10.2 Use Multiplication Properties of Exponents

Learning objectives

Learning objectives

• Simplify expressions with exponents

• Simplify expressions using the Product Property of Exponents

• Simplify expressions using the Power Property of Exponents

• Simplify expressions using the Product to a Power Property

• Simplify expressions by applying several properties

• Multiply monomials

Simplify Expressions with Exponents

Remember that an exponent indicates repeated multiplication of the same quantity. For example, ${2}^{4}$ means to multiply four factors of $2,$ so ${2}^{4}$ means $2\cdot{}2\cdot{}2\cdot{}2.$ This format is known as exponential notation.

In the expression ${a}^{m},$ the exponent tells us how many times we use the base $a$ as a factor.

Before we begin working with variable expressions containing exponents, let’s simplify a few expressions involving only numbers.

Example 10.11

Solution

1. Interpret the exponents. A positive exponent tells how many equal factors are multiplied, so $5^3$ uses three factors of $5$ and $9^1$ uses one factor of $9$.
2. Simplify part ⓐ. The repeated product is $5\cdot{}5\cdot{}5=125$.
3. Simplify part ⓑ. A single factor of $9$ is already in simplest form, so $9^1=9$.

Takeaway: A positive exponent counts repeated factors, so ${5}^{3}=125$ and ${9}^{1}=9$.

Example 10.12

Solution

1. Interpret both squares. Squaring a number multiplies it by itself, whether the number is a fraction or a decimal.
2. Square the fraction in part ⓐ. Multiplying numerator by numerator and denominator by denominator gives $\frac{7\cdot{}7}{8\cdot{}8}=\frac{49}{64}$.
3. Square the decimal in part ⓑ. Multiplying $0.74$ by $0.74$ gives $0.5476$, so the two results are $\frac{49}{64}$ and $0.5476$.

Takeaway: Squaring a fraction squares its numerator and denominator, while squaring a decimal multiplies it by itself, giving $\frac{49}{64}$ and $0.5476$.

Example 10.13

Solution

1. Identify the two bases. In ${(-3)}^{4}$, the parentheses make $-3$ the entire base, while in ${-3}^{4}$ only $3$ is raised to the fourth power.
2. Simplify part ⓐ. Four negative factors give a positive result, so ${(-3)}^{4}=(-3)(-3)(-3)(-3)=81$.
3. Simplify part ⓑ. First compute $3^4=81$ and then take its opposite, so ${-3}^{4}=-81$.
4. Compare the results. The different placements of the parentheses make the negative sign part of the base in part ⓐ but not in part ⓑ.

Takeaway: Parentheses determine whether the negative sign is raised to the power, so ${(-3)}^{4}=81$ but ${-3}^{4}=-81$.

Simplify Expressions Using the Product Property of Exponents

You have seen that when you combine like terms by adding and subtracting, you need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different, too. We’ll derive the properties of exponents by looking for patterns in several examples. All the exponent properties hold true for any real numbers, but right now we will only use whole number exponents.

First, we will look at an example that leads to the Product Property.

The base stayed the same and we added the exponents. This leads to the Product Property for Exponents.

An example with numbers helps to verify this property.

Example 10.14

Solution

1. Identify the common base. Both factors in ${x}^{5}\cdot{x}^{7}$ have base $x$, so the Product Property of Exponents applies.
2. Add the exponents. The rule ${x}^{m}\cdot{x}^{n}={x}^{m+n}$ gives ${x}^{5+7}={x}^{12}$.
3. Check the product by expanding factors. Writing the expression as $x\cdot{}x\cdot{}x\cdot{}x\cdot{}x$ times $x\cdot{}x\cdot{}x\cdot{}x\cdot{}x\cdot{}x$ confirms that the same base occurs $12$ times.
4. State the simplified result. The figure highlights the combined exponent $5+7$, and the product simplifies to ${x}^{12}$.

Takeaway: Multiply powers with the same base by adding their exponents, but only when the bases match exactly.

Example 10.15

Solution

1. Make the missing exponent explicit. Rewrite $b$ as ${b}^{1}$ so the expression becomes ${b}^{4}\cdot{b}^{1}$.
2. Apply the Product Property of Exponents. Because both factors have base $b$, add the exponents: ${b}^{4+1}={b}^{5}$.
3. Check by counting factors. The product contains four factors of $b$ from the first term and one factor of $b$ from the second, giving five factors in all.
4. State the simplified result. The figure highlights the exponent $4+1$, and the expression simplifies to ${b}^{5}$.

Takeaway: A variable with no written exponent has exponent $1$, which makes it eligible for the product rule.

Example 10.16

Solution

1. Identify the common base. The two factors $2^7$ and $2^9$ both have base $2$, so their exponents may be added.
2. Apply the Product Property of Exponents. Using ${a}^{m}\cdot{a}^{n}={a}^{m+n}$, the product becomes ${2}^{7+9}={2}^{16}$.
3. Check the exponent count. Repeated multiplication would contain seven factors of $2$ and nine more, for a total of sixteen factors of $2$.
4. State the simplified result. The figure highlights the combined exponent $7+9$, and the expression simplifies to ${2}^{16}$.

Takeaway: The product of powers with an identical base keeps that base and uses the sum of the exponents.

Example 10.17

Solution

1. Identify the common base. Both factors in ${y}^{17}\cdot{y}^{23}$ have base $y$.
2. Add every exponent. The Product Property gives ${y}^{17+23}={y}^{40}$.
3. Check by counting repeated factors. A complete expansion contains $17+23=40$ copies of $y$, so no additional factors remain.
4. State the simplified result. The figure highlights the combined exponent $17+23$, and the product simplifies to ${y}^{40}$.

Takeaway: For any number of factors with the same base, add all the exponents in one sum.

We can extend the Product Property of Exponents to more than two factors.

Example 10.18

Solution

1. Group factors by their common base. The expression ${x}^{3}\cdot{x}^{4}\cdot{x}^{2}$ contains only powers of $x$.
2. Add the exponents. Applying the Product Property repeatedly gives ${x}^{3+4+2}={x}^{9}$.
3. Check the total number of factors. Three factors contribute $3$, $4$, and $2$ copies of $x$, for $9$ copies altogether.
4. State the simplified result. The figure highlights the exponent sum, and the product simplifies to ${x}^{9}$.

Takeaway: When multiplying powers of one base, include every exponent in the sum rather than stopping after two factors.

Simplify Expressions Using the Power Property of Exponents

Now let’s look at an exponential expression that contains a power raised to a power. See if you can discover a general property.

We multiplied the exponents. This leads to the Power Property for Exponents.

An example with numbers helps to verify this property.

Example 10.19

Solution

1. Apply the Power Property to part ⓐ. For ${({x}^{5})}^{7}$, multiply the exponents to obtain ${x}^{5\cdot{}7}={x}^{35}$.
2. Apply the Power Property to part ⓑ. For ${({3}^{6})}^{8}$, multiply the exponents to obtain ${3}^{6\cdot{}8}={3}^{48}$.
3. Check each exponent product. The first product is $5\cdot{}7=35$, and the second is $6\cdot{}8=48$; the figures display these exponent products in red.
4. State both simplified results. Part ⓐ is ${x}^{35}$ and part ⓑ is ${3}^{48}$.

Takeaway: In a power of a power, keep the outer base and multiply the inner and outer exponents.

Simplify Expressions Using the Product to a Power Property

We will now look at an expression containing a product that is raised to a power. Look for a pattern.

The exponent applies to each of the factors. This leads to the Product to a Power Property for Exponents.

An example with numbers helps to verify this property:

Example 10.20

Solution

1. Distribute the outer exponent. Rewrite ${(-11x)}^{2}$ as ${(-11)}^{2}{x}^{2}$ because the exponent applies to both factors inside the parentheses.
2. Evaluate the numerical power. Using $(-a)^n=a^n$ when $n$ is even, ${(-11)}^{2}=(-11)(-11)=121$.
3. Keep the variable exponent unchanged. The factor $x$ becomes $x^2$, since its exponent is $1$ and $2\cdot{}1=2$.
4. State the simplified result. The figure highlights the two exponents, and the expression simplifies to $121{x}^{2}$.

Takeaway: An outer power acts on every factor inside the parentheses, including both the coefficient and the variable.

Example 10.21

Solution

1. Distribute the outer exponent. Rewrite ${(3xy)}^{3}$ as ${3}^{3}{x}^{3}{y}^{3}$ by applying the Product to a Power Property to every factor.
2. Evaluate the numerical factor. The coefficient becomes ${3}^{3}=3\cdot{}3\cdot{}3=27$.
3. Check the variable powers. Each variable has exponent $1$ inside the parentheses, so multiplying by the outer exponent $3$ gives $x^3$ and $y^3$.
4. State the simplified result. The figure shows the exponents of $x$ and $y$ in red, and the expression simplifies to $27{x}^{3}{y}^{3}$.

Takeaway: To raise a product to a power, raise every factor inside the parentheses to that same power.

Simplify Expressions by Applying Several Properties

We now have three properties for multiplying expressions with exponents. Let’s summarize them and then we’ll do some examples that use more than one of the properties.

Example 10.22

Solution

1. Apply the power property inside each factor. Multiplying the inner and outer exponents gives ${({x}^{2})}^{6}={x}^{12}$ and ${({x}^{5})}^{4}={x}^{20}$.
2. Combine the powers of the same base. The product property adds exponents of like bases, so $x^{12}\cdot{x}^{20}=x^{12+20}$.
3. State the simplified expression. The combined exponent is $32$, giving the final result $x^{32}$.

Takeaway: For powers of powers, multiply the exponents, so ${({x}^{2})}^{6}{({x}^{5})}^{4}={x}^{12}{x}^{20}={x}^{32}$.

Example 10.23

Solution

1. Distribute the outer square. The power of a product applies to each factor, so ${(-7{x}^{3}{y}^{4})}^{2}={(-7)}^{2}{({x}^{3})}^{2}{({y}^{4})}^{2}$.
2. Simplify the numerical factor. The square of $-7$ is $49$, since an even power makes the sign positive.
3. Multiply the variable exponents. The power property gives $(x^3)^2=x^6$ and $(y^4)^2=y^8$, so the result is $49{x}^{6}{y}^{8}$.

Takeaway: Squaring a product squares every factor, so ${(-7{x}^{3}{y}^{4})}^{2}=49{x}^{6}{y}^{8}$.

Example 10.24

Solution

1. Raise the first parenthesized product. The outside square applies to both factors in $6n$, giving ${(6n)}^{2}=6^{2}n^{2}=36n^2$.
2. Rewrite the whole product. This produces $36n^2\cdot{}4n^3$, with the second exponent still applying only to $n$.
3. Multiply the numerical coefficients. The constants give $36\cdot{}4=144$.
4. Combine the powers of $n$. Since $n^2$ and $n^3$ have the same base, add their exponents to obtain the final result $144n^5$.

Takeaway: An exponent outside parentheses applies to every factor inside, so $(6n)^2(4n^3)=36n^2\cdot4n^3=144n^5$.

Example 10.25

Solution

1. Raise every factor to its outer power. Distributing the powers gives $3^4(p^2)^4q^4\cdot{}2^3p^3(q^2)^3$.
2. Simplify each powered factor. The numerical powers are $3^4=81$ and $2^3=8$, while the power property gives $p^8q^4\cdot{}p^3q^6$.
3. Multiply like bases. Adding exponents for $p$ gives $8+3=11$, and adding exponents for $q$ gives $4+6=10$.
4. Combine the numerical coefficients. Since $81\cdot{}8=648$, the simplified product is $648p^{11}q^{10}$.

Takeaway: Raise each factor inside a product to the same power, then multiply coefficients and add exponents of like bases, giving $648p^{11}q^{10}$.

Multiply Monomials

Since a monomial is an algebraic expression, we can use the properties for simplifying expressions with exponents to multiply the monomials.

Example 10.26

Solution

1. Separate coefficients and variable bases. Rearranging by the commutative property gives $4\cdot(-5)\cdot{x}^{2}\cdot{x}^{3}$.
2. Multiply the coefficients. The product $4\cdot(-5)$ is $-20$, so the sign becomes negative.
3. Add the exponents of $x$. The powers $x^2$ and $x^3$ combine as $x^{2+3}=x^5$, giving the final product $-20{x}^{5}$.

Takeaway: Multiplying monomials multiplies coefficients and adds exponents of the same variable, so $(4x^2)(-5x^3)=-20x^5$.

Example 10.27

Solution

1. Separate the coefficients and variable bases. Rearranging the factors gives $\frac{3}{4}\cdot{}12\cdot{c}^{3}\cdot{}c\cdot{}d\cdot{}d^2$.
2. Simplify the numerical coefficient. The product $\frac{3}{4}\cdot{}12$ equals $9$ after cancelling the common factor of $4$.
3. Add exponents for each variable. The powers of $c$ give $c^{3+1}=c^4$, and the powers of $d$ give $d^{1+2}=d^3$.
4. State the product. Combining the simplified coefficient with the variable powers gives $9{c}^{4}{d}^{3}$.

Takeaway: Multiplying monomials multiplies numerical coefficients and adds exponents of matching variables, so $(\frac{3}{4}c^3d)(12cd^2)=9c^4d^3$.

Section 10.2 Exercises

Section 10.2 Exercises

Simplify Expressions with Exponents

In the following exercises, simplify each expression with exponents.

Simplify Expressions Using the Product Property of Exponents

In the following exercises, simplify each expression using the Product Property of Exponents.

Simplify Expressions Using the Power Property of Exponents

In the following exercises, simplify each expression using the Power Property of Exponents*.*

Simplify Expressions Using the Product to a Power Property

In the following exercises, simplify each expression using the Product to a Power Property.

Simplify Expressions by Applying Several Properties

In the following exercises, simplify each expression.

Multiply Monomials

In the following exercises, multiply the following monomials.

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

1. Simplify:

1. ⓐ $\,{4}^{3}$

2. ⓑ $\,{11}^{1}$

2. Simplify:

1. ⓐ $\,{3}^{4}$

2. ⓑ $\,{21}^{1}$

3. Simplify:

1. ⓐ $\,{(\frac{5}{8})}^{2}$

2. ⓑ $\,{(0.67)}^{2}$

4. Simplify:

1. ⓐ $\,{(\frac{2}{5})}^{3}$

2. ⓑ $\,{(0.127)}^{2}$

5. Simplify:

1. ⓐ $\,{(-2)}^{4}$

2. ⓑ $\,{-2}^{4}$

1. Simplify:

1. ⓐ $\,{(-8)}^{2}$

2. ⓑ $\,{-8}^{2}$

2. Simplify: ${x}^{7}\cdot{x}^{8}.$

3. Simplify: ${x}^{5}\cdot{x}^{11}.$

4. Simplify: ${p}^{9}\cdot{}p.$

5. Simplify: $m\cdot{m}^{7}.$

1. Simplify: $6\cdot{6}^{9}.$

2. Simplify: ${9}^{6}\cdot{9}^{9}.$

3. Simplify: ${y}^{24}\cdot{y}^{19}.$

4. Simplify: ${z}^{15}\cdot{z}^{24}.$

5. Simplify: ${x}^{7}\cdot{x}^{5}\cdot{x}^{9}.$

1. Simplify: ${y}^{3}\cdot{y}^{8}\cdot{y}^{4}.$${}^{}$

2. Simplify:

1. ⓐ $\,{({x}^{7})}^{4}$

2. ⓑ $\,{({7}^{4})}^{8}$

3. Simplify:

1. ⓐ $\,{({x}^{6})}^{9}$

2. ⓑ $\,{({8}^{6})}^{7}$

4. Simplify: ${(-14x)}^{2}.$

5. Simplify: ${(-12a)}^{2}.$

1. Simplify: ${(-4xy)}^{4}.$

2. Simplify: ${(6xy)}^{3}.$

3. Simplify: ${({x}^{4})}^{3}{({x}^{7})}^{4}.$

4. Simplify: ${({y}^{9})}^{2}{({y}^{8})}^{3}.$

5. Simplify: ${(-8{x}^{4}{y}^{7})}^{3}.$

1. Simplify: ${(-3{a}^{5}{b}^{6})}^{4}.$

2. Simplify: ${(7n)}^{2}(2{n}^{12}).$

3. Simplify: ${(4m)}^{2}(3{m}^{3}).$

4. Simplify: ${({u}^{3}{v}^{2})}^{5}{(4u{v}^{4})}^{3}.$

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

1. Multiply: $(7{x}^{7})(-8{x}^{4}).$

2. Multiply: $(-9{y}^{4})(-6{y}^{5}).$

3. Multiply: $(\frac{4}{5}\,{m}^{4}{n}^{3})(15m{n}^{3}).$

4. Multiply: $(\frac{2}{3}\,{p}^{5}q)(18{p}^{6}{q}^{7}).$

5. ${4}^{5}$

1. ${(\frac{1}{2})}^{2}$

2. ${-5}^{4}$

3. ${-10}^{4}$

4. ${(-\frac{2}{3})}^{3}$

5. $-{0.5}^{2}$

1. ${x}^{3}\cdot{x}^{6}$

2. $a\cdot{a}^{4}$

3. ${3}^{5}\cdot{3}^{9}$

4. $z\cdot{z}^{2}\cdot{z}^{3}$

5. ${x}^{a}\cdot{x}^{2}$

1. ${y}^{a}\cdot{y}^{b}$

2. ${({u}^{4})}^{2}$

3. ${({y}^{5})}^{4}$

4. ${({10}^{2})}^{6}$

5. ${({x}^{15})}^{6}$

1. ${({x}^{2})}^{y}$

2. ${({5}^{x})}^{y}$

3. ${(5a)}^{2}$

4. ${(-6m)}^{3}$

5. ${(4rs)}^{2}$

1. ${(4xyz)}^{4}$

2. ${({x}^{2})}^{4}\cdot{({x}^{3})}^{2}$

3. ${({a}^{2})}^{6}\cdot{({a}^{3})}^{8}$

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

5. ${(5a)}^{2}{(2a)}^{3}$

1. ${(2{m}^{6})}^{3}$

2. ${(10{x}^{2}y)}^{3}$

3. ${(-2{a}^{3}{b}^{2})}^{4}$

4. ${(\frac{2}{3}\,{x}^{2}y)}^{3}$

5. ${(8{a}^{3})}^{2}{(2a)}^{4}$

1. ${(10{p}^{4})}^{3}{(5{p}^{6})}^{2}$

2. ${(\frac{1}{2}\,{x}^{2}{y}^{3})}^{4}{(4{x}^{5}{y}^{3})}^{2}$

3. ${(3{m}^{2}n)}^{2}{(2m{n}^{5})}^{4}$

4. $(12{x}^{2})(-5{x}^{4})$

5. $(-8{u}^{6})(-9u)$

1. $(\frac{1}{5}\,{r}^{8})(20{r}^{3})$

2. $(4{a}^{3}b)(9{a}^{2}{b}^{6})$

3. $(\frac{4}{7}\,x{y}^{2})(14x{y}^{3})$

4. $(\frac{2}{3}\,{x}^{2}y)(\frac{3}{4}\,x{y}^{2})$

5. Email Janet emails a joke to six of her friends and tells them to forward it to six of their friends, who forward it to six of their friends, and so on. The number of people who receive the email on the second round is ${6}^{2},$ on the third round is ${6}^{3},$ as shown in the table. How many people will receive the email on the eighth round? Simplify the expression to show the number of people who receive the email.

1. Use the Product Property for Exponents to explain why $x\cdot{}x={x}^{2}.$

2. Jorge thinks ${(\frac{1}{2})}^{2}$ is $1.$ What is wrong with his reasoning?