SACE Board SACE 2023-style practice: Calculator-free. Write the expression 5ln 2 - 12ln 16 + ln 8 in the form ln k.

Use the power law on each coefficient, then the product and quotient laws. Power law: 5ln 2 = ln 2^5 = ln 32, and 12ln 16 = ln 16^1/2 = ln 4. So the expression becomes ln 32 - ln 4 + ln 8. Combine: $ln 32 - ln 4 + ln 8 = ln\!((32 × 8)/(4)) = ln\!((256)/(4)) = ln 64.$ 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 Board SACE 2021-style practice: Calculator-free. Express _a 8 + _a 5 - _a 2 as a single logarithm, then evaluate it when a = 20.

Combine with the product and quotient laws: $_a 8 + _a 5 - _a 2 = _a\!((8 × 5)/(2)) = _a 20.$ When a = 20, _20 20 = 1. Marks: one for the single logarithm _a 20, one for evaluating _20 20 = 1 using _a a = 1.

§-Syllabus dot point

SAMath MethodsSyllabus dot point

How do the laws of logarithms let us simplify and rearrange logarithmic expressions?

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

The product, quotient and power laws of logarithms, how they follow from the index laws, the change-of-base rule, and how to use them to simplify and solve.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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:

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:

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

The change-of-base rule

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

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

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 practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20233 marksCalculator-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 20223 marksCalculator-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 20212 marksCalculator-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 $\log_a a = 1$ .