Grade 10 — Mathematics 10C › C10U05 Exponents

Grade 10 › C10U05 Exponents › L5
GRADE 10 · ALBERTAOUTCOME AN3

L5 · Simplifying Expressions with Rational Exponents

Learning goal

“I can read a^(m/n) in both of its radical dresses, choose the root-first route to keep numbers small, convert general rational exponents in both directions, and apply all five exponent laws when the exponents are simple fractions.”

Root-first route — evaluating a^(m/n) by taking the root (denominator) before the power (numerator): it shrinks the base early, keeping every number hand-sized. The professional's default in a no-technology unit.
Power-first route — the other dress: a^(m/n) = the nth root of (a^m). Always correct, often heavy: the base grows before the root shrinks it.
Simplest form — this unit's canonical dress: positive exponents, power form unless a value or a radical is requested, exact values as whole numbers or single fractions.
Worked example

Evaluate 8^(2/3) - twice. First, read the exponent with L4's mapping, extended: the denominator (3) is the index - a cube root is involved; the numerator (2) is a power - something gets squared. Route 1 (root first): 8^(2/3) = (cube root of 8)^2 = 2^2 = 4. The base shrank from 8 to 2 before any squaring. Route 2 (power first): 8^(2/3) = cube root of (8^2) = cube root of 64 = 4. Same value - but the route passed through 64. Verdict, stated plainly: in a no-technology unit, take the root first whenever the root is exact. The dresses are equal (both are the same number, by the laws); the workloads are not.

Check yourself — explain it

For 27^(2/3), why is the root-first route easier even though both routes are correct?

Paired problem — example still visible

Evaluate 16^(3/4) -> 8 (root first: fourth root of 16 = 2; then 2^3 = 8)

Faded completion — last step is yours

"32^(2/5) by the root-first route: the fifth root of 32 is ___; then raise it to the 2nd power: ___^2 = ___" -> 2; 2; 4

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