Hey, remember indices (or powers)? Those little numbers that float above other numbers or variables? Of course you do, so let's revise the various laws we've learnt about and practice using them!

Rule Recap

The product rule: $a^m\times a^n=a^{m+n}$am×an=am+n
The quotient rule: $a^m\div a^n=a^{m-n}$am÷an=am−n
The zero index rule:$a^0=1$a0=1
The power of a power rule: $\left(a^m\right)^n=a^{mn}$(am)n=amn
The negative index rule: $a^{-m}=\frac{1}{a^m}$a−m=1am
The fractional index rule: $a^{\frac{m}{n}}=\sqrt[n]{a^m}$amn=ⁿ√am
The fractional base rule: $\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}$(ab)m=ambm

Now let's put them to the test in the following problems! Remember to be mindful of your order of operations when doing these problems.

Examples

QUESTION 1

Express in the simplest index form: $\left(\frac{2^0}{4^2}\right)^2$(2042)2

Think: We have too many powers so let's first see if you can express it without the brackets. Also, index form means we need to leave the powers rather than evaluate them.

Do:

$\left(\frac{2^0}{4^2}\right)^2$(2042)2	$=$=	$\frac{2^{0\times2}}{4^{2\times2}}$20×242×2
	$=$=	$\frac{2^0}{4^4}$2044
	$=$=	$\frac{1}{4^4}$144

Question 2

Simplify: $k^{\frac{1}{2}}\times\left(-2k^3\right)^2$k12×(−2k3)2

Think: When a negative number/variable is squared, it becomes positive again

Do:

$k^{\frac{1}{2}}\times\left(-2k^3\right)^2$`k`12×(−2`k`3)2	$=$=	$k^{\frac{1}{2}}\times\left(-2\times k^3\right)^2$`k`12×(−2×`k`3)2
	$=$=	$k^{\frac{1}{2}}\times\left(-2\right)^2\times k^{3\times2}$`k`12×(−2)2×`k`3×2
	$=$=	$k^{\frac{1}{2}}\times4\times k^6$`k`12×4×`k`6
	$=$=	$4k^{\frac{1}{2}+6}$4`k`12+6
	$=$=	$4k^{\frac{13}{2}}$4`k`132

question 3

Express in positive index form: $\frac{s^{-4}}{s^2\times s^{-5}}$s−4s2×s−5

Think: Simplify all powers before trying to convert negative indices into their positive counterparts

Do:

$\frac{s^{-4}}{s^2\times s^{-5}}$`s`−4`s`2×`s`−5	$=$=	$\frac{s^{-4}}{s^{2+\left(-5\right)}}$`s`−4`s`2+(−5)
	$=$=	$\frac{s^{-4}}{s^{-3}}$`s`−4`s`−3
	$=$=	$s^{-4-\left(-3\right)}$`s`−4−(−3)
	$=$=	$s^{-1}$`s`−1
	$=$=	$\frac{1}{s}$1`s`

More Worked Examples

Question 4

Simplify the following, giving your answer with a positive index: $2^2\times2^2$22×22

Question 5

Simplify $p^7\div p^3\times p^5$p7÷p3×p5.

Question 6

Simplify the following, writing without negative indices.

$7p^4q^{-8}\times4p^{-4}q^{-5}$7p4q−8×4p−4q−5

Revision of Index Laws