# Properties of indices and logarithms

> Edexcel International GCSE Further Pure Mathematics · 4PM1 2016 Spec
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s1-properties-of-indices-and-logarithms/

This guide teaches you all index and logarithm properties required for Edexcel IGCSE Further Pure Math (4PM1), including simplification rules, change of base formula, and solving exponential equations of the form $a^x = b$.

**Prerequisites:** [Basic integer exponent rules (GCSE Mathematics)](https://www.owlsprep.com/study/edexcel-gcse-math-indices-basics/); [Definition of logarithms](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s1-logarithm-basics/)

## Learning objectives

- Apply laws of indices for integer, negative and rational exponents
- Use core logarithm laws to simplify and rewrite logarithmic expressions
- Apply the change of base formula for logarithms, including the reciprocal special case
- Solve exponential equations of the form $a^x = b$ using logarithms

## Laws of Indices for Integer, Negative and Rational Exponents

**Laws of Indices** — The set of rules used to simplify expressions involving exponents, applicable for all rational exponents:
1. $a^0 = 1$ (for $a \neq 0$)
2. $a^{-n} = \frac{1}{a^n}$ (for $a \neq 0$)
3. $a^{m/n} = (\sqrt[n]{a})^m$ (for $a > 0$)

*Example:* $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$

These rules extend basic integer exponent rules to negative and fractional powers. Always simplify negative exponents first before handling fractional powers to avoid calculation errors.

**Worked example:** Simplify $\left(\frac{16x^{-4}}{81y^{8}}\right)^{-3/4}$, writing your answer with positive exponents only.

1. Apply the negative exponent rule to the outer power to flip the fraction inside the bracket:

   $$\left(\frac{81y^{8}}{16x^{-4}}\right)^{3/4}$$
2. Rewrite the negative inner exponent as a positive exponent:

   $$\left(\frac{81y^{8}x^{4}}{16}\right)^{3/4}$$
3. Apply the fractional exponent to each term individually:

   $$\frac{(\sqrt[4]{81})^3 \times (\sqrt[4]{y^8})^3 \times (\sqrt[4]{x^4})^3}{(\sqrt[4]{16})^3}$$
4. Evaluate each root and power to get the final simplified form:

   $$\frac{3^3 \times (y^2)^3 \times x^3}{2^3} = \frac{27x^3y^6}{8}$$

> **Exam tip:** Always present final answers with positive exponents unless the question explicitly asks otherwise, as marks are often deducted for negative exponents in final solutions.

*Calculator:* allowed

## Core Laws of Logarithms

**Laws of Logarithms** — Rules used to simplify or combine logarithmic expressions (for $a > 0$, $a \neq 1$, $x > 0$, $y > 0$):
1. $\log_a(xy) = \log_a x + \log_a y$
2. $\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$
3. $\log_a(x^k) = k\log_a x$
4. $\log_a a = 1$
5. $\log_a 1 = 0$

*Example:* $\log_2(8x) = \log_2 8 + \log_2 x = 3 + \log_2 x$

These laws are the inverse of the index laws, and you will need to recall all of them for the exam, as they are not provided on the formula sheet.

**Worked example:** Simplify $2\log_{10} 5 + \log_{10} 8 - \log_{10} 2$ to a single integer value.

1. Apply the power rule to the first term to move the coefficient inside the logarithm:

   $$\log_{10} 5^2 + \log_{10} 8 - \log_{10} 2 = \log_{10} 25 + \log_{10} 8 - \log_{10} 2$$
2. Combine the addition terms using the product rule:

   $$\log_{10}(25 \times 8) - \log_{10} 2 = \log_{10} 200 - \log_{10} 2$$
3. Combine the subtraction terms using the quotient rule:

   $$\log_{10}\left(\frac{200}{2}\right) = \log_{10} 100$$
4. Evaluate the logarithm to get the final integer value:

   $$\log_{10} 100 = 2$$

> **Exam tip:** Never split $\log_a(x + y)$ into $\log_a x + \log_a y$ — this is a very common error. The product rule only applies to products inside the logarithm, not sums.

*Calculator:* allowed

## Change of Base Formula for Logarithms

**Change of Base Formula** — A formula to rewrite a logarithm of base $a$ as a ratio of logarithms of a common base $b$, usually base 10 or natural log for calculator evaluation. The special case $\log_a b = \frac{1}{\log_b a}$ is derived by setting $x = b$ in the general formula.

*Notation:* \log_a x = \frac{\log_b x}{\log_b a}

*Example:* $\log_3 7 = \frac{\log_{10}7}{\log_{10}3} \approx 1.771$

This formula is given on your exam formula sheet, so you do not need to memorize it, but you must know how to apply it correctly, including the reciprocal special case.

**Worked example:** Evaluate $\log_4 27$ using base 2 logarithms, giving your answer as an exact simplified form.

1. Apply the change of base formula with base 2:

   $$\log_4 27 = \frac{\log_2 27}{\log_2 4}$$
2. Simplify the denominator, as $2^2 = 4$:

   $$\log_2 4 = 2$$
3. Rewrite the numerator using the power rule for logarithms:

   $$\log_2 3^3 = 3\log_2 3$$
4. Combine terms to get the final exact value:

   $$\frac{3\log_2 3}{2}$$

> **Exam tip:** When asked for an exact value, avoid using base 10 or natural log on your calculator, as these will give you a decimal approximation. Choose a base that simplifies the denominator to an integer where possible.

*Calculator:* allowed

## Solving Exponential Equations of the Form $a^x = b$

To solve for an unknown exponent in an equation of the form $a^x = b$, take the logarithm of both sides of the equation, then apply logarithm laws to isolate $x$. You can use any base for the logarithm, but using base $a$ will simplify the calculation.

**Worked example:** Solve $3^{2x - 1} = 20$, giving your answer to 3 significant figures.

1. Take log base 10 of both sides of the equation:

   $$\log_{10}(3^{2x - 1}) = \log_{10} 20$$
2. Apply the power rule to bring the exponent down:

   $$(2x - 1)\log_{10} 3 = \log_{10} 20$$
3. Isolate the bracket term by dividing both sides by $\log_{10} 3$:

   $$2x - 1 = \frac{\log_{10}20}{\log_{10}3} \approx 2.727$$
4. Solve for $x$ and round to 3 significant figures:

   $$2x = 3.727 \implies x \approx 1.86$$

**Check your understanding**

1. What is the first step to solve $5^{x + 2} = 12$?

   - Divide both sides by 5
   - Take logarithm of both sides
   - Subtract 2 from both sides

   *Answer:* Take logarithm of both sides

   *Why:* Correct! Taking the logarithm of both sides lets you use the power rule to bring the unknown exponent down to the level of the equation.

> **Exam tip:** Always check your answer by substituting it back into the original exponential equation to verify it is correct, especially if you have time left at the end of the exam.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Writing $a^{-n} = -a^n$
  - Why it fails: Confusing negative exponents with negative coefficients. Negative exponents indicate a reciprocal, not a negative value.
  - Correct: Rewrite $a^{-n}$ as $\frac{1}{a^n}$ for $a \neq 0$.
- **Wrong:** Splitting $\log_a(x + y)$ into $\log_a x + \log_a y$
  - Why it fails: Misapplying the product rule, which only applies to products, not sums, inside logarithms.
  - Correct: Leave $\log_a(x + y)$ as is unless you can factor the expression inside the logarithm into a product.
- **Wrong:** Incorrectly applying the change of base formula as $\log_a x = \frac{\log_b a}{\log_b x}$
  - Why it fails: Swapping the numerator and denominator of the ratio.
  - Correct: Remember that the base of the original logarithm appears in the denominator of the change of base ratio.
- **Wrong:** Forgetting that $a^0 = 1$ only applies when $a \neq 0$
  - Why it fails: Ignoring the domain constraint, as $0^0$ is undefined.
  - Correct: Always state that $a \neq 0$ when applying the $a^0 = 1$ rule if the base includes a variable.
- **Wrong:** Rounding intermediate values when solving exponential equations
  - Why it fails: Premature rounding introduces errors in subsequent calculation steps, leading to inaccurate final answers.
  - Correct: Keep all intermediate values in your calculator memory and only round the final answer to the required number of significant figures or decimal places.

## Cheatsheet

| Concept | Rule | Key Note |
| --- | --- | --- |
| Index Laws | $a^0 = 1$ | $a \neq 0$ |
| Index Laws | $a^{-n} = 1/a^n$ | $a \neq 0$ |
| Index Laws | $a^{m/n} = (\sqrt[n]{a})^m$ | $a > 0$ |
| Logarithm Laws | $\log_a(xy) = \log_a x + \log_a y$ | $x,y > 0, a>0, a\neq1$ |
| Logarithm Laws | $\log_a(x/y) = \log_a x - \log_a y$ | $x,y > 0, a>0, a\neq1$ |
| Logarithm Laws | $\log_a(x^k) = k\log_a x$ | $x > 0, a>0, a\neq1$ |
| Change of Base | $\log_a x = \log_b x / \log_b a$ | Given on formula sheet |
| Change of Base Special Case | $\log_a b = 1/\log_b a$ | Derived from general formula |
| Solve $a^x = b$ | $x = \log_a b = \log b / \log a$ | Use any base for calculation |

## What's next

Now that you have mastered the properties of indices and logarithms, you are ready to apply these rules to more advanced topics in Edexcel IGCSE Further Pure Math. You will frequently use these laws when simplifying algebraic expressions, solving exponential and logarithmic equations, and working with sequences and series. Be sure to practice both simplification problems and equation-solving questions to build fluency, as these concepts are often tested alongside other topics like coordinate geometry and trigonometry. Next, you can explore related topics to reinforce your knowledge and prepare for more complex exam questions.

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