Algebraic indices and equal-base equations
Extend the laws of powers to variable expressions and solve exponential equations by rewriting a common base.
Core: positive, zero and negative integer indices and simple equal-base equations. Extended also uses fractional indices and more involved exponent equations. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Use zero and negative integer powers.
- Solve a simple linear equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is x⁴ × x³?
What you will learn
- Simplify products and quotients of algebraic powers.
- Solve equal-base index equations without logarithms.
- Use fractional indices in Extended work.
Treat coefficients and variable powers separately
The product (3x²)(4x⁵) is 12x⁷: multiply the coefficients and add the exponents of the common base x. In 15a⁶b²/(3a²b), divide coefficients and subtract exponents to get 5a⁴b, for a and b non-zero in the original quotient. Index laws combine powers of one base, not powers of unrelated variables.
A power of a product acts on every factor: (2x³)² = 4x⁶. A negative power is a reciprocal: x⁻³ = 1/x³ for x ≠ 0. In x² + x³ there is a sum, so the product rule does not apply. Factorising x²(1 + x) is a different, valid operation.
Rewrite an exponential equation with one base
To solve 2ˣ = 32, write 32 = 2⁵. Equal powers of a positive base other than one have equal exponents, giving x = 5. For 4ˣ = 8, rewrite both sides using base two: 2²ˣ = 2³, so 2x = 3 and x = 3/2.
Extended equations may have variable exponents on both sides. In 3²ˣ⁻¹ = 27ˣ⁻², write 27 = 3³, giving 2x − 1 = 3x − 6. The solution is x = 5. The equal-base step is justified because powers of three are one-to-one; this does not apply to base one.
Extended: connect fractional powers to roots
For a positive variable x, x¹⁄² = √x and x²⁄³ = (∛x)². A negative fractional power also takes a reciprocal. Thus x⁻¹⁄² = 1/√x for x > 0. In these lessons, positive bases keep all fractional-power manipulations within their real domains.
The quotient (6x³⁄²)/(2x¹⁄²) is 3x for x > 0. Subtract the fractional exponents exactly before evaluating. With (x⁴)¹⁄², unrestricted real x gives x², whereas (x²)¹⁄² gives |x|. Stating a positive-variable assumption prevents an invalid cancellation of a square root's sign.
Pause and explain
Solve 2ˣ⁺¹ = 8.
Worked example
Simplify (3a²b⁻¹)²/(9a), for a ≠ 0 and b ≠ 0.
Show the worked solution
- Square all factors in the numerator: 9a⁴b⁻².
- Divide by 9a to obtain a³b⁻².
- Write with positive indices: a³/b². The original restrictions remain.
Answer a³/b², a ≠ 0 and b ≠ 0
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Simplify 12x⁵/(3x⁻²), for x ≠ 0.
Give me a hint
Subtract the entire denominator exponent, including its sign.
Compare my reasoning
- Divide the coefficients: 12/3 = 4.
- The index is 5 − (−2) = 7.
- The result is 4x⁷, with the original non-zero restriction.
4x⁷, x ≠ 0
Look for these in your work
- Divided the coefficients.
- Subtracted a negative exponent correctly.
Extended: solve 5ˣ⁺¹ = 25ˣ⁻¹.
Give me a hint
Write 25 as 5² before equating exponents.
Compare my reasoning
- The right side is 5²ˣ⁻².
- Equal bases give x + 1 = 2x − 2.
- Thus x = 3; both sides equal 5⁴ = 625.
x = 3
Look for these in your work
- Multiplied the exponent of the rewritten base.
- Checked the original equation.
A simplified storage model doubles once per hour from 64 MB. After t hours its size is 64 × 2ᵗ MB. When does it reach 1024 MB?
Give me a hint
Divide out the starting value before comparing powers.
Compare my reasoning
- 64 × 2ᵗ = 1024 gives 2ᵗ = 16.
- Write 16 = 2⁴, so t = 4.
- The model reaches 1024 MB after four hours; substituting checks the amount.
4 hours
Look for these in your work
- Separated the initial amount from the growth factor.
- Matched powers rather than using linear growth.
Common mistakes
- Adding coefficients when multiplying powers.
- Forgetting that subtracting a negative exponent increases the index.
- Equating exponents before rewriting a common base.
What can you explain now?
Simplify (2x²)³.
Compare with the explanation
8x⁶
Cube the numerical factor and multiply the variable's nested indices: 2³x²×³.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Simplify an expression with a negative index, then solve an equal-base equation. State any non-zero or positive-variable assumptions.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.