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# 6.1: Product Rules for Exponents

Difficulty Level: At Grade Created by: CK-12

Suppose you have the expression:

xxxxxxxxxyyyyyxxxx\begin{align*}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\end{align*}

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

### Watch This

James Sousa: Exponential Notation

### Guidance

In the expression x3\begin{align*}x^3\end{align*}, the x\begin{align*}x\end{align*} is called the base and the 3\begin{align*}3\end{align*} 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:

• x3=xxx\begin{align*}x^3=x\cdot x \cdot x\end{align*}
• 24=2222=16\begin{align*}2^4=2\cdot 2 \cdot 2 \cdot 2=16\end{align*}.

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.

am×an=(a×a××a) (a×a××a)  m factorsn factorsam×an=(a×a×a×a) m+n factorsam×an=am+n\begin{align*}& 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}}\end{align*}

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

(ab)n=(ab)×(ab)××(ab)n factors(ab)n=(a×a××a)×(b×b××b) n factors n factors(ab)n=anbn\begin{align*}&(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}}\end{align*}

#### Example A

Evaluate 32×33\begin{align*}3^2 \times 3^3\end{align*}.

Solution:

32×3332+335The base is 3.Keep the base of 3 and add the exponents.This answer is in exponential form.\begin{align*}& 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.}\end{align*}

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

35=3×3×3×3×335=24332×33=35=243\begin{align*}& 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}\end{align*}

#### Example B

Evaluate (x3)(x6)\begin{align*}(x^3) (x^6)\end{align*}.

Solution:

(x3)(x6)x3+6x9(x3)(x6)=x9The base is x.Keep the base of x and add the exponents.The answer is in exponential form.\begin{align*}& (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}\end{align*}

#### Example C

Evaluate y5y2\begin{align*}y^5 \cdot y^2\end{align*}.

Solution:

y5y2y5+2y7y5y2=y7The base is y.Keep the base of y and add the exponents.The answer is in exponential form.\begin{align*}& 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}\end{align*}

#### Example D

Evaluate 5x2y33xy2\begin{align*}5x^2 y^3 \cdot 3xy^2\end{align*}.

Solution:

5x2y33xy215(x2y3)(xy2)15x2+1y3+215x3y55x2y33xy2=15x3y5The bases are x and y.Multiply the coefficients - 5×3=15. Keep the base of x and y and addthe exponents of the same base. If a base does not have a writtenexponent, it is understood as 1.The answer is in exponential form.\begin{align*}& 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}\end{align*}

#### Concept Problem Revisited

xxxxxxxxxyyyyyxxxx\begin{align*}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\end{align*} can be rewritten as x9y5x4\begin{align*} x^9 y^5 x^4\end{align*}. Then, you can use the rules of exponents to simplify the expression to x13y5\begin{align*}x^13 y^5\end{align*}. 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 25\begin{align*}2^5\end{align*}, ‘2’ is the base. In the expression (3y)4\begin{align*}(-3y)^4\end{align*}, ‘3y\begin{align*}-3y\end{align*}’ 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 25\begin{align*}2^5\end{align*}, ‘5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25=2×2×2×2×2\begin{align*}2^5=2 \times 2 \times 2 \times 2 \times 2\end{align*}
In the expression (3y)4\begin{align*}(-3y)^4\end{align*}, ‘4’ is the exponent. It means to multiply 3y\begin{align*}-3y\end{align*} times itself 4 times as shown here: (3y)4=3y×3y×3y×3y\begin{align*}(-3y)^4=-3y \times -3y \times -3y \times -3y\end{align*}.
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\begin{align*}(-3x)^3\end{align*}

2. (5x2y4)3\begin{align*}(5x^2 y^4)^3\end{align*}

3. (23×32)2\begin{align*}(2^3 \times 3^2)^2\end{align*}

4. (3x2y5)3\begin{align*}(3x^{-2} y^5)^3\end{align*}

1.

(3x)2(3)1×2(x)1×2(3)2(x)29x2(3x)2=9x2The base is 3x.Keep the base of 3x and multiply the exponents of each factor ofthe base by 2.Simplify. Apply the exponents to each factor of the base.The answer is in exponential form.\begin{align*}& (-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}\end{align*}

2.

Prime causes double exponent: use braces to clarify\begin{align*}& (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}}\end{align*}

3.

\begin{align*}& (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.}\end{align*}
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.
\begin{align*}& 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}\end{align*}

4.

\begin{align*}& (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}}\end{align*}

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Date Created:
Jan 16, 2013