UK Secondary (7-11)
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Multiplication law with variable bases and negative powers

We've already learnt about the Index Multiplication Law, which states $a^x\times a^y=a^{x+y}$ax×ay=ax+y, as well as the negative index law, which states $a^{-x}=\frac{1}{a^x}$ax=1ax. Now we are going to combine these rules to simplify expressions which involve multiplication and negative indices.

Consider the expression: $e^7\times e^{-4}$e7×e4

Notice the following: 

  • There is multiplication and the bases are the same (we can apply the multiplication law)
  • One of the powers is negative (we can express the second term with a positive power if we wish)

When negative powers are involved, this opens up choices in how we go about trying to simplify the expression.

With the above example, I have two choices:

One Approach: Add the powers immediately as the bases are the same and we are multiplying 

$e^7\times e^{-4}$e7×e4 $=$= $e^{7+\left(-4\right)}$e7+(4)  
  $=$= $e^{7-4}$e74 (recall that a plus and minus sign next to each other result in a minus)
  $=$= $e^3$e3  


Another Approach: First express the second term with a positive power

$e^7\times e^{-4}$e7×e4 $=$= $e^7\times\frac{1}{e^4}$e7×1e4  
  $=$= $\frac{e^7}{e^4}$e7e4  
  $=$= $e^{7-4}$e74 (subtract the powers using the division rule)
  $=$= $e^3$e3  

Of course, which way you go about it is completely up to you.



Question 1

Simplify the expression, expressing in positive index form: $q^2\times q^{-7}$q2×q7.

Think: $q^2\times q^{-7}=q^{2+\left(-7\right)}$q2×q7=q2+(7)

Do: $q^{2+\left(-7\right)}=q^{-5}$q2+(7)=q5 (Now using the negative index law)

                    = $\frac{1}{q^5}$1q5


Question 2

Express $2y^9\times3y^{-5}$2y9×3y5 with a positive index.

  1. Give your answer in its simplest form.

Question 3

Express $p^{-2}q^3$p2q3 as a fraction without negative indices.

Question 4

Simplify the following, writing without negative indices.



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