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 $a^{-x}=\frac{1}{a^x}$a−x=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×e−4

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

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

 

Examples

Question 1

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

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

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

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

 

Question 2

Question 3

Question 4