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Study GuideYear 11Level 2 · StandardHSCVCE

Free Year 11 Index and Logarithm Laws

An HSC/VCE-style study guide that sets out the index laws, the definition of a logarithm and the logarithm laws side by side. Worked examples cover simplifying index expressions, fractional and negative powers, evaluating logarithms without a calculator, combining logs, solving exponential equations with ln, and solving log equations while checking the domain, with a six-question self-check and answers.

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Year
Year 11
Subject
Mathematics
Topic
Exponentials & Logarithms
Difficulty
Level 2 · Standard
Estimated time
25 minutes
Curriculum
Australian Curriculum
Answers
Not applicable
Format
PDF (A4) + print

Students will practise

  • simplifying expressions using the index laws, including fractional and negative indices
  • converting between index form and logarithmic form
  • simplifying and expanding expressions using the logarithm laws
  • solving exponential and logarithmic equations, checking that solutions are in the domain

Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Index and Logarithm Laws

Ready for more? Move on to Year 12 Mathematics.

Study guide

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practise with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

Year 11 students (typically ages 16–17) working on exponentials & logarithms. It is pitched at level 2 · standard.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 25 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Explore more Year 11 mathematics resources.

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Year 11 · Mathematics · Exponentials & Logarithms

Index and Logarithm Laws

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What you need to know

Index laws describe how powers combine. In each law the base must be the same: am × an = am + n, am ÷ an = am − n, (am)n = amn, (ab)n = anbn, a0 = 1, a−n = , and a1/n = ⁿ√a, so am/n = (ⁿ√a)m.

A logarithm is the answer to the question "what power?". logₐ x = y means a[sup y] = x. For example log₂ 8 = 3 because 23 = 8. The base a must be positive and not equal to 1, and you can only take the log of a positive number. Two bases get special names: log₁₀ x is often written log x, and logₑ x (base e ≈ 2.718) is the natural logarithm, written ln x.

Logarithm lawIndex law it comes from
logₐ (xy) = logₐ x + logₐ yam × an = am + n
logₐ () = logₐ x − logₐ yam ÷ an = am − n
logₐ (xn) = n logₐ x(am)n = amn
logₐ a = 1 and logₐ 1 = 0a1 = a and a0 = 1
logₐ x = (change of base)lets a calculator evaluate any base

Simplify (2x3)4 ÷ 4x5

  1. (2x3)4 = 24 × x12 = 16x12
  2. 16x12 ÷ 4x5 = 4x12 − 5 = 4x[sup 7]

Evaluate 82/3 and 27−1/3 without a calculator

  1. 82/3 = (³√8)2 = 22 = 4
  2. 27−1/3 = = = [frac 1/3]

Evaluate log₂ 32, log₅ () and log₁₀ 0.001

  1. log₂ 32: 25 = 32, so log₂ 32 = 5
  2. log₅ (): 5−2 = , so log₅ () = −2
  3. log₁₀ 0.001: 10−3 = 0.001, so log₁₀ 0.001 = −3

Simplify log₃ 54 − log₃ 2, and expand ln (x2√y)

  1. log₃ 54 − log₃ 2 = log₃ () = log₃ 27 = 3 (because 33 = 27)
  2. ln (x2√y) = ln x2 + ln y1/2 = 2 ln x + [frac 1/2] ln y

Solve 3x = 20, correct to 2 decimal places

  1. Take ln of both sides: ln 3x = ln 20
  2. Bring the power down: x ln 3 = ln 20
  3. x = = = 2.73
  4. Check: 32.73 ≈ 20.1, which is close to 20.

Solve log₂ x + log₂ (x − 2) = 3

  1. Combine into one log: log₂ [x(x − 2)] = 3
  2. Rewrite in index form: x(x − 2) = 23 = 8
  3. x2 − 2x − 8 = 0, so (x − 4)(x + 2) = 0, giving x = 4 or x = −2.
  4. Check the domain: log₂ x needs x > 0, so x = −2 is rejected. The only solution is x = 4.
Common mistake: there is no law for the log of a sum. log (x + y) is not log x + log y, and log x × log y is not log (xy). The laws only turn products, quotients and powers inside a log into sums, differences and multiples outside it. Another common mistake: forgetting to reject solutions that make the inside of a log zero or negative. Always substitute each solution back into the original equation before writing your final answer.

Check yourself

  1. Simplify x5 × x−2 ÷ x.
  2. Evaluate 163/4.
  3. Evaluate log₃ 81.
  4. Write log 2 + 2 log 5 as a single logarithm.
  5. Solve 2x = 10, correct to 2 decimal places.
  6. Solve log₅ (2x + 1) = 2.

Answers: 1. x5 − 2 − 1 = x2 2. (⁴√16)3 = 23 = 8 3. 4 (because 34 = 81) 4. log 2 + log 25 = log 50 5. x = ≈ 3.32 6. 2x + 1 = 25, so x = 12

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Created by
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Last reviewed
1 October 2026
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