1.04 Revision of index laws

Index laws

Here is a summary of the index laws we know so far:

 

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 examples

example 1

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.

example 2

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.

 

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:

 

Practice questions

Question 1

Question 2

Question 3

Question 4

 

Fractional bases with negative indices

When raising a fractional base to a negative power we can combine the individual rules we have seen.

Example 3

Express the following with a positive index: $\left(\frac{a}{b}\right)^{-3}$(ab​)−3

Think: We want to combine the rules for raising fractions with the rule for negative indices.

That is we want to use the rules $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$(ab​)n=anbn​ and $a^{-n}=\frac{1}{a^n}$a−n=1an​.

Do:

 

We can see that $\left(\frac{a}{b}\right)^{-3}=\frac{b^3}{a^3}$(ab​)−3=b3a3​$=$=$\left(\frac{b}{a}\right)^3$(ba​)3

Reflect:  What has happened is we have found the reciprocal of the expression in the question, and turned the power into a positive. Using this trick will save a lot of time!

 

 

Practice questions

Question 5

Question 6

Question 7