Middle Years

Lesson

What happens if we want to divide one term by another and when we perform the subtraction and we are left with a power of $0$0? For example,

$x^5\div x^5$x5÷x5 |
$=$= | $x^{5-5}$x5−5 |

$=$= | $x^0$x0 |

To think about what value we can assign to the term $x^0$`x`0, let's write this division problem as the fraction $\frac{x^5}{x^5}$`x`5`x`5. Since the numerator and denominator are the same, the fraction simplifies to $1$1. Notice that this will also be the case with $\frac{k^{20}}{k^{20}}$`k`20`k`20 or any expression where we are dividing like bases whose powers are the same.

So the result we arrive at by using index laws is $x^0$`x`0, and the result we arrive at by simplifying fractions is $1$1. This must mean that $x^0=1$`x`0=1.

There is nothing special about $x$`x`, so we can extend this observation to any base. This result is summarised by the zero power law.

The zero power law

For any base $a$`a`,

$a^0=1$`a`0=1

This says that taking the zeroth power of any number will always result in $1$1.

So far we have looked at expressions of the form $\frac{a^m}{a^n}$`a``m``a``n` where $m>n$`m`>`n` and where $m=n$`m`=`n`, and how to simplify them using the division rule and also the zero power rule.

But what happens when $m$`m` is smaller than $n$`n`? For example, if we simplified $a^3\div a^5$`a`3÷`a`5 using the division law, we would get $a^{-2}$`a`−2. So what does a negative index mean? Let's expand the example to find out:

Remember that when we are simplifying fractions, we are looking to cancel out common factors in the numerator and denominator. Remember that any number divided by itself is $1$1.

So using the second approach, we can also express $a^3\div a^5$`a`3÷`a`5 with a positive index as $\frac{1}{a^2}$1`a`2. The result is summarised by the negative index law.

Negative index law

For any base $a$`a`,

$a^{-x}=\frac{1}{a^x}$`a`−`x`=1`a``x`, $a\ne0$`a`≠0.

That is, when raising a base to a negative power:

- Take the reciprocal of the expression
- Turn the power into a negative

Evaluate $\left(6\times19\right)^0$(6×19)0.

Express $6^{-10}$6−10 with a positive index.

Simplify $\frac{\left(5^2\right)^9\times5^6}{5^{40}}$(52)9×56540, giving your answer in the form $a^n$`a``n`.