Mathematics (0580) | Indices II

Study Notes: Indices II (Algebra and Graphs Section)

Hello there! Welcome to Indices II. You’ve already met the basic rules of indices (or powers) in earlier chapters. Now, we are going to unlock the door to the really powerful stuff: what happens when your index is zero, negative, or even a fraction!

Understanding these rules is essential for manipulating complex algebraic expressions quickly and accurately. This chapter acts as a crucial bridge to many advanced topics in IGCSE Mathematics, especially when dealing with graphs of exponential functions.

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Quick Review: The Three Classic Rules

Before we dive into the new concepts, let’s quickly remind ourselves of the core rules for indices, which apply regardless of whether the index is positive, negative, or fractional. Let \(a\) and \(b\) be bases, and \(m\) and \(n\) be indices:

• Multiplication Rule: When multiplying terms with the same base, you add the indices.
  Example: \(a^m \times a^n = a^{m+n}\)
• Division Rule: When dividing terms with the same base, you subtract the indices.
  Example: \(a^m \div a^n = a^{m-n}\)
• Power of a Power Rule: When raising a power to another power, you multiply the indices.
  Example: \((a^m)^n = a^{mn}\)

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Section 1: The Zero Index (The Simplest Rule!)

This rule is one of the most satisfying because it gives such a simple result.

What is the Rule?

Any non-zero number raised to the power of zero is always 1.

Rule 4: The Zero Index
$$\mathbf{a^0 = 1 \quad (where \ a \neq 0)}$$

Why does this happen? (Understanding the concept)

Imagine using the Division Rule:

If we divide \(5^3\) by \(5^3\):
$$5^3 \div 5^3 = \frac{5 \times 5 \times 5}{5 \times 5 \times 5} = 1$$

But using the Division Rule for indices:

$$5^3 \div 5^3 = 5^{3-3} = 5^0$$

Since both results must be the same, it means \(5^0\) must equal 1.

Example:
\((100)^0 = 1\)
\((x^2y^3)^0 = 1\)
Be careful! \(-3^0 \neq 1\). Only the 3 is raised to the power of zero, so \(-3^0 = -(3^0) = -1\).

Key Takeaway:

If you see a power of 0, the answer is immediately 1 (unless the base is 0, which is undefined).

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Section 2: The Negative Index (The Reciprocal Rule)

A negative index does NOT mean the final answer is negative. It means you need to use the reciprocal of the base.

What is the Rule?

A number raised to a negative power is equal to the reciprocal of that number raised to the positive power.

Rule 5: The Negative Index
$$\mathbf{a^{-n} = \frac{1}{a^n}}$$

Analogy: The Elevator Trick

Think of the fraction line as the ground floor. If a term has a negative index, it’s unhappy being on that 'floor' and wants to move. Changing the sign of the index means moving the term across the fraction line (reciprocal).

• If \(a^{-n}\) is in the numerator, move it to the denominator to make the index positive: \(\frac{a^{-n}}{1} = \frac{1}{a^n}\)
• If \(a^{-n}\) is in the denominator, move it to the numerator to make the index positive: \(\frac{1}{a^{-n}} = a^n\)

Step-by-Step Examples:

Example 1: Numbers
$$3^{-2}$$

1. Recognise the negative index. This means reciprocal.
2. Move the term to the denominator and change the index sign: \(\frac{1}{3^2}\)
3. Calculate: \(\frac{1}{9}\)

Example 2: Algebra (Extended Syllabus Focus)
Simplify \(12a^5 \div 3a^{-2}\)

1. Separate coefficients and indices: \((12 \div 3) \times (a^5 \div a^{-2})\)
2. Simplify the coefficients: \(4\)
3. Use the Division Rule (subtract indices): \(a^{5 - (-2)}\)
4. Simplify: \(4a^{5+2} = 4a^7\)

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Section 3: The Fractional Index (The Root Rule)

This is where we connect powers directly to roots (like square roots and cube roots). This rule is covered in the Extended syllabus (E1.7 and E2.4).

The Single Fractional Index \(\mathbf{a^{1/n}}\)

If the index is a unit fraction (1/n), the power is equivalent to taking the \(n\)th root of the base.

Rule 6a: The Root Rule
$$\mathbf{a^{1/n} = \sqrt[n]{a}}$$

Analogy: The Root is Underground

When you look at a fraction \(m/n\), remember that the denominator \(n\) is the 'root'—and roots grow underground (in the bottom of the fraction). The numerator \(m\) is the normal 'power'.

Example 1: Find the value of \(81^{1/4}\)
$$81^{1/4} = \sqrt[4]{81}$$

We are looking for a number that, when multiplied by itself four times, equals 81. That number is 3.
Answer: 3

Example 2: Write \(\sqrt{6}\) in index form.
Remember, a square root has an invisible index of 2.
Answer: \(6^{1/2}\)

The General Fractional Index \(\mathbf{a^{m/n}}\)

When the index is a general fraction \(m/n\), you apply both the root and the power.

Rule 6b: Combined Fractional Index
$$\mathbf{a^{m/n} = (\sqrt[n]{a})^m \quad or \quad \sqrt[n]{a^m}}$$

Tip: It’s usually easier to calculate the root first, and then apply the power. Roots usually make the number smaller, making the subsequent power calculation easier.

Step-by-Step Example:

Work out \(27^{2/3}\)

1. Identify the denominator (root) and numerator (power): Root = 3, Power = 2.
2. Calculate the root first: \(\sqrt[3]{27}\) (The cube root of 27 is 3).
3. Apply the power: \((3)^2\)
4. Simplify: 9

Combining Negative and Fractional Indices

What about \(16^{-3/4}\)? This uses both the negative rule (reciprocal) and the fractional rule (root/power).

1. Apply the Negative Rule first (Reciprocal): \(\frac{1}{16^{3/4}}\)
2. Apply the Fractional Rule (Root first, then power): \(\frac{1}{(\sqrt[4]{16})^3}\)
3. Calculate the root: \(\sqrt[4]{16} = 2\)
4. Calculate the power: \(\frac{1}{(2)^3} = \frac{1}{8}\)

Key Takeaway:

Fractional indices mean roots. If you see \(a^{m/n}\), remember: The root is below, the power is above!

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Section 4: Simplifying Complex Expressions

In the Algebra section (E2.4), you will need to apply these rules to simplify expressions involving variables.

Rule Reminder: All the original rules still apply, just with the new indices.

Example 1: Simplifying with Negative Indices

Simplify \((5x^3)^2 \times x^{-4}\)

1. Apply the Power of a Power rule to the bracket: \((5)^2 \times (x^3)^2 = 25x^6\)
2. Rewrite the whole expression: \(25x^6 \times x^{-4}\)
3. Apply the Multiplication Rule (add indices): \(25x^{6 + (-4)}\)
4. Simplify: \(25x^2\)

Example 2: Combining Division and Negative Indices

Simplify \(\frac{2y^2}{5y^5} \times (10y^{-2})\)

1. Rewrite as a single fraction: \(\frac{2y^2 \times 10y^{-2}}{5y^5}\)
2. Multiply the numerator: \(\frac{20y^{2 + (-2)}}{5y^5} = \frac{20y^0}{5y^5}\)
3. Apply the Zero Index Rule (\(y^0 = 1\)): \(\frac{20 \times 1}{5y^5}\)
4. Simplify the coefficients and use the negative index concept: \(\frac{4}{y^5}\) (Or, using the division rule: \(4y^{0-5} = 4y^{-5}\). Both forms are usually acceptable unless specified.)

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Section 5: Solving Exponential Equations (Extended Syllabus)

An exponential equation is one where the unknown variable (\(x\)) is in the index, for example, \(3^x = 9\).

To solve these, the key technique is to express both sides of the equation with the same base.

Step-by-Step Process:

Example 1: Basic Exponential Equation
Solve \(2^x = 32\)

1. Identify the base (2). Try to write 32 as a power of 2.
2. We know \(2 \times 2 \times 2 \times 2 \times 2 = 32\), so \(32 = 2^5\).
3. Rewrite the equation: \(2^x = 2^5\).
4. Since the bases are the same, the indices must be equal.
   Answer: \(x = 5\)

Example 2: Using Negative Indices
Solve \(4^x = \frac{1}{64}\)

1. Identify the base (4). Write 64 as a power of 4: \(64 = 4^3\).
2. Rewrite the right side using the negative index rule: \(\frac{1}{64} = 4^{-3}\).
3. Set the equation: \(4^x = 4^{-3}\).
4. Equate the indices.
   Answer: \(x = -3\)

Example 3: Using Fractional Indices (A common exam style question)

Solve \(25^x = 5\)

1. Find a common base. Both 25 and 5 can be written using base 5.
2. Convert 25: \(25 = 5^2\).
3. Rewrite the equation: \((5^2)^x = 5^1\)
4. Apply the Power of a Power Rule: \(5^{2x} = 5^1\)
5. Equate the indices: \(2x = 1\)
6. Solve for \(x\).
   Answer: \(x = 1/2\) (This makes sense, as \(25^{1/2} = \sqrt{25} = 5\)).

Example 4: Solving involving different powers (Syllabus example: \(5^{x+1} = 25^x\))

Solve \(5^{x+1} = 25^x\)

1. Find the common base: 5. Convert 25 to \(5^2\).
2. Rewrite the equation: \(5^{x+1} = (5^2)^x\)
3. Simplify the right side: \(5^{x+1} = 5^{2x}\)
4. Equate the indices: \(x + 1 = 2x\)
5. Solve the linear equation:
   \(1 = 2x - x\)
   Answer: \(x = 1\)

Key Takeaway:

To solve equations where the variable is in the index, always aim to make the bases identical. Once the bases are the same, you can set the indices equal to each other.