Practice Maths

Introducing Logarithms and Logarithmic Laws

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Key Terms

Definition
loga(x) = y ⇔ ay = x   (the logarithm is the exponent).
Common log
log10(x) is written log(x).   Natural log: loge(x) = ln(x).
Log Law 1 (Product)
loga(AB) = loga(A) + loga(B)
Log Law 2 (Quotient)
loga(A/B) = loga(A) − loga(B)
Log Law 3 (Power)
loga(An) = n loga(A)
Change of base
loga(x) = log(x) / log(a)
Special values
loga(1) = 0;   loga(a) = 1;   loga(ax) = x;   aloga(x) = x
📚 QCAA Formula Sheet
loga(AB) = logaA + logaB
loga(A/B) = logaA − logaB
loga(An) = n logaA
Change of base: logax = logbx / logba
Exponential form Logarithmic form Reading
23 = 8log2(8) = 3"log base 2 of 8 is 3"
102 = 100log(100) = 2"log of 100 is 2"
e1 = eln(e) = 1"natural log of e is 1"
50 = 1log5(1) = 0"log base 5 of 1 is 0"
3−2 = 1/9log3(1/9) = −2"log base 3 of 1/9 is −2"
x y 0 1 2 4 1 2 y = x (0, 1) (1, 0) (2, 1) y = 2x y = log₂(x)
Hot Tip When evaluating loga(x), ask: "What power do I raise a to in order to get x?" For example, log2(32) = ? Ask: 2 to what power gives 32? Since 25 = 32, log2(32) = 5.

Worked Example 1 — Evaluating logarithms

Evaluate: (a) log3(81)   (b) log10(0.001)   (c) log5(1/25)

(a) Ask: 3? = 81. Since 34 = 81, log3(81) = 4

(b) Ask: 10? = 0.001 = 10−3. So log(0.001) = −3

(c) Ask: 5? = 1/25 = 5−2. So log5(1/25) = −2

Worked Example 2 — Applying logarithmic laws

Simplify: log(8) + log(125)   using log10

Apply Product Law: log(8) + log(125) = log(8 × 125) = log(1000) = log(103) = 3

Simplify: 3log(4) − log(8)

Apply Power Law first: 3log(4) = log(43) = log(64)

Then Quotient Law: log(64) − log(8) = log(64/8) = log(8) = log(23) = 3log(2) ≈ 0.903

Introduction

Logarithms were invented to make calculation easier — before calculators, scientists used log tables to turn multiplication into addition. Today, logs appear in every branch of science: measuring earthquake magnitude, sound intensity, pH, and population growth. Understanding logarithms means understanding the inverse of exponential growth.

What is a Logarithm?

The logarithm answers the question: "What exponent do I need?" If 25 = 32, then log2(32) = 5. The base of the log must match the base of the exponential. Think of it as the inverse operation: exponentials build up, logarithms tear down.

The formal definition: loga(x) = y ⇔ ay = x   (where a > 0, a ≠ 1, x > 0)

Two special bases are used most often:

  • Base 10 (common log): written log(x). Used in pH, Richter scale, decibels.
  • Base e ≈ 2.718 (natural log): written ln(x). Used in calculus and continuous growth models.

Special Values

These follow directly from the definition and are worth memorising:

  • loga(1) = 0 — because a0 = 1 for any base
  • loga(a) = 1 — because a1 = a
  • loga(ax) = x — logs and exponentials cancel
  • aloga(x) = x — exponentials and logs cancel

The Three Logarithmic Laws

These laws come directly from the laws of indices. They let you expand, contract, or simplify logarithmic expressions.

Law 1 — Product Law: loga(AB) = logaA + logaB

This mirrors the index law am × an = am+n.

Example: log2(4 × 8) = log2(4) + log2(8) = 2 + 3 = 5. Check: 25 = 32 = 4 × 8 ✓

Law 2 — Quotient Law: loga(A/B) = logaA − logaB

This mirrors the index law am / an = am−n.

Example: log3(81/9) = log3(81) − log3(9) = 4 − 2 = 2. Check: 32 = 9 = 81/9 ✓

Law 3 — Power Law: loga(An) = n logaA

This mirrors the index law (am)n = amn.

Example: log2(84) = 4 log2(8) = 4 × 3 = 12. Check: 212 = 4096 = 84 ✓

Change of Base

Calculators only have log (base 10) and ln (base e) buttons. To evaluate any other base, use:

loga(x) = log(x) / log(a)   =   ln(x) / ln(a)

Worked Example 1 — Expanding a log expression

Expand: log3(9x2/y)

Step 1: Quotient law: log3(9x2) − log3(y)

Step 2: Product law on first term: log3(9) + log3(x2) − log3(y)

Step 3: Power law and evaluate: 2 + 2log3(x) − log3(y)

Answer: 2 + 2log3(x) − log3(y)

Worked Example 2 — Contracting a log expression

Write as a single log: 3log(x) + log(2) − log(x2)

Step 1: Power law: log(x3) + log(2) − log(x2)

Step 2: Product law: log(2x3) − log(x2)

Step 3: Quotient law: log(2x3/x2) = log(2x)

Answer: log(2x)

💡 Key Reminder: You can only apply the product/quotient law when the two logs have the same base. Never combine log2(x) + log3(x) directly — they have different bases!

Summary

loga(x) = y means ay = x. The three laws (product, quotient, power) all mirror index laws. Change of base formula allows any log to be evaluated on a calculator. Key special values: loga(1) = 0, loga(a) = 1, loga(ax) = x.

Mastery Practice

See Answers ➔
  1. Fluency

    Convert each statement between exponential and logarithmic form.

    1. (a) 43 = 64  →  write as a log
    2. (b) 10−3 = 0.001  →  write as a log
    3. (c) log5(125) = 3  →  write as an exponential
    4. (d) log2(1/8) = −3  →  write as an exponential
  2. Fluency

    Evaluate each logarithm without a calculator.

    1. (a) log2(16)
    2. (b) log3(27)
    3. (c) log10(10 000)
    4. (d) log5(1)
    5. (e) log4(1/16)
    6. (f) log7(7)
  3. Fluency

    Use the special values to evaluate each expression without a calculator.

    1. (a) log6(65)
    2. (b) 3log3(11)
    3. (c) log2(2−4)
    4. (d) 10log(7)
  4. Fluency

    Expand each expression using the logarithmic laws.

    1. (a) log2(8x)
    2. (b) log(100/x)
    3. (c) log3(x4)
    4. (d) log(5x3y)
  5. Understanding

    Write each expression as a single logarithm.

    1. (a) log(3) + log(4)
    2. (b) log2(20) − log2(5)
    3. (c) 3log(x) + log(2)
    4. (d) 2log(x) − log(y) + log(z)
  6. Understanding

    Simplify each expression to a number or simple form.

    Strategy: Apply power law first, then product/quotient law, then evaluate.
    1. (a) log2(4) + log2(8)
    2. (b) log(1000) − 2log(10)
    3. (c) log3(272)
    4. (d) 2log5(25) + log5(5)
  7. Understanding

    Use the change of base formula to evaluate correct to 3 decimal places.

    1. (a) log2(10)
    2. (b) log3(50)
    3. (c) log7(200)
  8. Understanding

    Expand fully using the logarithmic laws.

    1. (a) log2(8x3/y2)
    2. (b) ln(√(x/(x+1)))
  9. Problem Solving

    Solving logarithmic equations.

    Strategy: Write the equation in exponential form ay = x to isolate the unknown.
    1. (a) log2(x) = 5
    2. (b) log3(x + 1) = 4
    3. (c) log(x) + log(x − 3) = 1
  10. Problem Solving

    Prove and apply.

    1. (a) Given that loga(2) = p and loga(3) = q, express loga(72) in terms of p and q.
    2. (b) If log2(3) = k, express log2(36) in terms of k.
    3. (c) Show that loga(b) × logb(a) = 1 using the change of base formula.