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? For example,
\displaystyle x^5\div x^5 | \displaystyle = | \displaystyle x^{5-5} |
\displaystyle = | \displaystyle 0 |
To think about what value we can assign to the term x^0, let's write this division problem as the fraction \dfrac{x^5}{x^5}. Since the numerator and denominator are the same, the fraction simplifies to 1. Notice that this will also be the case with \dfrac{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, and the result we arrive at by simplifying fractions is 1. This must mean that x^0=1.
There is nothing special about x, so we can extend this observation to any base. This result is summarised by the zero power law.
For any base a, a^0=1. This says that taking the zeroth power of any number will always result in 1.
Evaluate \left(6\times 19\right)^0.
For any base a,a^0=1
This says that taking the zeroth power of any number will always result in 1.
So far we have looked at expressions of the form \dfrac{a^m}{a^n}where m>n and where m=n, and how to simplify them using the division rule and also the zero power rule.
But what happens when m is smaller than n? For example, if we simplified a^3\div a^5 using the division law, we would get 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.
So using the second approach, we can also express a^3\div a^5 with a positive index as \dfrac{1}{a^2}. The result is summarised by the negative index law.
For any base a, \, a^{-x}=\dfrac{1}{a^x},\,a\neq 0.
That is, when raising a base to a negative power:
Take the reciprocal of the expression
Turn the power into a positive
Express 6^{-10} with a positive index.
Simplify \dfrac{\left(5^2\right)^9\times 5^6}{5^{40}}, giving your answer in the form a^n.
For any base a,a^{-x}=\dfrac{1}{a^x},a\neq 0That is, when raising a base to a negative power:
Take the reciprocal of the expression
Turn the power into a positive