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.
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
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.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources