We've already learnt about the Exponent Multiplication Law, which states $a^x\times a^y=a^{x+y}$ax×ay=ax+y, as well as the negative exponent 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 exponents.
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 exponential 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 exponent law)
= $\frac{1}{q^5}$1q5
Question 2
Express $2y^9\times3y^{-5}$2y9×3y−5 with a positive exponent.
Give your answer in its simplest form.
Question 3
Express $p^{-2}q^3$p−2q3 as a fraction without negative exponents.
Question 4
Simplify the following, writing without negative exponents.
$5p^2q^{-4}\times8p^{-2}q^6$5p2q−4×8p−2q6