Expressions with fractional bases and integer powers
In this chapter we are going to look at how to raise fractions to a power.
If we consider something like $\left(\frac{1}{2}\right)^2$(12)2 we know this expands to $\frac{1}{2}\times\frac{1}{2}=\frac{1\times1}{2\times2}=\frac{1}{4}$12×12=1×12×2=14. Similarly, a slightly harder expression like $\left(\frac{2}{3}\right)^3$(23)3 expands to $\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}$23×23×23 giving us $\frac{2\times2\times2}{3\times3\times3}$2×2×23×3×3. So we can see that $\left(\frac{2}{3}\right)^3=\frac{2^3}{3^3}$(23)3=2333.
This can be generalised to give us the following rule:
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 is negative then we also use the fact $a^{-n}=\frac{1}{a}$a−n=1a. This gives us the following rule:
$\left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^n$(ab)−n=(ba)n
Practice questions
Question 1
Simplify the following using index laws, giving your answer as a fully simplified fraction: $\left(\frac{1}{4}\right)^2$(14)2
Question 2
Simplify the following using index laws, giving your answer in simplest index form: $\left(\frac{23}{41}\right)^8$(2341)8
Question 3
Simplify the following using index laws, giving your answer as a fully simplified fraction: $\left(\frac{3}{5}\right)^{-2}$(35)−2