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# 6.1: Product Rules for Exponents

Difficulty Level: At Grade Created by: CK-12

Suppose you have the expression:

$x\cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot x \cdot x \cdot x \cdot x$

How could you write this expression in a more concise way?

### Guidance

In the expression $x^3$ , the $x$ is called the base and the $3$ is called the exponent . Exponents are often referred to as powers . When an exponent is a positive whole number, it tells you how many times to multiply the base by itself. For example:

• $x^3=x\cdot x \cdot x$
• $2^4=2\cdot 2 \cdot 2 \cdot 2=16$ .

There are many rules that have to do with exponents (often called the Laws of Exponents ) that are helpful to know so that you can work with expressions and equations that involve exponents more easily. Here you will learn two rules that have to do with exponents and products.

RULE: To multiply two terms with the same base, add the exponents.

$& a^m \times a^n = \underleftrightarrow{(a \times a \times \ldots \times a)} \ \underleftrightarrow{(a \times a \times \ldots \times a)}\\& \qquad \qquad \qquad \qquad \ {\color{red}\downarrow} \qquad \qquad \qquad \quad \ {\color{red}\downarrow}\\& \qquad \qquad \qquad {\color{red} m \ \text{factors}} \qquad \qquad {\color{red} n \ \text{factors}}\\& a^m \times a^n = \underleftrightarrow{(a \times a \times a \ldots \times a)}\\& \qquad \qquad \qquad \qquad \ {\color{red}\downarrow}\\& \qquad \qquad \qquad {\color{red} m+n \ \text{factors}}\\& a^m \times a^n=a^{{\color{red}m+n}}$

RULE: To raise a product to a power, raise each of the factors to the power.

$&(ab)^n=\underleftrightarrow{(ab) \times (ab) \times \ldots \times (ab)}\\& \qquad \qquad \qquad \qquad {\color{red}\downarrow}\\& \qquad \qquad \qquad {\color{red}n} \ {\color{red}\text{factors}}\\& (ab)^n=\underleftrightarrow{(a \times a \times \ldots \times a)} \times \underleftrightarrow{(b \times b \times \ldots \times b)}\\& \qquad \qquad \qquad \quad {\color{red}\downarrow} \qquad \qquad \qquad \qquad {\color{red}\downarrow}\\& \qquad \qquad \quad \ {\color{red}n} \ {\color{red}\text{factors}} \qquad \qquad \ {\color{red}n} \ {\color{red}\text{factors}}\\& (ab)^n=a^{{\color{red}n}} b^{{\color{red}n}}$

#### Example A

Evaluate $3^2 \times 3^3$ .

Solution:

$& 3^2 \times 3^3 && \text{The base is} \ 3'.\\& 3^{2+3} && \text{Keep the base of} \ 3' \ \text{and add the exponents.}\\& 3^{\color{red}5} && \text{This answer is in exponential form.}$

The answer can be taken one step further. The base is numerical so the term can be evaluated.

$& 3^5=3 \times 3 \times 3 \times 3 \times 3\\& {\color{red}3^5}={\color{red}243}\\& \boxed{3^2 \times 3^3 = 3^5=243}$

#### Example B

Evaluate $(x^3) (x^6)$ .

Solution:

$& (x^3)(x^6) && \text{The base is} \ x.\\& x^{3+6} && \text{Keep the base of} \ x \ \text{and add the exponents.}\\& x^{\color{red}9} && \text{The answer is in exponential form.}\\& \boxed{(x^3)(x^6)=x^9}$

#### Example C

Evaluate $y^5 \cdot y^2$ .

Solution:

$& y^5 \cdot y^2 && \text{The base is} \ y.\\& y^{5+2} && \text{Keep the base of} \ y \ \text{and add the exponents.}\\& y^{\color{red}7} && \text{The answer is in exponential form.}\\& \boxed{y^5 \cdot y^2=y^7}$

#### Example D

Evaluate $5x^2 y^3 \cdot 3xy^2$ .

Solution:

$& 5x^2 y^3 \cdot 3xy^2 && \text{The bases are} \ x \ \text{and} \ y.\\& 15(x^2 y^3)(xy^2) && \text{Multiply the coefficients -} \ 5 \times 3=15. \ \text{Keep the base of} \ x \ \text{and} \ y \ \text{and add}\\& && \text{the exponents of the same base. If a base does not have a written}\\& && \text{exponent, it is understood as} \ 1'.\\& 15x^{2+1} y^{3+2}\\& 15x^{\color{red}3} y^{\color{red}5} && \text{The answer is in exponential form.}\\& \boxed{5x^2 y^3 \cdot 3xy^2=15x^3y^5}$

#### Concept Problem Revisited

$x\cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot x \cdot x \cdot x \cdot x$ can be rewritten as $x^9 y^5 x^4$ . Then, you can use the rules of exponents to simplify the expression to $x^13 y^5$ . This is certainly much quicker to write!

### Vocabulary

Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers. Some examples are: In the expression $2^5$ , ‘2’ is the base. In the expression $(-3y)^4$ , ‘ $-3y$ ’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times to multiply the base times itself. Some examples are:
In the expression $2^5$ , ‘5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: $2^5=2 \times 2 \times 2 \times 2 \times 2$
In the expression $(-3y)^4$ , ‘4’ is the exponent. It means to multiply $-3y$ times itself 4 times as shown here: $(-3y)^4=-3y \times -3y \times -3y \times -3y$ .
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents when dealing with exponential expressions.

### Guided Practice

Evaluate each of the following:

1. $(-3x)^3$

2. $(5x^2 y^4)^3$

3. $(2^3 \times 3^2)^2$

4. $(3x^{-2} y^5)^3$

1.

$& (-3x)^2 && \text{The base is} \ -3x'.\\& (-3)^{1 \times 2} \cdot (x)^{1 \times 2} && \text{Keep the base of} \ -3x' \ \text{and multiply the exponents of each factor of}\\& && \text{the base by} \ 2.\\& (-3)^{\color{red}2} \cdot (x)^{\color{red}2} && \text{Simplify. Apply the exponents to each factor of the base.}\\& {\color{red}9}x^{\color{red}2} && \text{The answer is in exponential form.}\\& \boxed{(-3x)^2=9x^2}$

2.

$& (5x^2 y^4)^3 && \text{The base is} \ 5x^2 y^4'.\\& (5)^{1 \times 3} \cdot (x)^{2 \times 3} \cdot (y)^{4 \times 3} && \text{Keep the base of} \ 5x^2 y^4' \ \text{and multiply the exponents of each}\\& && \text{factor of the base by} \ 3.\\& (5)^{\color{red}3} \cdot (x)^{\color{red}6} \cdot (y)^{\color{red}12} && \text{Simplify. Apply the exponents to each factor of the base.}\\& {\color{red}125} x^{\color{red}6} y^{\color{red}12} && \text{The answer is in exponential form.}\\& \boxed{(5x^2 y^4)^3 = 125x^6 y^{12}}$

3.

$& (2^3 \times 3^3)^2 && \text{The base is} \ 2^3 \times 3^2'.\\& (2)^{3 \times 2} \cdot (3)^{2 \times 2} && \text{Keep the base of} \ 2^3 \times 3^2' \ \text{and multiply the exponents of each}\\& && \text{factor of the base by} \ 2.\\& 2^{\color{red}6} \times 3^{\color{red}4} && \text{Simplify. Apply the exponents to each factor of the base.}\\& 2^{\color{red}6} \times 3^{\color{red}4} && \text{The answer is in exponential form.}$
The answer can be taken one step further. The base of each factor is numerical so each term can be evaluated. The final answer will be the product of the two answers.
$& 2^6=2 \times 2 \times 2 \times 2 \times 2 \times 2\\& 2^6 = 64\\& 3^4=3 \times 3 \times 3 \times 3\\& 3^4 = 81\\& {\color{red}64 \times 81}={\color{red}5184}\\& \boxed{(2^3 \times 3^3)^2=2^6 \times 3^6=5184}$

4.

$& (3x^{-2} y^5)^3 && \text{The base is} \ 3x^{-2} y^5'.\\& (3)^{1 \times 3} \cdot (x)^{-2 \times 3} \cdot (y)^{5 \times 3} && \text{Keep the base of} \ `3x^{-2} y^5' \ \text{and multiply the exponents of each}\\& && \text{factor of the base by} \ 3.\\& 3^{\color{red}3} \cdot x^{{\color{red}-6}} \cdot y^{{\color{red}15}} && \text{Simplify. Apply the exponents to each factor of the base.}\\& {\color{red}27} x^{{\color{red}-6}} y^{\color{red}15} && \text{Write} \ x \ \text{in the denominator with a positive exponent.}\\& \frac{{\color{red}27} y^{{\color{red}15}}}{x^{\color{red}6}} && \text{The answer has been simplified and is in exponential form.}\\& \boxed{(3x^{-2} y^5)^3 = \frac{27 y^{15}}{x^6}}$

## Date Created:

Jan 16, 2013

Apr 29, 2014
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