer

A logarithm is the inverse of an exponential: $\log_a x = y$ means $a^y = x$ . The three laws are $\log_a(xy)=\log_a x+\log_a y$ , $\log_a\!\left(\tfrac{x}{y}\right)=\log_a x-\log_a y$ , and $\log_a(x^n)=n\log_a x$ . They turn multiplication into addition and powers into multipliers.

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What this dot point is asking
The three laws
The change-of-base rule
Common errors
Solving equations with the laws
Why it matters
Exam-style questions

What this dot point is asking

A logarithm answers the question "to what power must the base be raised to get this number?" The definition is:

Logarithm

\log_a x = y \quad\Longleftrightarrow\quad a^y = x, \qquad a>0,\ a\ne 1,\ x>0.

The logarithm and the exponential are inverse operations.

The natural logarithm $\ln x = \log_e x$ uses base $e\approx 2.718$ and is the version that appears throughout calculus. The common logarithm $\log x=\log_{10} x$ uses base $10$ and is what older scientific scales were built on. Whichever base you use, the three laws below have exactly the same form, so a result proved for $\ln$ holds for every base.

The three laws

Because logs are inverse to indices, each index law has a matching log law:

Laws of logarithms

\log_a(xy)=\log_a x+\log_a y \qquad\text{(product law)}

\log_a\!\left(\frac{x}{y}\right)=\log_a x-\log_a y \qquad\text{(quotient law)}

\log_a(x^n)=n\log_a x \qquad\text{(power law)}

Two useful special values: $\log_a 1=0$ (since $a^0=1$ ) and $\log_a a=1$ (since $a^1=a$ ).

Combining logs into one

Worked example

Write $2\log x + \log y - \tfrac12\log z$ as a single logarithm.

Use the power law to move coefficients into exponents:

2\log x=\log x^2, \qquad \tfrac12\log z=\log z^{1/2}=\log\sqrt{z}.

Then combine with the product and quotient laws:

\log x^2+\log y-\log\sqrt{z}=\log\!\left(\frac{x^2 y}{\sqrt{z}}\right).

Splitting a single log into parts

Worked example

Expand $\ln\!\left(\dfrac{x^3\sqrt{y}}{z^2}\right)$ into separate logarithms.

Apply the quotient law to separate numerator and denominator, then the product and power laws:

\ln\!\left(\frac{x^3\sqrt{y}}{z^2}\right)=\ln(x^3\sqrt{y})-\ln z^2=\ln x^3+\ln y^{1/2}-\ln z^2.

Bring the indices down with the power law:

=3\ln x+\tfrac12\ln y-2\ln z.

This expanded form is exactly what you want before differentiating a complicated logarithm, because each term differentiates in one step.

The change-of-base rule

To evaluate a logarithm in any base using a calculator (which has only $\ln$ and $\log_{10}$ ):

Change of base

\log_a x=\frac{\log_b x}{\log_b a}=\frac{\ln x}{\ln a}.

For example, $\log_2 10=\dfrac{\ln 10}{\ln 2}\approx \dfrac{2.3026}{0.6931}\approx 3.322$ .

Evaluating with change of base

Evaluate $\log_5 40$ to three decimal places.

Step 1: Recognise that the base is not available on the calculator

A calculator provides $\ln$ (base $e$ ) and $\log_{10}$ (base $10$ ), but not $\log_5$ . The change-of-base rule converts any logarithm into a ratio of two logarithms in a base the calculator can handle:

\log_5 40=\frac{\ln 40}{\ln 5}.

The base you choose for both numerator and denominator does not matter as long as it is the same; here we use $\ln$ throughout.

Step 2: Evaluate each logarithm and divide

Using a calculator:

\ln 40 \approx 3.6889, \qquad \ln 5 \approx 1.6094.

Dividing gives:

\log_5 40 = \frac{3.6889}{1.6094} \approx 2.292.

Final answer: $\log_5 40 \approx 2.292$ . (Check: $5^{2.292}\approx 40$ , confirming the answer.)

Common errors

Inventing a "law" for the log of a sum

There is no rule that simplifies $\log(x+y)$ - it does not equal $\log x+\log y$ (that is the product law, for $\log(xy)$ ). Similarly $\dfrac{\log x}{\log y}\ne \log x-\log y$ ; the quotient law applies to $\log\!\left(\tfrac{x}{y}\right)$ , the log of a quotient, not the quotient of two logs. Only products, quotients and powers inside a single log can be split.

Match each law to its index law

If you forget a log law, recall its index partner: $a^m a^n=a^{m+n}$ gives the product law, $\tfrac{a^m}{a^n}=a^{m-n}$ gives the quotient law, and $(a^m)^n=a^{mn}$ gives the power law. The structure is identical with multiplication replaced by addition.

Solving equations with the laws

The laws let you collapse several logarithms into one, after which you remove the log entirely. If $\log_a M=\log_a N$ then $M=N$ (the function is one-to-one); and if $\log_a M=c$ then $M=a^c$ (rewrite in exponential form). For example, to solve $\log_2 x+\log_2(x-2)=3$ , combine the left side to $\log_2\big(x(x-2)\big)=3$ , rewrite as $x(x-2)=2^3=8$ , and solve $x^2-2x-8=0$ to get $x=4$ or $x=-2$ . The domain (both arguments positive) forces $x>2$ , so only $x=4$ is valid. Checking the domain after combining is essential, because the combining step can introduce solutions that the original logs do not allow.

Why it matters

The log laws are the toolkit for the rest of Topic 4: they are essential for differentiating logs, sketching log graphs, and - above all - for solving exponential equations, where taking logs of both sides and applying the power law is the standard move. The same one-to-one property is what lets you equate arguments, a step examiners expect to see justified rather than assumed.

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-style3 marks

Calculator-free. Write the expression

5\ln 2 - \tfrac{1}{2}\ln 16 + \ln 8

in the form

\ln k

Show worked answer →

Use the power law on each coefficient, then the product and quotient laws.

Power law: $5\ln 2 = \ln 2^5 = \ln 32$ , and $\tfrac{1}{2}\ln 16 = \ln 16^{1/2} = \ln 4$ .

So the expression becomes $\ln 32 - \ln 4 + \ln 8$ . Combine:

\ln 32 - \ln 4 + \ln 8 = \ln\!\left(\frac{32 \times 8}{4}\right) = \ln\!\left(\frac{256}{4}\right) = \ln 64.

So $k = 64$ . Marks: one for the power law on both coefficients, one for combining, one for $\ln 64$ . Watch the subtracted middle term: $16^{1/2} = 4$ goes in the denominator.

SACE 2022-style3 marks

Calculator-free. Solve the equation

\ln 16 - 2\ln x = 0

for

x > 0

Show worked answer →

Use the power law: $2\ln x = \ln x^2$ , so the equation becomes $\ln 16 - \ln x^2 = 0$ , i.e. $\ln 16 = \ln x^2$ .

Since $\ln$ is one-to-one, equal logarithms mean equal arguments: $x^2 = 16$ , so $x = 4$ or $x = -4$ .

The domain $x > 0$ rules out $x = -4$ , so $x = 4$ .

Marks: one for the power law, one for equating arguments to get $x^2 = 16$ , and one for selecting $x = 4$ using the domain. Forgetting to discard $x = -4$ is the common error.

SACE 2021-style2 marks

Calculator-free. Express

\log_a 8 + \log_a 5 - \log_a 2

as a single logarithm, then evaluate it when

a = 20

Show worked answer →

Combine with the product and quotient laws:

\log_a 8 + \log_a 5 - \log_a 2 = \log_a\!\left(\frac{8 \times 5}{2}\right) = \log_a 20.

When $a = 20$ , $\log_{20} 20 = 1$ .

Marks: one for the single logarithm $\log_a 20$ , one for evaluating $\log_{20} 20 = 1$ using $\log_a a = 1$ .

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