Indices
UK Secondary (7-11)
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Manipulate single terms that involve negative powers
Lesson

We've already learnt the division law which states: $\frac{a^m}{a^n}=a^{m-n}$aman=amn.

But what happens in a case where the power in the denominator is greater than the power in the numerator? For example, if we simplify $\frac{x^8}{x^{10}}$x8x10 using the division law, we get $x^{8-10}=x^{-2}$x810=x2. We are left with a negative power. But what does this mean? Let's expand this example to find out.

 

Worked Example

Simplify the following using index laws, expressing your answer with a positive index: $x^8\div x^{10}$x8÷​x10.

Think: Let's write this in expanded form.

Using the division law, we would get $x^{8-10}=x^{-2}$x810=x2. We can see in the picture above that when we cancel out the common factors, the numerator is $1$1 and the denominator is $x^2$x2. So the negative power of $x^{-2}$x2 can be expressed with a positive index as $\frac{1}{x^2}$1x2.

Do: $\frac{x^8}{x^{10}}=\frac{1}{x^2}$x8x10=1x2

Reflect: We can see that, in general, $a^{-n}$an is the reciprocal (that is, the flipped version) of $a^n$an. So $a^{-n}$an actually represents a fraction, not a negative number.

 

The negative index law

$a^{-n}=\frac{1}{a^n}$an=1an, where $a\ne0$a0.

 

Common Fractions

Consider the number represented by $2^{-3}$23. What is this number?

Well we can rewrite it in positive index form by taking the reciprocal: $2^{-3}=\frac{1}{2^3}$23=123.

But $\frac{1}{2^3}=\frac{1}{8}$123=18. So we have found another way to represent the fraction $\frac{1}{8}$18, which is $2^{-3}=\frac{1}{8}$23=18.

 

Examples

Question 1

Express $3^{-1}$31 as a fraction in simplest form.

Question 2

Express $2^{-8}$28 in the form $\frac{1}{x^y}$1xy, using positive indices.

Question 3

Express $\frac{1}{6^4}$164 with a negative index.

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