Module 8 (M8) - Algebra - Indices

Before reading this guide, it may be useful to look at M7 Indices.

Rules of Indices when working with the same base

We can add a further rule for fractional indices -

Example

Evaluate \( 25^{\frac{1}{2}}\)

Solution:

\(25^{\frac{1}{2}} = \sqrt{25} = 5\)

Example

Evaluate \(81^{\frac{1}{4}}\times 9^{-1}\)

Solution

\( 81^{\frac{1}{4}} = \sqrt[4]{81} = 3\)

\( 9^{-1} = \frac{1}{9}\)

\(81^{\frac{1}{4}} \times 9^{-1} = 3 \times \frac{1}{9} = \frac{3}{9} = \frac{1}{3}\)

Answer

\(\frac{1}{3}\)

Example

Evaluate \(100^{–\frac{3}{2}}\)

Solution

\(100^{–\frac{3}{2}} = \frac{1}{100^{\frac{3}{2}}} = \frac{1}{\sqrt{(100)}^3} = \frac{1}{1000}\)

Example

Simplify \((a^4)^{–\frac{1}{2}} \times a^{\frac{3}{2}} \)

Solution

\((a^4)^{\mathbf{–\frac{1}{2}}}\times a^{\frac{3}{2}}\)

Use rule 3 – multiply indices to raise a further power

\((a^{\mathbf{4}})^{\mathbf {–\frac{1}{2}}}\times a^{\frac{3}{2}} = a^{\mathbf{-2}} \times a^{\mathbf{\frac{3}{2}}}\)

Now use rule 1 – add the indices (\(= -2 + \frac{3}{2} = -{\frac{1}{2}}\))

\(= a^{–\frac{1}{2}}\)

Example

Solve the equation \( 4^{-y} = \sqrt{16^3}\)

Solution

Rewrite Right Hand Side (RHS) in powers of 4

\(\sqrt{16^3} = 4^3\)

Equate Right Hand Side and Left Hand Side (LHS)

\(4^{-y} = 4^3\)

Answer: \(\mathbf {y = -3}\)