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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 4.77 min answer

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

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  1. What this dot point is asking
  2. The three laws
  3. The change-of-base rule
  4. Common errors
  5. 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⁡ex\ln x = \log_e x uses base e≈2.718e\approx 2.718 and is the version that appears throughout calculus.

The three laws

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

Two useful special values: log⁡a1=0\log_a 1=0 (since a0=1a^0=1) and log⁡aa=1\log_a a=1 (since a1=aa^1=a).

The change-of-base rule

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

For example, log⁡210=ln⁡10ln⁡2≈2.30260.6931≈3.322\log_2 10=\dfrac{\ln 10}{\ln 2}\approx \dfrac{2.3026}{0.6931}\approx 3.322.

Common errors

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.

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.

2017 SACE Stage 23 marksWrite the expression 5 ln 2 - (1/2) ln 16 + ln 8 in the form ln k.
Show worked answer →

Use the power law (coefficient becomes an exponent) on each term, then the product and quotient laws to combine.

Power law: 5 ln 2 = ln(2^5) = ln 32, and (1/2) ln 16 = ln(16^(1/2)) = ln 4.

So the expression becomes ln 32 - ln 4 + ln 8.

Combine with the quotient and product laws: ln 32 - ln 4 + ln 8 = ln( (32 times 8) / 4 ) = ln(256 / 4) = ln 64.

So the expression equals ln 64, i.e. k = 64. Marks: one for applying the power law to both coefficients, one for combining with the product/quotient laws, and one for the final value ln 64. Watch the sign on the middle term - it is subtracted, so 16^(1/2) = 4 goes in the denominator.

2017 SACE Stage 23 marksSolve the equation ln 16 - 2 ln x = 0 for x > 0.
Show worked answer →

First use the power law to combine the logarithms: 2 ln x = ln(x^2), so the equation becomes ln 16 - ln(x^2) = 0.

Rearrange: ln 16 = ln(x^2). Since the logarithm function is one-to-one, equal logarithms mean equal arguments, so x^2 = 16.

Solving gives x = 4 or x = -4. The domain restriction x > 0 (required because ln x is only defined for positive x) rules out x = -4.

Therefore x = 4. Marks: one for applying the power law, one for equating arguments to get x^2 = 16, and one for selecting the valid solution x = 4 using the domain x > 0. Forgetting to discard x = -4 is the common error here.