Indices
UK Secondary (7-11)
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Expressions with algebraic fraction bases and integer powers
Lesson

We previously looked at how to raise fractions to a power. In this chapter we are going to extend that concept to include algebraic fractional bases.

Raising a fraction to a power

For any base number of the form $\frac{a}{b}$ab, and any number $n$n as a power,

$\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$(ab)n=anbn

If $n$n happens to be negative, then we also use the fact $a^{-n}=\frac{1}{a}$an=1a. This gives us the following rule:

$\left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^n$(ab)n=(ba)n

 

For example, the expression $\left(\frac{x}{y}\right)^2$(xy)2 can be expanded in the following way: $\frac{x}{y}\times\frac{x}{y}=\frac{x\times x}{y\times y}$xy×xy=x×xy×y$=$=$\frac{x^2}{y^2}$x2y2. This shows that $\left(\frac{x}{y}\right)^2=\frac{x^2}{y^2}$(xy)2=x2y2

Similarly, an expression like $\left(\frac{3x}{2y}\right)^3$(3x2y)3 expands to $\frac{3x\times3x\times3x}{2y\times2y\times2y}=\frac{27x^3}{8y^3}$3x×3x×3x2y×2y×2y=27x38y3. This shows that $\left(\frac{3x}{2y}\right)^3=\frac{27x^3}{8y^3}$(3x2y)3=27x38y3.

Practice questions

Question 1

Rewrite the following using an index law.

  1. $\left(\frac{a}{b}\right)^3=\frac{\left(\editable{}\right)^{\editable{}}}{\left(\editable{}\right)^{\editable{}}}$(ab)3=()()

Question 2

Simplify, and evaluate where possible, the following expression:

$\left(\frac{8}{b}\right)^{-2}$(8b)2

Question 3

Simplify, and evaluate where possible, the following expression:

$\left(\frac{3n}{6}\right)^3$(3n6)3

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