We've already learnt the division law which states: $\frac{a^m}{a^n}=a^{m-n}$aman=am−n.
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 $x^{8-10}=x^{-2}$x8−10=x−2. 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 exponent laws, expressing your answer with a positive exponent: $x^8\div x^{10}$x8÷x10.
Think: Let's write this in expanded form.
Using the division law, we would get $x^{8-10}=x^{-2}$x8−10=x−2. 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 $x^{-2}$x−2 can be expressed with a positive exponent 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, $a^{-n}$a−n is the reciprocal (that is, the flipped version) of $a^n$an. So $a^{-n}$a−n actually represents a fraction, not a negative number.
$a^{-n}=\frac{1}{a^n}$a−n=1an, where $a\ne0$a≠0.
Common Fractions
Consider the number represented by $2^{-3}$2−3. What is this number?
Well we can rewrite it in positive exponential form by taking the reciprocal: $2^{-3}=\frac{1}{2^3}$2−3=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}$2−3=18.
Examples
Question 1
Express $3^{-1}$3−1 as a fraction in simplest form.
Question 2
Express $2^{-8}$2−8 in the form $\frac{1}{x^y}$1xy, using positive exponents.
Question 3
Express $\frac{1}{6^4}$164 with a negative exponent.