Mathematics • Year 10 • Unit 1 • Lesson 13
Fractional Indices in the Real World
Apply Lesson 13's fractional-index rules to formula-driven contexts: side lengths of cubes from volumes, geometry of squares, scaling in physics, and the same index laws extended to fractions. Then explain in your own words why a^(m/n) is more powerful than ⁿ√(a^m).
1. Word problems
Each problem uses a^(1/n) = ⁿ√a or a^(m/n) = (ⁿ√a)^m. Show how you converted to radical form (or back) at each step.
1.1 — Side length of a cubic water tank. A cubic rainwater tank has volume V = 1000 L = 1 m³. The side length s of a cube is given by s = V^(1/3).
(a) Calculate s in metres.
(b) A larger cubic tank has V = 8 m³. Calculate its side length.
(c) Why is the (1/3) index the natural way to write a cube root in a formula? 3 marks
1.2 — Diagonal of a square solar panel. A square solar panel has area A = 144 cm². The side length is s = A^(1/2).
(a) Calculate s in cm using the fractional-index rule.
(b) The diagonal of a square is d = s × 2^(1/2). Calculate d, leaving 2^(1/2) in exact form (i.e. √2).
(c) Why might an installer prefer the exact form 12√2 cm over a rounded decimal? 3 marks
1.3 — Quadrupling a quantity using a (3/2) index. In senior physics, the Kepler period formula uses T = a^(3/2) where a is the semi-major axis in astronomical units (AU) and T is in years.
(a) For a = 4 AU, evaluate T = 4^(3/2). Show both the root step and the power step.
(b) For a = 9 AU, evaluate T = 9^(3/2). 3 marks
1.4 — Index laws with fractional indices. The lesson states that all index laws apply to fractional indices too.
(a) Simplify x^(5/6) ÷ x^(1/3), leaving your answer with a single fractional index.
(b) Simplify (x^(1/2))⁴ using the power-of-a-power law.
(c) A student claims 8^(2/3) = (8²)^(1/3). Verify by evaluating both sides. 3 marks
1.5 — A composite fraction with both root and power. Evaluate (27/8)^(2/3).
(a) Apply the rule (a/b)^n = a^n / b^n to split the fraction.
(b) Take the cube root of 27 and of 8, then square each.
(c) Combine to give the final fraction. 3 marks
2. Explain your thinking
This question is about communication, not just numbers. Use full sentences. 4 marks
2.1 A friend says: "Why bother with fractional indices when I can just write the root symbol?" Using everything from Lesson 13, explain (i) what fractional indices let you do that radicals alone do not (think about combining with the index laws), (ii) why the lesson states "the denominator gives the root, the numerator gives the power" — what does each part of the fraction represent? — and (iii) why a fractional index like (3/2) appears in real physics formulas (Kepler's third law from Q1.3 is your hook). Refer to "all index laws apply" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cubic tank side lengths
(a) s = 1^(1/3) = ³√1 = 1 m.
(b) s = 8^(1/3) = ³√8 = 2 m (since 2³ = 8).
(c) The (1/3) index is natural because volume = side³, so to undo cubing we need the cube root — and a^(1/3) is exactly the cube root by the fractional-index rule. It also lets the formula stay one symbol and combine with other indices.
1.2 — Solar panel diagonal
(a) s = 144^(1/2) = √144 = 12 cm.
(b) d = 12 × √2 = 12√2 cm (≈ 16.97 cm decimal).
(c) The exact form 12√2 carries no rounding error, so any further calculations (cuts, brackets, framing lengths) stay precise; the decimal version loses precision at every step.
1.3 — Kepler periods
(a) T = 4^(3/2) = (√4)³ = 2³ = 8 years.
(b) T = 9^(3/2) = (√9)³ = 3³ = 27 years.
The fractional index neatly captures "square root, then cube" in one operation.
1.4 — Index laws with fractional indices
(a) x^(5/6) ÷ x^(1/3) = x^(5/6 − 2/6) = x^(3/6) = x^(1/2) (= √x).
(b) (x^(1/2))⁴ = x^((1/2)×4) = x² = x².
(c) LHS: 8^(2/3) = (³√8)² = 2² = 4. RHS: (8²)^(1/3) = 64^(1/3) = ³√64 = 4. Verified.
1.5 — (27/8)^(2/3)
(a) (27/8)^(2/3) = 27^(2/3) / 8^(2/3).
(b) 27^(2/3) = (³√27)² = 3² = 9. 8^(2/3) = (³√8)² = 2² = 4.
(c) Answer: 9/4 (= 2.25).
2.1 — Explain your thinking (sample response)
(i) Fractional indices let you combine roots with the rest of the index toolkit — multiplication, division, power-of-a-power — using a single common notation. With radicals alone, you cannot easily simplify x^(5/6) ÷ x^(1/3); written as fractional indices it becomes a one-line subtraction. (ii) The denominator of the fraction is the root index (the "n" in ⁿ√), and the numerator is the power that follows the root, so a^(m/n) literally reads as "take the n-th root of a, then raise to the m-th power". (iii) In Kepler's third law T = a^(3/2), the (3/2) index encodes "square root, then cube" as a single operation — and because all index laws apply to fractional indices, physicists can manipulate the formula algebraically (e.g. solving for a by raising both sides to the (2/3) power) without ever switching to radical notation.
Marking: 1 for explaining algebraic power (combining with index laws), 1 for correctly naming denominator-root / numerator-power, 1 for a real-formula example (Kepler or similar), 1 for using "all index laws apply".