Fractional Indices (Roots & Powers)

Unit 2 • Lesson 03 of 05

Fractional Indices (Roots & Powers)

Grade 11 Mathematics • G.C.E. O/L Sri Lanka Local Syllabus

Learning Objectives

  • Interpret unit fractional exponents (\(a^{\frac{1}{n}}\)) as \(n^{\text{th}}\) roots (\(\sqrt[n]{a}\)).
  • Evaluate general fractional powers (\(a^{\frac{m}{n}}\)) and negative fractional powers (\(a^{-\frac{m}{n}}\)).
  • Use prime factorization to simplify fractional exponents without a calculator.

Fractional Index Rules

Unit Fractional Index
\[a^{\frac{1}{n}} = \sqrt[n]{a}\]
General Fractional Index
\[a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}\]
Negative Fractional Index
\[a^{-\frac{m}{n}} = \frac{1}{\left(\sqrt[n]{a}\right)^m}\]
Exam Strategy: In \(a^{\frac{m}{n}}\), the denominator \(n\) is the root and the numerator \(m\) is the power. Always express the base in prime index form first (e.g., write \(27\) as \(3^3\), or \(32\) as \(2^5\)) so the denominator cancels cleanly.

Step-by-Step Worked Examples

Example 1: Evaluate \(27^{\frac{2}{3}}\)
  1. Write \(27\) as a power of its prime factor (\(27 = 3^3\)): \[(3^3)^{\frac{2}{3}}\]
  2. Multiply the exponents using \((a^m)^n = a^{mn}\): \[3^{3 \times \frac{2}{3}} = 3^2 = 9\]
Example 2: Evaluate \(\left(\frac{8}{27}\right)^{-\frac{2}{3}}\)
  1. Apply the negative index rule to invert the fraction: \[\left(\frac{27}{8}\right)^{\frac{2}{3}}\]
  2. Write numerator and denominator in prime index form (\(27 = 3^3\) and \(8 = 2^3\)): \[\left(\frac{3^3}{2^3}\right)^{\frac{2}{3}}\]
  3. Distribute the fractional exponent \(\frac{2}{3}\) to both numerator and denominator: \[\frac{3^{3 \times \frac{2}{3}}}{2^{3 \times \frac{2}{3}}} = \frac{3^2}{2^2} = \frac{9}{4} = 2\frac{1}{4}\]

Practice Drill

1. Evaluate: \(32^{\frac{3}{5}}\)
Show Step-by-Step Answer

Step 1: Express \(32\) in prime index form (\(32 = 2^5\)):
\((2^5)^{\frac{3}{5}}\)

Step 2: Multiply the exponents:
\(2^{5 \times \frac{3}{5}} = 2^3 = \mathbf{8}\)

2. Evaluate: \(16^{-\frac{3}{4}} \times 81^{\frac{1}{4}}\)
Show Step-by-Step Answer

Step 1: Express \(16\) and \(81\) as prime powers (\(16 = 2^4\) and \(81 = 3^4\)):
\((2^4)^{-\frac{3}{4}} \times (3^4)^{\frac{1}{4}}\)

Step 2: Multiply exponents:
\(2^{-3} \times 3^1 = \frac{1}{2^3} \times 3 = \frac{1}{8} \times 3 = \mathbf{\frac{3}{8}}\)

3. Evaluate: \(\left(\frac{25}{49}\right)^{-\frac{1}{2}}\)
Show Step-by-Step Answer

Step 1: Invert the fraction to remove the negative sign:
\(\left(\frac{49}{25}\right)^{\frac{1}{2}} = \sqrt{\frac{49}{25}}\)

Step 2: Take the square root of numerator and denominator:
\(\frac{7}{5} = \mathbf{1\frac{2}{5}}\)

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