<meta http-equiv="refresh" content="1; url=/nojavascript/"> Exponent Properties Involving Products | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Second Edition Go to the latest version.

# 8.1: Exponent Properties Involving Products

Created by: CK-12

## Learning Objectives

• Use the product of a power property.
• Use the power of a product property.
• Simplify expressions involving product properties of exponents.

## Introduction

Back in chapter 1, we briefly covered expressions involving exponents, like $3^5$ or $x^3$. In these expressions, the number on the bottom is called the base and the number on top is the power or exponent. The whole expression is equal to the base multiplied by itself a number of times equal to the exponent; in other words, the exponent tells us how many copies of the base number to multiply together.

Example 1

Write in exponential form.

a) $2 \cdot 2$

b) $(-3)(-3)(-3)$

c) $y \cdot y \cdot y \cdot y \cdot y$

d) $(3a)(3a)(3a)(3a)$

Solution

a) $2 \cdot 2 = 2^2$ because we have 2 factors of 2

b) $(-3)(-3)(-3) = (-3)^3$ because we have 3 factors of (-3)

c) $y \cdot y \cdot y \cdot y \cdot y = y^5$ because we have 5 factors of $y$

d) $(3a)(3a)(3a)(3a)=(3a)^4$ because we have 4 factors of $3a$

When the base is a variable, it’s convenient to leave the expression in exponential form; if we didn’t write $x^7$, we’d have to write $x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$ instead. But when the base is a number, we can simplify the expression further than that; for example, $2^7$ equals $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$, but we can multiply all those 2’s to get 128.

Let’s simplify the expressions from Example 1.

Example 2

Simplify.

a) $2^2$

b) $(-3)^3$

c) $y^5$

d) $(3a)^4$

Solution

a) $2^2 = 2 \cdot 2 =4$

b) $(-3)^3 = (-3)(-3)(-3)=-27$

c) $y^5$ is already simplified

d) $(3a)^4 = (3a)(3a)(3a)(3a) = 3 \cdot 3 \cdot 3 \cdot 3 \cdot a \cdot a \cdot a \cdot a = 81 a^4$

Be careful when taking powers of negative numbers. Remember these rules:

$(negative \ number) \cdot (positive \ number) = negative \ number\! \\(negative \ number) \cdot (negative \ number) = positive \ number$

So even powers of negative numbers are always positive. Since there are an even number of factors, we pair up the negative numbers and all the negatives cancel out.

$(-2)^6 = (-2)(-2)(-2)(-2)(-2)(-2) = \underbrace{ (-2)(-2) }_{+4} \cdot \underbrace{ (-2)(-2) }_{+4} \cdot \underbrace{ (-2)(-2) }_{+4} = +64$

And odd powers of negative numbers are always negative. Since there are an odd number of factors, we can still pair up negative numbers to get positive numbers, but there will always be one negative factor left over, so the answer is negative:

$(-2)^5 = (-2)(-2)(-2)(-2)(-2) = \underbrace{(-2)(-2)}_{+4} \cdot \underbrace{(-2)(-2)}_{+4} \cdot \underbrace{(-2)}_{-2} = -32$

## Use the Product of Powers Property

So what happens when we multiply one power of $x$ by another? Let’s see what happens when we multiply $x$ to the power of 5 by $x$ cubed. To illustrate better, we’ll use the full factored form for each:

$\underbrace{(x \cdot x \cdot x \cdot x \cdot x)}_{x^5} \cdot \underbrace{(x \cdot x \cdot x)}_{x^3} = \underbrace{(x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x)}_{x^8}$

So $x^5 \times x^3 = x^8$. You may already see the pattern to multiplying powers, but let’s confirm it with another example. We’ll multiply $x$ squared by $x$ to the power of 4:

$\underbrace{(x \cdot x)}_{x^2} \cdot \underbrace{(x \cdot x \cdot x \cdot x)}_{x^4} = \underbrace{(x \cdot x \cdot x \cdot x \cdot x \cdot x)}_{x^6}$

So $x^2 \times x^4 = x^6$. Look carefully at the powers and how many factors there are in each calculation. $5 \ x$’s times $3 \ x$’s equals $(5 + 3) = 8 \ x$’s. $2 \ x$’s times $4 \ x$’s equals $(2 + 4) = 6 \ x$’s.

You should see that when we take the product of two powers of $x$, the number of $x$’s in the answer is the total number of $x$’s in all the terms you are multiplying. In other words, the exponent in the answer is the sum of the exponents in the product.

Product Rule for Exponents: $x^n \cdot x^m = x^{(n+m)}$

There are some easy mistakes you can make with this rule, however. Let’s see how to avoid them.

Example 3

Multiply $2^2 \cdot 2^3$.

Solution

$2^2 \cdot 2^3 = 2^5 = 32$

Note that when you use the product rule you don’t multiply the bases. In other words, you must avoid the common error of writing $2^2 \cdot 2^3 = 4^5$. You can see this is true if you multiply out each expression: 4 times 8 is definitely 32, not 1024.

Example 4

Multiply $2^2 \cdot 3^3$.

Solution

$2^2 \cdot 3^3 = 4 \cdot 27 = 108$

In this case, we can’t actually use the product rule at all, because it only applies to terms that have the same base. In a case like this, where the bases are different, we just have to multiply out the numbers by hand—the answer is not $2^5$ or $3^5$ or $6^5$ or anything simple like that.

## Use the Power of a Product Property

What happens when we raise a whole expression to a power? Let’s take $x$ to the power of 4 and cube it. Again we’ll use the full factored form for each expression:

$(x^4)^3 & = x^4 \times x^4 \times x^4 \qquad 3 \ factors \ of \ \{x \ to \ the \ power \ 4\} \\(x \cdot x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x) & = x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x = x^{12}$

So $(x^4)^3 = x^{12}$. You can see that when we raise a power of $x$ to a new power, the powers multiply.

Power Rule for Exponents: $(x^n)^m = x^{(n\cdot m)}$

If we have a product of more than one term inside the parentheses, then we have to distribute the exponent over all the factors, like distributing multiplication over addition. For example:

$(x^2y)^4 = (x^2)^4 \cdot (y)^4 = x^8y^4.$

Or, writing it out the long way:

$(x^2 y)^4 = (x^2 y) (x^2 y) (x^2 y) (x^2 y) & = (x \cdot x \cdot y) (x \cdot x \cdot y) (x \cdot x \cdot y) (x \cdot x \cdot y)\\& = x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y = x^8 y^4$

Note that this does NOT work if you have a sum or difference inside the parentheses! For example, $(x+y)^2 \neq x^2+y^2$. This is an easy mistake to make, but you can avoid it if you remember what an exponent means: if you multiply out $(x+y)^2$ it becomes $(x+y)(x+y)$, and that’s not the same as $x^2+y^2$. We’ll learn how we can simplify this expression in a later chapter.

The following video from YourTeacher.com may make it clearer how the power rule works for a variety of exponential expressions:

Example 5

Simplify the following expressions.

a) $3^5 \cdot 3^7$

b) $2^6 \cdot 2$

c) $(4^2)^3$

Solution

When we’re just working with numbers instead of variables, we can use the product rule and the power rule, or we can just do the multiplication and then simplify.

a) We can use the product rule first and then evaluate the result: $3^5 \cdot 3^7 = 3^{12}=531441$.

OR we can evaluate each part separately and then multiply them: $3^5 \cdot 3^7 = 243 \cdot 2187 = 531441$.

b) We can use the product rule first and then evaluate the result: $2^6 \cdot 2 = 2^7 = 128$.

OR we can evaluate each part separately and then multiply them: $2^6 \cdot 2 = 64 \cdot 2 = 128$.

c) We can use the power rule first and then evaluate the result: $(4^2)^3 = 4^6 = 4096$.

OR we can evaluate the expression inside the parentheses first, and then apply the exponent outside the parentheses: $(4^2)^3 = (16)^3 = 4096$.

Example 6

Simplify the following expressions.

a) $x^2 \cdot x^7$

b) $(y^3)^5$

Solution

When we’re just working with variables, all we can do is simplify as much as possible using the product and power rules.

a) $x^2 \cdot x^7 = x^{2+7} = x^9$

b) $(y^3)^5 = y^{3 \times 5}= y^{15}$

Example 7

Simplify the following expressions.

a) $(3x^2 y^3) \cdot (4xy^2)$

b) $(4xyz) \cdot (x^2 y^3) \cdot (2y z^4)$

c) $(2a^3b^3)^2$

Solution

When we have a mix of numbers and variables, we apply the rules to each number and variable separately.

a) First we group like terms together: $(3x^2y^3) \cdot (4xy^2) = (3 \cdot 4) \cdot (x^2 \cdot x) \cdot (y^3 \cdot y^2)$

Then we multiply the numbers or apply the product rule on each grouping: $=12 x^3y^5$

b) Group like terms together: $(4xy z) \cdot (x^2 y^3) \cdot (2y z^4) = (4 \cdot 2) \cdot (x \cdot x^2) \cdot (y \cdot y^3 \cdot y) \cdot (z \cdot z^4)$

Multiply the numbers or apply the product rule on each grouping: $= 8x^3 y^5 z^5$

c) Apply the power rule for each separate term in the parentheses: $(2a^3b^3)^2 = 2^2 \cdot (a^3)^2 \cdot (b^3)^2$

Multiply the numbers or apply the power rule for each term $=4a^6 b^6$

Example 8

Simplify the following expressions.

a) $(x^2)^2 \cdot x^3$

b) $(2x^2y) \cdot (3xy^2)^3$

c) $(4a^2 b^3)^2 \cdot (2ab^4)^3$

Solution

In problems where we need to apply the product and power rules together, we must keep in mind the order of operations. Exponent operations take precedence over multiplication.

a) We apply the power rule first: $(x^2)^2 \cdot x^3 = x^4 \cdot x^3$

Then apply the product rule to combine the two terms: $x^4 \cdot x^3 = x^7$

b) Apply the power rule first: $(2x^2 y) \cdot (3xy^2)^3 = (2x^2y) \cdot (27x^3y^6)$

Then apply the product rule to combine the two terms: $(2x^2 y) \cdot (27 x^3 y^6) = 54x^5y^7$

c) Apply the power rule on each of the terms separately: $(4a^2 b^3)^2 \cdot (2ab^4)^3 = (16a^4 b^6) \cdot (8a^3 b^{12})$

Then apply the product rule to combine the two terms: $(16a^4 b^6) \cdot (8a^3 b^{12}) = 128a^7 b^{18}$

## Review Questions

Write in exponential notation:

1. $4 \cdot 4 \cdot 4 \cdot 4 \cdot 4$
2. $3x \cdot 3x \cdot 3x$
3. $(-2a)(-2a)(-2a)(-2a)$
4. $6 \cdot 6 \cdot 6 \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y$
5. $2 \cdot x \cdot y \cdot 2 \cdot 2 \cdot y \cdot x$

Find each number.

1. $5^4$
2. $(-2)^6$
3. $(0.1)^5$
4. $(-0.6)^3$
5. $(1.2)^2+5^3$
6. $3^2 \cdot (0.2)^3$

Multiply and simplify:

1. $6^3 \cdot 6^6$
2. $2^2 \cdot 2^4 \cdot 2^6$
3. $3^2 \cdot 4^3$
4. $x^2 \cdot x^4$
5. $(-2y^4)(-3y)$
6. $(4a^2)(-3a)(-5a^4)$

Simplify:

1. $(a^3)^4$
2. $(xy)^2$
3. $(3a^2b^3)^4$
4. $(-2xy^4z^2)^5$
5. $(-8x)^3(5x)^2$
6. $(4a^2)(-2a^3)^4$
7. $(12xy)(12xy)^2$
8. $(2xy^2)(-x^2y)^2(3x^2y^2)$

Feb 23, 2012

Mar 05, 2014