New Zealand
Level 7 - NCEA Level 2

# Revision of Index Laws

Lesson

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=amn
• 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: $a^{-m}=\frac{1}{a^m}$am=1am
• The fractional index rule: $a^{\frac{m}{n}}=\sqrt[n]{a^m}$amn=nam
• 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:

 $k^{\frac{1}{2}}\times\left(-2k^3\right)^2$k12​×(−2k3)2 $=$= $k^{\frac{1}{2}}\times\left(-2\times k^3\right)^2$k12​×(−2×k3)2 $=$= $k^{\frac{1}{2}}\times\left(-2\right)^2\times k^{3\times2}$k12​×(−2)2×k3×2 $=$= $k^{\frac{1}{2}}\times4\times k^6$k12​×4×k6 $=$= $4k^{\frac{1}{2}+6}$4k12​+6 $=$= $4k^{\frac{13}{2}}$4k132​

##### question 3

Express in positive index form: $\frac{s^{-4}}{s^2\times s^{-5}}$s4s2×s5

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

Do:

 $\frac{s^{-4}}{s^2\times s^{-5}}$s−4s2×s−5​ $=$= $\frac{s^{-4}}{s^{2+\left(-5\right)}}$s−4s2+(−5)​ $=$= $\frac{s^{-4}}{s^{-3}}$s−4s−3​ $=$= $s^{-4-\left(-3\right)}$s−4−(−3) $=$= $s^{-1}$s−1 $=$= $\frac{1}{s}$1s​

#### 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}$7p4q8×4p4q5

### Outcomes

#### M7-6

Manipulate rational, exponential, and logarithmic algebraic expressions

#### 91261

Apply algebraic methods in solving problems