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:
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.
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
Express $2y^9\times3y^{-5}$2y9×3y−5 with a positive index.
Give your answer in its simplest form.
Express $p^{-2}q^3$p−2q3 as a fraction without negative indices.
Simplify the following, writing without negative indices.
$5p^2q^{-4}\times8p^{-2}q^6$5p2q−4×8p−2q6
Manipulate rational, exponential, and logarithmic algebraic expressions
Apply algebraic methods in solving problems