International Mathematics (0607) | Indices II

🔥 IGCSE Mathematics (0607) Study Notes: Indices II (Algebra) 🔥

Welcome to Indices II! If you’ve already mastered the basic index laws, this chapter is where we apply those rules to trickier situations, involving zero, negative, and fractional indices, especially when working with algebraic expressions.

Understanding these concepts is vital because it allows us to simplify complex algebraic fractions and solve powerful exponential equations. Don't worry if this seems tricky at first—we’ll break down each index type step-by-step!

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1. Quick Review: The Fundamental Index Laws

These three rules are the building blocks for everything else we do. Remember that the base (\(a\)) must be the same for the first two rules to work!

Multiplication Rule (Adding Powers)

When multiplying powers with the same base, you add the exponents.

\( a^m \times a^n = a^{m+n} \)

Example: \( x^3 \times x^5 = x^{3+5} = x^8 \)

Division Rule (Subtracting Powers)

When dividing powers with the same base, you subtract the exponents.

\( a^m \div a^n = a^{m-n} \)

Example: \( y^7 \div y^2 = y^{7-2} = y^5 \)

Power of a Power Rule (Multiplying Powers)

When raising a power to another power, you multiply the exponents.

\( (a^m)^n = a^{mn} \)

Example: \( (w^4)^3 = w^{4 \times 3} = w^{12} \)

Key Takeaway: Check the base first! If the bases match, you can use these simple addition, subtraction, or multiplication rules for the indices.

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2. Special Indices: Zero and Negative

2.1 The Zero Index

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

\( a^0 = 1 \), where \( a \ne 0 \)

Why? Think about the division rule: \( 5^3 \div 5^3 \). By division rule, this is \( 5^{3-3} = 5^0 \). But we know that anything divided by itself is 1. Therefore, \( 5^0 = 1 \).

• Example: \( 100^0 = 1 \)
• Example (Algebraic): \( (3xy)^0 = 1 \)
• Example (Careful!): \( 3x^0 \). Only the \( x \) is raised to the power 0, so \( 3 \times x^0 = 3 \times 1 = 3 \).

2.2 The Negative Index

A negative index does not mean the number is negative. It means you must take the reciprocal (flip the base) to make the index positive.

\( a^{-n} = \frac{1}{a^n} \)

🧠 Memory Aid: Think of the negative sign as a "ticket to the basement" (the denominator) or a "flip ticket." Once you flip, the index becomes positive!

• Example 1: Find the value of \( 7^{-2} \).
  \( 7^{-2} = \frac{1}{7^2} = \frac{1}{49} \)
• Example 2: Simplify \( x^{-5} \).
  \( x^{-5} = \frac{1}{x^5} \)
• Example 3 (Fractions): If the negative power is in the denominator, it moves up to the numerator:
  \( \frac{1}{y^{-3}} = y^3 \)

Key Takeaway: \( a^0=1 \). A negative index means "flip it" to make the index positive. The negative index rule is essential for simplifying algebraic fractions.

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3. Fractional Indices (Extended Content E2.4/E1.7)

Fractional indices are simply another way of writing roots. The denominator of the fraction tells you which root to take.

3.1 The Unit Fraction Index

An index of \(\frac{1}{n}\) means take the \(n\)-th root of the base.

\( a^{\frac{1}{n}} = \sqrt[n]{a} \)

• Example 1: \( 9^{\frac{1}{2}} \) means square root of 9: \( \sqrt{9} = 3 \)
• Example 2: \( 8^{\frac{1}{3}} \) means cube root of 8: \( \sqrt[3]{8} = 2 \)
• Example 3 (Syllabus Example E1.7): \( 64^{\frac{1}{2}} = \sqrt{64} = 8 \)
• Example 4 (Syllabus Example E1.7): \( 16^{\frac{1}{4}} = \sqrt[4]{16} = 2 \)

3.2 The General Fractional Index

When the numerator is not 1, we combine the root and the power.

\( a^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{(a^m)} \)

Method: It is usually easiest to do the root first (using the denominator) and then the power second (using the numerator).

Step-by-step Example: Calculate \( 27^{\frac{2}{3}} \)

1. Denominator is 3, so find the cube root: \( \sqrt[3]{27} = 3 \)
2. Numerator is 2, so raise the result to the power of 2: \( 3^2 = 9 \)
3. Therefore, \( 27^{\frac{2}{3}} = 9 \).

3.3 Combining Negative and Fractional Indices

If you have a negative fractional index, apply the negative rule (reciprocal/flip) first, then apply the fractional index rule.

\( a^{-\frac{m}{n}} = \frac{1}{a^{\frac{m}{n}}} \)

Step-by-step Example: Calculate \( 8^{-\frac{2}{3}} \)

1. Apply the negative rule (Flip it!): \( 8^{-\frac{2}{3}} = \frac{1}{8^{\frac{2}{3}}} \)
2. Find the cube root of 8: \( \sqrt[3]{8} = 2 \)
3. Raise the result to the power of 2: \( 2^2 = 4 \)
4. Final Answer: \( \frac{1}{4} \)

Key Takeaway: The denominator is the root, the numerator is the power. Always simplify by doing the root first if possible!

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4. Simplifying Algebraic Expressions (C2.4 / E2.4)

When simplifying expressions that include indices, remember to treat the coefficients (numbers) and the variables (letters) separately, and use the index rules we just covered.

4.1 Products Involving Indices

Remember that the rules apply to ALL parts of the product within the bracket.

Example 1 (Power of a Power): Simplify \( (5x^3)^2 \)

Apply the square to both the coefficient (5) and the variable (\(x^3\)):

Step 1: Square the coefficient: \( 5^2 = 25 \)

Step 2: Use the Power of a Power rule for the variable: \( (x^3)^2 = x^{3 \times 2} = x^6 \)

Result: \( 25x^6 \)

Example 2 (Multiplication with Negative Indices): Simplify \( 6x^3y^4 \times 5x^{-3}y^{-3} \)

1. Multiply the coefficients: \( 6 \times 5 = 30 \)
2. Group the \( x \) terms and add indices: \( x^3 \times x^{-3} = x^{3+(-3)} = x^0 \)
3. Group the \( y \) terms and add indices: \( y^4 \times y^{-3} = y^{4-3} = y^1 = y \)
4. Since \( x^0 = 1 \), the expression simplifies to \( 30 \times 1 \times y \).

Result: \( 30y \)

4.2 Division Involving Indices

Use the division rule (subtraction of powers) and divide the coefficients.

Example 3 (Division with Negative Indices): Simplify \( 12a^5 \div 3a^{-2} \)

1. Divide the coefficients: \( 12 \div 3 = 4 \)
2. Subtract the indices: \( a^5 \div a^{-2} = a^{5 - (-2)} \)
3. Be very careful with the double negative: \( 5 - (-2) = 5 + 2 = 7 \)

Result: \( 4a^7 \)

Key Takeaway: Always handle numbers and letters separately. Remember that subtracting a negative index results in addition!

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5. Solving Simple Exponential Equations (C2.4 / E2.4)

An exponential equation is one where the unknown variable is in the index (exponent). We solve simple versions of these by making the bases the same.

The Core Strategy: If \( a^x = a^y \), then \( x = y \). The moment the bases are identical, we can forget the bases and equate the powers.

Step-by-step Process

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

1. Check if the larger number (32) can be written as a power of the smaller base (2).
   \( 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32 \).
2. Rewrite the equation with the same base:
   \( 2^x = 2^5 \)
3. Equate the indices: \( x = 5 \)

Example 2 (Algebra in the Index): Solve \( 3^{2x-1} = 27 \)

1. Rewrite the larger number (27) using the base 3:
   \( 27 = 3^3 \).
2. Rewrite the equation:
   \( 3^{2x-1} = 3^3 \)
3. Equate the indices:
   \( 2x - 1 = 3 \)
4. Solve the linear equation:
   \( 2x = 4 \)
   \( x = 2 \)

Example 3 (Bases that need rewriting): Solve \( 5^{x+1} = 25^x \) (Syllabus type E2.4)

1. Identify the common base: Both 5 and 25 can use base 5.
   Rewrite 25 as \( 5^2 \).
2. Substitute back into the equation, remembering brackets:
   \( 5^{x+1} = (5^2)^x \)
3. Use the Power of a Power rule on the right side:
   \( 5^{x+1} = 5^{2x} \)
4. Equate the indices and solve:
   \( x + 1 = 2x \)
   \( 1 = 2x - x \)
   \( x = 1 \)