iGCSE (2021 Edition)

# 1.02 Index laws

Lesson

### Raising a power to a power

Consider the expression $\left(5^2\right)^3$(52)3. What is the resulting power of base $5$5? To find out, have a look at the expanded form of the expression:

 $\left(5^2\right)^3$(52)3 $=$= $\left(5\times5\right)^3$(5×5)3 $=$= $\left(5\times5\right)\times\left(5\times5\right)\times\left(5\times5\right)$(5×5)×(5×5)×(5×5) $=$= $5^6$56

In the expanded form, we can see that $5$5 is multiplied by itself $6$6 times, which was the $5^2$52 multiplied by itself $3$3 times. So we got the total index as $3\times2$3×2.

The index laws
• When multiplying powers with the same base, add the indices.

$a^m\times a^n=a^{m+n}$am×an=am+n

• When dividing powers with the same base, subtract the indices.

$\frac{a^m}{a^n}=a^{m-n}$aman=amn

• When raising a power to a power, multiply the indices.

$\left(a^m\right)^n=a^{m\times n}$(am)n=am×n

• When raising a product to a power, we can rewrite this as a product of powers.

$\left(ab\right)^m=a^mb^m$(ab)m=ambm

• Any non-zero number raised to the power of zero is equal to 1.

$a^0=1,\text{for }a\ne0$a0=1,for a0

Combining one or more of these laws, together with the order of operations, we can simplify more complicated expressions involving powers. Let's look at some examples to see how.

#### Worked example

Simplify the expression $8^2\times8^9\div8^3$82×89÷​83.

Think: This expression contains three terms and two operations. We can perform one operation on two of the terms, then performing the remaining operation on the resulting expression.

Do: Let's begin with the multiplication, then do the division.

 $8^2\times8^9\div8^3$82×89÷​83 $=$= $8^{2+9}\div8^3$82+9÷​83 $=$= $8^{11}\div8^3$811÷​83 $=$= $8^{11-3}$811−3 $=$= $8^8$88

Reflect: We combined the first two terms using the multiplication law. Once we had evaluated the new power we then used the division law to complete the simplification. Alternatively, we could have started by simplifying the last two terms using the division law, then completing the process with the multiplication law, as follows.

 $8^2\times8^9\div8^3$82×89÷​83 $=$= $8^2\times8^{9-3}$82×89−3 $=$= $8^2\times8^6$82×86 $=$= $8^{2+6}$82+6 $=$= $8^8$88

So by the two approaches we get the same answer. This can be useful when one approach clearly involves less work than the other.

#### Worked example

Simplify the expression $\frac{\left(20^3\right)^5}{\left(20^0\right)^7}$(203)5(200)7 using index laws.

Think: Right away we can see that simplifying this expression will involve the power of a power law and the zero power law. Once we complete the simplification we need to make sure our answer has a positive power.

Do:

 $\frac{\left(20^3\right)^5}{\left(20^0\right)^7}$(203)5(200)7​ $=$= $\frac{\left(20^3\right)^5}{1^7}$(203)517​ Using the zero power law $=$= $\frac{\left(20^3\right)^5}{1}$(203)51​ Simplify $1^7$17 $=$= $\left(20^3\right)^5$(203)5 Simplify the quotient $=$= $20^{3\times5}$203×5 Using the power of a power law $=$= $20^{15}$2015 Evaluate the product in the index

Reflect: We could have used the power of a power law for both the numerator and the denominator from the beginning, which would give the step $\frac{20^{15}}{20^0}$2015200. And from there we could use the zero power law to simplify further. In this way we can be creative with the order that we use each law.

## Negative bases

The same rules apply when we are dealing with negative bases, we just need to take care if we are asked to evaluate. We know that the product of two negative numbers is positive, and the product of a positive and a negative number is negative. This means we need to be extra careful when evaluating powers of negative bases.

Watch out!

A negative base raised to an even power will evaluate to a positive answer.

• For example $\left(-3\right)^4=3^4$(3)4=34$=$=$81$81
• $\left(-3\right)^4\ne-3^4$(3)434

A negative base raised to a odd power will evaluate to a negative answer.

• For example $\left(-2\right)^5=-2^5$(2)5=25$=$=$-32$32
• $\left(-2\right)^5=-2^5$(2)5=25

## Fractional bases

When raising a fractional base, we apply the power to both the numerator and the denominator.

Let's consider a simple example like $\left(\frac{1}{2}\right)^2$(12)2. This expands to $\frac{1}{2}\times\frac{1}{2}=\frac{1\times1}{2\times2}$12×12=1×12×2, which evaluates to $\frac{1}{4}$14 or $\frac{1^2}{2^2}$1222.

Similarly, a slightly harder expression like $\left(\frac{2}{3}\right)^3$(23)3 expands to $\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}$23×23×23 giving us $\frac{2\times2\times2}{3\times3\times3}$2×2×23×3×3. So we can see that $\left(\frac{2}{3}\right)^3=\frac{2^3}{3^3}$(23)3=2333.

This can be generalised to give us the following rule:

Raising a fraction to a power

For any base number of the form $\frac{a}{b}$ab, and any number $n$n as a power,

$\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$(ab)n=anbn

#### Practice questions

##### Question 1

Fill in the blank to make the equation true.

1. $11^{11}\times11^{\editable{}}=11^{19}$1111×11=1119

##### Question 2

Simplify the following using index laws, giving your answer in simplest index form: $\left(\frac{29}{41}\right)^7$(2941)7

##### Question 3

Using index laws, evaluate $\left(-4\right)^{11}\div\left(-4\right)^7$(4)11÷​(4)7.

##### Question 4

Simplify $\frac{\left(17^5\right)^8}{17^{32}}$(175)81732, giving your answer in index form.

### Outcomes

#### 0606C4.1A

Perform simple operations with indices.