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Multiplication and division law for integer base and negative powers


We've already looked at algebraic terms and numbers with negative index terms, as well as how to re-write them as positive index terms. Now let's look at how we can apply the multiplication law and the division law to terms with numerical bases.


Multiplication law: $a^m\times a^n=a^{m+n}$am×an=am+n

Division law: $a^m\div a^n=a^{m-n}$am÷​an=amn

Luckily for us the way we use these laws does not change. The only difference now is that we can begin or end with terms that have negative powers.

Recall that raising a base to the power $-1$1 corresponds to taking the reciprocal of the base. For example, $3^{-1}=\frac{1}{3}$31=13 and $15^{-1}=\frac{1}{15}$151=115.

Similarly, raising a base to any negative power $-n$n is the same as taking the reciprocal of the base and raising that to the power of $n$n. For example, $3^{-4}=\frac{1}{3^4}$34=134 and $15^{-2}=\frac{1}{15^2}$152=1152.


Worked example

Simplify the expression $7^8\times7^{-5}$78×75.

Think: Both terms have the same base, so we can use the multiplication law to add the powers. Notice in this case we will be adding a negative number.


$7^8\times7^{-5}$78×75 $=$= $7^{8+\left(-5\right)}$78+(5)
  $=$= $7^{8-5}$785
  $=$= $7^3$73

Reflect: In the second line we have the expression $7^{8-5}$785, which should remind us of an application of the division law. By working backwards we can see that $7^{8-5}=\frac{7^8}{7^5}$785=7875, and by breaking up this fraction we have $\frac{7^8}{7^5}=7^8\times\frac{1}{7^5}$7875=78×175. Since we can rewrite reciprocal terms using negative powers, we arrive back at $7^8\times\frac{1}{7^5}=7^8\times7^{-5}$78×175=78×75, our original expression.


Practice questions

Question 1

Rewrite $10^{-10}\times10^4$1010×104 in the form $a^n$an.

Question 2

Rewrite $10^{-10}\div10^4$1010÷​104 in the form $a^n$an.

Question 3

Evaluate $5^2\div5^{-2}$52÷​52.

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