Indices II
Section: Algebra and graphs 1 | Syllabus: Cambridge IGCSE Mathematics (0580)
Simplifying Algebraic Terms with Indices
The same laws of indices used for pure numbers apply to algebraic terms too. When a term has both a coefficient and a power, the coefficient and the index are handled separately.
- When multiplying or dividing algebraic terms, deal with the coefficients and the indices separately
- Multiply/divide the coefficients as ordinary numbers
- Add the indices when multiplying powers of the same letter; subtract them when dividing
- A reciprocal of a power can always be written using a negative index: (1/y^n) = y^-n
Worked Example: Multiplying and Dividing Algebraic Terms
- Question: Simplify.
(i) 4x⁵ × 3x² (ii) 12a⁵ ÷ 3a⁻²- Step 1 (i): Multiply the coefficients: 4 × 3 = 12. Add the indices: x^5 × x^2 = x^7
- Answer (i): 12x⁷
- Step 1 (ii): Divide the coefficients: 12 ÷ 3 = 4. Subtract the indices: a^5 ÷ a^-2 = a^5-(-2) = a^7
- Answer (ii): 4a⁷
Worked Example: Writing an Expression as a Single Power
- Question: Write 1/(y × y × y) as a power of y.
- Step 1: y × y × y = y^3
- Step 2: (1/y^3) = y^-3
- Answer: y⁻³
Common Mistakes
MistakeMultiplying the indices as well as the coefficients, e.g. simplifying 4x⁵ × 3x² as 12x¹⁰
Fixindices are added when multiplying powers of the same base, not multiplied: 4x⁵ × 3x² = 12x⁷
MistakeMaking a sign error when dividing by a negative index, e.g. simplifying a⁵ ÷ a⁻² as a³
Fixdividing subtracts the second index from the first: 5 - (-2) = 7, so a⁵ ÷ a⁻² = a⁷
Solving Simple Exponential Equations
When the unknown appears as an index, write the other side of the equation as a power of the same base first - then the indices themselves must be equal.
- If a^x = a^n for the same base a, then x = n
- A fraction can often be rewritten as a negative power of the base, e.g. 1/9 = 3⁻²
Worked Example: Solving for an Index
- Question: Solve.
(i) 2ˣ = 64 (ii) 3ˣ = 1/9- Step 1 (i): Write 64 as a power of 2: 64 = 2^6
- Answer (i): x = 6
- Step 1 (ii): Write 1/9 as a power of 3: (1/9) = 3^-2
- Answer (ii): x = -2
Common Mistakes
MistakeNot recognising that a fraction can be written as a negative power, and getting stuck on equations like 3ˣ = 1/9
Fixrewrite the fraction using the same base first: 1/9 = 1/3² = 3⁻², so x = -2
MistakeSetting x equal to the number on the other side, e.g. solving 2ˣ = 64 as x = 64
Fixx is the index, not the whole value - rewrite 64 as a power of 2 first (64 = 2⁶), then x equals that index, 6
Fractional and Negative Indices in Algebraic Expressions
Extended Only
A fractional or negative index applied to a whole algebraic term uses the same power-of-a-product law as before - the power applies to the coefficient and the letter part separately.
- Apply the outer power to both the coefficient and the variable part: (k x^n)^p = k^p × x^np
- Deal with a fractional index in two stages: the denominator of the fraction is a root, and the numerator is a power
Worked Example: Combining Fractional and Negative Indices
- Question: Simplify (8x^6)^-(2/3).
- Step 1: Apply the power to each factor: (8x^6)^-(2/3) = 8^-(2/3) × (x^6)^-(2/3)
- Step 2: Evaluate the number part: 8^-(2/3) = (8^(1/3))^-2 = 2^-2 = (1/4)
- Step 3: Evaluate the letter part: (x^6)^-(2/3) = x^6 × -(2/3) = x^-4
- Step 4: Combine: (1/4) × x^-4 = (1/4x^4)
- Answer: 1/(4x⁴)
Common Mistakes
MistakeApplying the outer power to only the coefficient or only the variable, not both
Fixa power applied to a product distributes to every factor: (8x^6)^-(2/3) needs the power applied to both 8 and x⁶ separately
MistakeOnly taking the root, or only applying the outer power, but not both parts of a fractional index
Fixa fractional index like -2/3 means both a root (denominator, cube root) and a power (numerator, squared then reciprocated) - both steps are needed
Solving Exponential Equations by Matching Bases
Extended Only
When both sides of an equation are powers with different bases, rewriting one side so both bases match turns the problem into an ordinary linear equation in the exponents.
- Rewrite one side so both sides share the same base, using the laws of indices
- Once the bases match, set the two exponents equal to each other and solve the resulting equation
Worked Example: Matching Bases to Solve an Equation
- Question: Solve. 5^x = 125^1-3x
- Step 1: Write 125 as a power of 5: 125 = 5^3, so 125^1-3x = 5^3(1-3x)
- Step 2: The bases now match, so equate the exponents: x = 3(1-3x) = 3 - 9x
- Step 3: Solve: x + 9x = 3 ⇒ 10x = 3 ⇒ x = (3/10)
- Answer: x = 3/10
Worked Example: A Second Matching-Bases Equation
- Question: Solve. 3^x+2 = 9^x
- Step 1: Write 9 as a power of 3: 9 = 3^2, so 9^x = 3^2x
- Step 2: Equate the exponents: x+2 = 2x
- Step 3: Solve: 2 = 2x - x = x
- Answer: x = 2
Common Mistakes
MistakeEquating the exponents before rewriting both sides with the same base, e.g. jumping straight from 5^x = 125^1-3x to x = 1-3x
Fixthe bases must be identical before the exponents can be equated - 125 must first be rewritten as 5³
MistakeLosing a bracket when expanding the new exponent, e.g. simplifying 3(1-3x) as 3-3x instead of 3-9x
Fixmultiply every term inside the bracket by the outer number: 3(1-3x) = 3(1) - 3(3x) = 3 - 9x
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