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Guides d'étude > Intermediate Algebra

Exponent Rules

7.1 Learning Objectives

  • Define and identify notation and terminology for exponents
    • Identify the components of a term containing integer exponents
    • Evaluate expressions containing integer exponents
  • The product rule
    • Use the product rule to multiply exponential expressions
    • Use the quotient rule to divide exponential expressions
  • The power rule
    • Use the power rule to simplify expressions with exponents raised to powers
  • Negative and Zero Exponent Rules
    • Define and use the zero exponent rule
    • Define and use the negative exponent rule
  • Find the Power of a Product and a Quotient
    • Simplify an expression with a product raised to a power
    • Simplify an expression with a quotient raised to a power
    • Combine all the exponents rules to simplify expressions
 
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7.1.1 Anatomy of exponential terms

We use exponential notation to write repeated multiplication. For example [latex]10\cdot10\cdot10[/latex] can be written more succinctly as [latex]10^{3}[/latex]. The 10 in [latex]10^{3}[/latex] is called the base. The 3 in [latex]10^{3}[/latex] is called the exponent. The expression [latex]10^{3}[/latex] is called the exponential expression. Knowing the names for the parts of an exponential expression or term will help you learn how to perform mathematical operations on them.

[latex]\text{base}\rightarrow10^{3\leftarrow\text{exponent}}[/latex]

[latex]10^{3}[/latex] is read as “10 to the third power” or “10 cubed.” It means [latex]10\cdot10\cdot10[/latex], or 1,000. [latex]8^{2}[/latex] is read as “8 to the second power” or “8 squared.” It means [latex]8\cdot8[/latex], or 64. [latex]5^{4}[/latex] is read as “5 to the fourth power.” It means [latex]5\cdot5\cdot5\cdot5[/latex], or 625. [latex]b^{5}[/latex] is read as “b to the fifth power.” It means [latex]{b}\cdot{b}\cdot{b}\cdot{b}\cdot{b}[/latex]. Its value will depend on the value of b. The exponent applies only to the number that it is next to. Therefore, in the expression [latex]xy^{4}[/latex], only the y is affected by the 4. [latex]xy^{4}[/latex] means [latex]{x}\cdot{y}\cdot{y}\cdot{y}\cdot{y}[/latex]. The x in this term is a coefficient of y. If the exponential expression is negative, such as [latex]−3^{4}[/latex], it means [latex]–\left(3\cdot3\cdot3\cdot3\right)[/latex] or [latex]−81[/latex]. If [latex]−3[/latex] is to be the base, it must be written as [latex]\left(−3\right)^{4}[/latex], which means [latex]−3\cdot−3\cdot−3\cdot−3[/latex], or 81. Likewise, [latex]\left(−x\right)^{4}=\left(−x\right)\cdot\left(−x\right)\cdot\left(−x\right)\cdot\left(−x\right)=x^{4}[/latex], while [latex]−x^{4}=–\left(x\cdot x\cdot x\cdot x\right)[/latex]. You can see that there is quite a difference, so you have to be very careful! The following examples show how to identify the base and the exponent, as well as how to identify the expanded and exponential format of writing repeated multiplication.

Example 7.1.A

Identify the exponent and the base in the following terms, then simplify:
  1. [latex]7^{2}[/latex]
  2. [latex]{\left(\frac{1}{2}\right)}^{3}[/latex]
  3. [latex]2x^{3}[/latex]
  4. [latex]\left(-5\right)^{2}[/latex]

Answer:

1) [latex]7^{2}[/latex] The exponent in this term is 2 and the base is 7. To simplify, expand the term: [latex]7^{2}=7\cdot{7}=49[/latex] 2) [latex]{\left(\frac{1}{2}\right)}^{3}[/latex] The exponent on this term is 3, and the base is [latex]\frac{1}{2}[/latex]. To simplify, expand the multiplication and remember how to multiply fractions: [latex]{\left(\frac{1}{2}\right)}^{3}=\frac{1}{2}\cdot{\frac{1}{2}}\cdot{\frac{1}{2}}=\frac{1}{8}[/latex] 3)  [latex]2x^{3}[/latex] The exponent on this term is 3, and the base is x, the 2 is not getting the exponent because there are no parentheses that tell us it is.  This term is in its most simplified form. 4) [latex]\left(-5\right)^{2}[/latex] The exponent on this terms is 2 and the base is [latex]-5[/latex]. To simplify, expand the multiplication: [latex]\left(-5\right)^{2}=-5\cdot{-5}=25[/latex]

In the following video you are provided more examples of applying exponents to various bases.

https://youtu.be/ocedY91LHKU

Evaluate expressions

Evaluating expressions containing exponents is the same as evaluating the linear expressions from earlier in the course. You substitute the value of the variable into the expression and simplify. You can use the order of operations to evaluate the expressions containing exponents. First, evaluate anything in Parentheses or grouping symbols. Next, look for Exponents, followed by Multiplication and Division (reading from left to right), and lastly, Addition and Subtraction (again, reading from left to right). So, when you evaluate the expression [latex]5x^{3}[/latex] if [latex]x=4[/latex], first substitute the value 4 for the variable x. Then evaluate, using order of operations.

Example 7.1.B

Evaluate the following expressions for the given value.
  1. [latex]5x^{3}[/latex] if [latex]x=4[/latex]
  2. [latex]\left(5x\right)^{3}[/latex] if [latex]x=4[/latex]
  3. [latex]x^{3}[/latex] if [latex]x=−4[/latex]
  4. [latex]3^x[/latex] if [latex]x = 4[/latex]
 

Answer: 1) Substitute 4 for the variable x.

[latex]5\cdot4^{3}[/latex]

Evaluate [latex]4^{3}[/latex]. Multiply.

[latex]5\left(4\cdot4\cdot4\right)=5\cdot64=320[/latex]

Answer

[latex-display]5x^{3}=320[/latex] when [latex]x=4[/latex-display]   2) [latex]\left(5x\right)^{3}[/latex] if [latex]x=4[/latex] Substitute 4 for the variable x. notice the how adding parentheses can change the outcome when you are simplifying terms with exponents.

[latex]\left(5\cdot4\right)3[/latex]

Multiply inside the parentheses, then apply the exponent—following the rules of PEMDAS.

[latex]20^{3}[/latex]

Evaluate [latex]20^{3}[/latex].

[latex]20\cdot20\cdot20=8,000[/latex]

Answer

[latex-display]\left(5x\right)3=8,000[/latex] when [latex]x=4[/latex-display]   3)  [latex]x^{3}[/latex] if [latex]x=−4[/latex]. Substitute [latex]−4[/latex] for the variable x.

[latex]\left(−4\right)^{3}[/latex]

Evaluate. Note how placing parentheses around the [latex]−4[/latex] means the negative sign also gets multiplied.

[latex]−4\cdot−4\cdot−4[/latex]

Multiply.

[latex]−4\cdot−4\cdot−4=−64[/latex]

Answer

[latex-display]x^{3}=−64[/latex] when [latex]x=−4[/latex-display] 4)  [latex]3^x[/latex] if [latex]x = 4[/latex] Substitute x = 4 into the exponent.  [latex]3^x=3^4=3\cdot3\cdot3\cdot3=81[/latex]

Answer

[latex-display]3^{x}=81[/latex] when [latex]x=4[/latex-display]

 
Caution Caution! Whether to include a negative sign as part of a base or not often leads to confusion. To clarify whether a negative sign is applied before or after the exponent, here is an example. What is the difference in the way you would evaluate these two terms?
  1. [latex]-{3}^{2}[/latex]
  2. [latex]{\left(-3\right)}^{2}[/latex]
To evaluate 1), you would apply the exponent to the three first, then apply the negative sign last, like this:

[latex]\begin{array}{c}-\left({3}^{2}\right)\\=-\left(9\right) = -9\end{array}[/latex]

To evaluate 2), you would apply the exponent to the 3 and the negative sign:

[latex]\begin{array}{c}{\left(-3\right)}^{2}\\=\left(-3\right)\cdot\left(-3\right)\\={ 9}\end{array}[/latex]

The key to remembering this is to follow the order of operations. The first expression does not include parentheses so you would apply the exponent to the integer 3 first, then apply the negative sign. The second expression includes parentheses, so hopefully you will remember that the negative sign also gets squared.

In the next sections, you will learn how to simplify expressions that contain exponents. Come back to this page if you forget how to apply the order of operations to a term with exponents, or forget which is the base and which is the exponent!

In the following video you are provided with examples of evaluating exponential expressions for a given number.

https://youtu.be/pQNz8IpVVg0

7.1.2 The product rule

Exponential notation was developed to write repeated multiplication more efficiently. There are times when it is easier or faster to leave the expressions in exponential notation when multiplying or dividing. Let’s look at rules that will allow you to do this. For example, the notation [latex]5^{4}[/latex] can be expanded and written as [latex]5\cdot5\cdot5\cdot5[/latex], or 625. And don’t forget, the exponent only applies to the number immediately to its left, unless there are parentheses. What happens if you multiply two numbers in exponential form with the same base? Consider the expression [latex]{2}^{3}{2}^{4}[/latex]. Expanding each exponent, this can be rewritten as [latex]\left(2\cdot2\cdot2\right)\left(2\cdot2\cdot2\cdot2\right)[/latex] or [latex]2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2[/latex]. In exponential form, you would write the product as [latex]2^{7}[/latex]. Notice that 7 is the sum of the original two exponents, 3 and 4. What about [latex]{x}^{2}{x}^{6}[/latex]? This can be written as [latex]\left(x\cdot{x}\right)\left(x\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\right)=x\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}\cdot{x}[/latex] or [latex]x^{8}[/latex]. And, once again, 8 is the sum of the original two exponents. This concept can be generalized in the following way:

The Product Rule for Exponents

For any number x and any integers a and b, [latex]\left(x^{a}\right)\left(x^{b}\right) = x^{a+b}[/latex]. To multiply exponential terms with the same base, add the exponents.

Example 7.1.C

Write each of the following products with a single base. Do not simplify further.
  1. [latex]{t}^{5}\cdot {t}^{3}[/latex]
  2. [latex]\left(-3\right)^{5}\cdot \left(-3\right)[/latex]
  3. [latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}[/latex]

Answer: Solution Use the product rule to simplify each expression.

  1. [latex]{t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}[/latex]
  2. [latex]{\left(-3\right)}^{5}\cdot \left(-3\right)={\left(-3\right)}^{5}\cdot {\left(-3\right)}^{1}={\left(-3\right)}^{5+1}={\left(-3\right)}^{6}[/latex]
  3. [latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}[/latex]
At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.
[latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}=\left({x}^{2}\cdot {x}^{5}\right)\cdot {x}^{3}=\left({x}^{2+5}\right)\cdot {x}^{3}={x}^{7}\cdot {x}^{3}={x}^{7+3}={x}^{10}[/latex]
Notice we get the same result by adding the three exponents in one step.
[latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[/latex]

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Caution! Do not try to apply this rule to sums. Think about the expression [latex]\left(2+3\right)^{2}[/latex]

Does [latex]\left(2+3\right)^{2}[/latex] equal [latex]2^{2}+3^{2}[/latex]?

No, it does not because of the order of operations!

[latex]\left(2+3\right)^{2}=5^{2}=25[/latex]

and

[latex]2^{2}+3^{2}=4+9=13[/latex]

Therefore, you can only use this rule when the numbers inside the parentheses are being multiplied (or divided, as we will see next).
In the following video you will see more examples of using the product rule for exponents to simplify expressions. https://youtu.be/P0UVIMy2nuI In our last product rule example we will show that an exponent can be an algebraic expression.  We can use the product rule for exponents no matter what the exponent looks like, as long as the base is the same.

Example 7.1.d

Multiply. [latex]x^{a+2}\cdot{x^{3a-9}}[/latex]

Answer: We have two exponentiated terms with the same base, so we can multiply them together. The product rule for exponents says that we can add the exponents. [latex-display]x^{a+2}\cdot{x^{3a-9}}=x^{(a+2)+(3a-9)}=x^{4a-7}[/latex-display] The expression can't be simplified any further.

Answer

[latex-display]x^{a+2}\cdot{x^{3a-9}}=x^{4a-7}[/latex-display]

7.1.3 Use the quotient rule to divide exponential expressions

Let’s look at dividing terms containing exponential expressions. What happens if you divide two numbers in exponential form with the same base? Consider the following expression.

[latex] \displaystyle \frac{{{4}^{5}}}{{{4}^{2}}}[/latex]

You can rewrite the expression as: [latex] \displaystyle \frac{4\cdot 4\cdot 4\cdot 4\cdot 4}{4\cdot 4}[/latex]. Then you can cancel the common factors of 4 in the numerator and denominator: [latex] \displaystyle [/latex] Finally, this expression can be rewritten as [latex]4^{3}[/latex] using exponential notation. Notice that the exponent, 3, is the difference between the two exponents in the original expression, 5 and 2. So, [latex] \displaystyle \frac{{{4}^{5}}}{{{4}^{2}}}=4^{5-2}=4^{3}[/latex]. Be careful that you subtract the exponent in the denominator from the exponent in the numerator. So, to divide two exponential terms with the same base, subtract the exponents.

The Quotient (Division) Rule for Exponents

For any non-zero number x and any integers a and b: [latex] \displaystyle \frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[/latex]

Example 7.1.e

Write each of the following products with a single base. Do not simplify further.
  1. [latex]\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}[/latex]
  2. [latex]\frac{{t}^{23}}{{t}^{15}}[/latex]
  3. [latex]\frac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}[/latex]

Answer: Use the quotient rule to simplify each expression.

  1. [latex]\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}={\left(-2\right)}^{14 - 9}={\left(-2\right)}^{5}[/latex]
  2. [latex]\frac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[/latex]
  3. [latex]\frac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}={\left(z\sqrt{2}\right)}^{5 - 1}={\left(z\sqrt{2}\right)}^{4}[/latex]

As we showed with the product rule, you may be given a quotient with an exponent that is an algebraic expression to simplify.  As long as the bases agree, you may use the quotient rule for exponents.

Example 7.1.f

Simplify. [latex]\frac{y^{x-3}}{y^{9-x}}[/latex]

Answer: We have a quotient whose terms have the same base so we can use the quotient rule for exponents. [latex-display]\frac{y^{x-3}}{y^{9-x}}=y^{(x-3)-(9-x)}=y^{(x-3-9+x)}=y^{2x-12}[/latex-display]

In the following video, you will we more examples of using the quotient rule for exponents. https://youtu.be/xy6WW7y_GcU

7.1.4 Raise powers to powers

Another word for exponent is power.  You have likely seen or heard an example such as [latex]3^5[/latex] can be described as 3 raised to the 5th power. In this section we will further expand our capabilities with exponents. We will learn what to do when a term with a power is raised to another power, and what to do when two numbers or variables are multiplied and both are raised to an exponent.  We will also learn what to do when numbers or variables that are divided are raised to a power.  We will begin by raising powers to powers. Let’s simplify [latex]\left(5^{2}\right)^{4}[/latex]. In this case, the base is [latex]5^2[/latex] and the exponent is 4, so you multiply [latex]5^{2}[/latex] four times: [latex]\left(5^{2}\right)^{4}=5^{2}\cdot5^{2}\cdot5^{2}\cdot5^{2}=5^{8}[/latex] (using the Product Rule—add the exponents). [latex]\left(5^{2}\right)^{4}[/latex] is a power of a power. It is the fourth power of 5 to the second power. And we saw above that the answer is [latex]5^{8}[/latex]. Notice that the new exponent is the same as the product of the original exponents: [latex]2\cdot4=8[/latex]. So, [latex]\left(5^{2}\right)^{4}=5^{2\cdot4}=5^{8}[/latex] (which equals 390,625, if you do the multiplication). Likewise, [latex]\left(x^{4}\right)^{3}=x^{4\cdot3}=x^{12}[/latex] This leads to another rule for exponents—the Power Rule for Exponents. To simplify a power of a power, you multiply the exponents, keeping the base the same. For example, [latex]\left(2^{3}\right)^{5}=2^{15}[/latex].

The Power Rule for Exponents

For any positive number x and integers a and b: [latex]\left(x^{a}\right)^{b}=x^{a\cdot{b}}[/latex]. Take a moment to contrast how this is different from the product rule for exponents found on the previous page.

Example 7.1.g

Write each of the following products with a single base. Do not simplify further.
  1. [latex]{\left({x}^{2}\right)}^{7}[/latex]
  2. [latex]{\left({\left(2t\right)}^{5}\right)}^{3}[/latex]
  3. [latex]{\left({\left(-3\right)}^{5}\right)}^{11}[/latex]

Answer: Use the power rule to simplify each expression.

  1. [latex]{\left({x}^{2}\right)}^{7}={x}^{2\cdot 7}={x}^{14}[/latex]
  2. [latex]{\left({\left(2t\right)}^{5}\right)}^{3}={\left(2t\right)}^{5\cdot 3}={\left(2t\right)}^{15}[/latex]
  3. [latex]{\left({\left(-3\right)}^{5}\right)}^{11}={\left(-3\right)}^{5\cdot 11}={\left(-3\right)}^{55}[/latex]

 In the following video you will see more examples of using the power rule to simplify expressions with exponents. https://youtu.be/VjcKU5rA7F8 Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.
Product Rule Power Rule
[latex]5^{3}\cdot5^{4}[/latex] =  [latex]5^{3+4}[/latex] = [latex]5^{7}[/latex] but [latex]\left(5^{3}\right)^{4}[/latex] = [latex]5^{3\cdot4}[/latex] = [latex]5^{12}[/latex]
[latex]x^{5}\cdot x^{2}[/latex] = [latex]x^{5+2}[/latex] = [latex]x^{7}[/latex] but [latex]\left(x^{5}\right)^{2}[/latex] =  [latex]x^{5\cdot2}[/latex] = [latex]x^{10}[/latex]
[latex]\left(3a\right)^{7}\cdot\left(3a\right)^{10} [/latex] = [latex]\left(3a\right)^{7+10} [/latex] = [latex]\left(3a\right)^{17}[/latex] but [latex]\left(\left(3a\right)^{7}\right)^{10} [/latex] = [latex]\left(3a\right)^{7\cdot10} [/latex] = [latex]\left(3a\right)^{70}[/latex]

7.1.5 Define and use the zero exponent rule

Return to the quotient rule. We worked with expressions for which  [latex]a>b[/latex] so that the difference [latex]a-b[/latex] would never be zero or negative.

The Quotient (Division) Rule for Exponents

For any non-zero number x and any integers a and b: [latex] \displaystyle \frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[/latex]
What would happen if [latex]a=b[/latex]? In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.

[latex]\frac{t^{8}}{t^{8}}=\frac{\cancel{t^{8}}}{\cancel{t^{8}}}=1[/latex]

If we were to simplify the original expression using the quotient rule, we would have
[latex]\frac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}[/latex]
If we equate the two answers, the result is [latex]{t}^{0}=1[/latex]. This is true for any nonzero real number, or any variable representing a real number.
[latex]{a}^{0}=1[/latex]
The sole exception is the expression [latex]{0}^{0}[/latex]. This appears later in more advanced courses, but for now, we will consider the value to be undefined.

The Zero Exponent Rule of Exponents

For any nonzero real number [latex]a[/latex], the zero exponent rule of exponents states that
[latex]{a}^{0}=1[/latex]

Example 7.1.h

Simplify each expression using the zero exponent rule of exponents.
  1. [latex]\Large\frac{{c}^{3}}{{c}^{3}}[/latex]
  2. [latex]\Large\frac{-3{x}^{5}}{{x}^{5}}[/latex]
  3. [latex]\Large\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}[/latex]
  4. [latex]\Large\frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}[/latex]

Answer: Use the zero exponent and other rules to simplify each expression.

  1. [latex]\Large\begin{array}\text{ }\frac{c^{3}}{c^{3}} \hfill& =c^{3-3} \\ \hfill& =c^{0} \\ \hfill& =1\end{array}[/latex]
  2. [latex]\Large\begin{array}{ccc}\hfill \frac{-3{x}^{5}}{{x}^{5}}& =& -3\cdot \frac{{x}^{5}}{{x}^{5}}\hfill \\ & =& -3\cdot {x}^{5 - 5}\hfill \\ & =& -3\cdot {x}^{0}\hfill \\ & =& -3\cdot 1\hfill \\ & =& -3\hfill \end{array}[/latex]
  3. [latex]\large\begin{array}{cccc}\hfill \frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}& =& \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{1+3}}\hfill &\text{Use the product rule in the denominator}.\hfill \\ & =&\frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}}\hfill & \text{Simplify}.\hfill \\ & =&{\left({j}^{2}k\right)}^{4 - 4}\hfill &\text{Use the quotient rule}.\hfill \\ & =& {\left({j}^{2}k\right)}^{0}\hfill&\text{Simplify}.\hfill \\ & =& 1& \end{array}[/latex]
  4. [latex]\large\begin{array}{cccc}\hfill \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& =&5{\left(r{s}^{2}\right)}^{2 - 2}\hfill &\text{Use the quotient rule}.\hfill \\ & =& 5{\left(r{s}^{2}\right)}^{0}\hfill &\text{Simplify}.\hfill \\ & =& 5\cdot 1\hfill &\text{Use the zero exponent rule}.\hfill \\ & =& 5\hfill &\text{Simplify}.\hfill \end{array}[/latex]

In the following video you will see more examples of simplifying expressions whose exponents may be zero. https://youtu.be/rpoUg32utlc

7.1.6 Define and use the negative exponent rule

Another useful result occurs if we relax the condition that [latex]a>b[/latex] in the quotient rule even further. For example, can we simplify [latex]\frac{{h}^{3}}{{h}^{5}}[/latex]? When [latex]a<b[/latex]—that is, where the difference [latex]a-b[/latex] is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, [latex]\frac{{h}^{3}}{{h}^{5}}[/latex].
[latex]\Large\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h}\hfill \\ & =& \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h}\hfill \\ & =& \frac{1}{h\cdot h}\hfill \\ & =& \frac{1}{{h}^{2}}\hfill \end{array}[/latex]
If we were to simplify the original expression using the quotient rule, we would have
[latex]\Large\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\hfill \\ & =& \text{ }{h}^{-2}\hfill \end{array}[/latex]
Putting the answers together, we have [latex]{h}^{-2}=\frac{1}{{h}^{2}}[/latex]. This is true for any nonzero real number, or any variable representing a nonzero real number. A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.
[latex]\begin{array}{ccc}{a}^{-n}=\frac{1}{{a}^{n}}& \text{and}& {a}^{n}=\frac{1}{{a}^{-n}}\end{array}[/latex]
We have shown that the exponential expression [latex]{a}^{n}[/latex] is defined when [latex]n[/latex] is a natural number, 0, or the negative of a natural number. That means that [latex]{a}^{n}[/latex] is defined for any integer [latex]n[/latex]. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer [latex]n[/latex].

The Negative Rule of Exponents

For any nonzero real number [latex]a[/latex] and natural number [latex]n[/latex], the negative rule of exponents states that
[latex]{a}^{-n}=\frac{1}{{a}^{n}}[/latex]

Example 7.1.i

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
  1. [latex]\Large\frac{{(2b) }^{3}}{{(2b) }^{10}}[/latex]
  2. [latex]\Large\frac{{z}^{2}\cdot z}{{z}^{4}}[/latex]
  3. [latex]\Large\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}[/latex]

Answer:

Solution

  1. [latex]\Large\frac{{(2b) }^{3}}{{(2b)}^{10}}={(2b)}^{3 - 10}={(2b) }^{-7}=\frac{1}{{(2b)}^{7}}[/latex]
  2. [latex]\Large\frac{{z}^{2}\cdot z}{{z}^{4}}=\frac{{z}^{2+1}}{{z}^{4}}=\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\frac{1}{z}[/latex]
  3. [latex]\Large\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4 - 8}={\left(-5{t}^{3}\right)}^{-4}=\frac{1}{{\left(-5{t}^{3}\right)}^{4}}[/latex]

In the following video you will see examples of simplifying expressions with negative exponents. https://youtu.be/Gssi4dBtAEI

7.1.7 Combine exponent rules to simplify expressions

In the next examples we will combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.

Example 7.1.j

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
  1. [latex]{b}^{2}\cdot {b}^{-8}[/latex]
  2. [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}[/latex]
  3. [latex]\frac{-7z}{{\left(-7z\right)}^{5}}[/latex]

Answer:

Solution

  1. [latex]{b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}[/latex]
  2. [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1[/latex]
  3. [latex]\frac{-7z}{{\left(-7z\right)}^{5}}=\frac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}={\left(-7z\right)}^{1 - 5}={\left(-7z\right)}^{-4}=\frac{1}{{\left(-7z\right)}^{4}}[/latex]

The following video shows more examples of how to combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents. https://youtu.be/EkvN5tcp4Cc

7.1.8 Finding the Power of a Product

To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider [latex]{\left(pq\right)}^{3}[/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors.
[latex]\begin{array}{ccc}\hfill {\left(pq\right)}^{3}& =& \stackrel{3\text{ factors}}{{\left(pq\right)\cdot \left(pq\right)\cdot \left(pq\right)}}\hfill \\ & =& p\cdot q\cdot p\cdot q\cdot p\cdot q\hfill \\ & =& \stackrel{3\text{ factors}}{{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{{q\cdot q\cdot q}}\hfill \\ & =& {p}^{3}\cdot {q}^{3}\hfill \end{array}[/latex]
In other words, [latex]{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}[/latex].

The Power of a Product Rule of Exponents

For any real numbers [latex]a[/latex] and [latex]b[/latex] and any integer [latex]n[/latex], the power of a product rule of exponents states that
[latex]{\left(ab\right)}^{n}={a}^{n}{b}^{n}[/latex]

Example 7.1.k

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
  1. [latex]{\left(a{b}^{2}\right)}^{3}[/latex]
  2. [latex]{\left(2^a{t}\right)}^{15}[/latex]
  3. [latex]{\left(-2{w}^{3}\right)}^{3}[/latex]
  4. [latex]\frac{1}{{\left(-7z\right)}^{4}}[/latex]
  5. [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}[/latex]

Answer: Use the product and quotient rules and the new definitions to simplify each expression.

  1. [latex]{\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}[/latex]
  2. [latex]{\left(2^a{t}\right)}^{15}={\left(2^a\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{a\cdot15}\cdot{t}^{15}=2^{15a}\cdot{t}^{15}[/latex]
  3. [latex]{\left(-2{w}^{3}\right)}^{3}={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=-8\cdot {w}^{3\cdot 3}=-8{w}^{9}[/latex]
  4. [latex]\frac{1}{{\left(-7z\right)}^{4}}=\frac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{1}{2,401{z}^{4}}[/latex]
  5. [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}={e}^{-14}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}[/latex]

We may even encounter In the following video,  we provide more examples of how to find the power of a product. https://youtu.be/p-2UkpJQWpo

7.1.9 Finding the Power of a Quotient

To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example.
[latex]{\left({e}^{-2}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}[/latex]
Let’s rewrite the original problem differently and look at the result.
[latex]\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}[/latex]
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.
[latex]\normalsize\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}}\hfill \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}[/latex]

The Power of a Quotient Rule of Exponents

For any real numbers [latex]a[/latex] and [latex]b[/latex] and any integer [latex]n[/latex], the power of a quotient rule of exponents states that
[latex]{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}[/latex]

Example 7.1.l

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
  1. [latex]{\left(\frac{4}{{z}^{11}}\right)}^{3}[/latex]
  2. [latex]{\left(\frac{p}{{q}^{3}}\right)}^{6}[/latex]
  3. [latex]{\left(\frac{-1}{{t}^{2}}\right)}^{27}[/latex]
  4. [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}[/latex]
  5. [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}[/latex]

Answer:

  1. [latex]\Large{\left(\frac{4}{{z}^{11}}\right)}^{3}=\frac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\frac{64}{{z}^{11\cdot 3}}=\frac{64}{{z}^{33}}[/latex]
  2. [latex]\Large{\left(\frac{p}{{q}^{3}}\right)}^{6}=\frac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\frac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\frac{{p}^{6}}{{q}^{18}}[/latex]
  3. [latex]\Large{\left(\frac{-1}{{t}^{2}}\right)}^{27}=\frac{{\left(-1\right)}^{27}}{{\left({t}^{2}\right)}^{27}}=\frac{-1}{{t}^{2\cdot 27}}=\frac{-1}{{t}^{54}}=-\frac{1}{{t}^{54}}[/latex]
  4. [latex]\Large{\left({j}^{3}{k}^{-2}\right)}^{4}={\left(\frac{{j}^{3}}{{k}^{2}}\right)}^{4}=\frac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\frac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\frac{{j}^{12}}{{k}^{8}}[/latex]
  5. [latex]\Large{\left({m}^{-2}{n}^{-2}\right)}^{3}={\left(\frac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\frac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\frac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\frac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\frac{1}{{m}^{6}{n}^{6}}[/latex]

The following video provides more examples of simplifying expressions using the power of a quotient and other exponent rules. https://youtu.be/BoBe31pRxFM

Summary

  • Evaluating expressions containing exponents is the same as evaluating any expression. You substitute the value of the variable into the expression and simplify.
  • The product rule for exponents: For any number x and any integers a and b, [latex]\left(x^{a}\right)\left(x^{b}\right) = x^{a+b}[/latex].
  • The quotient rule for exponents: For any non-zero number x and any integers a and b: [latex] \displaystyle \frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[/latex]
  • The power rule for exponents:
    1. For any nonzero numbers a and b and any integer x, [latex]\left(ab\right)^{x}=a^{x}\cdot{b^{x}}[/latex].
    2. For any number a, any non-zero number b, and any integer x, [latex] \displaystyle {\left(\frac{a}{b}\right)}^{x}=\frac{a^{x}}{b^{x}}[/latex]

Licenses & Attributions

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  • Simplify Basic Exponential Expressions. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
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  • Ex: Simplify Exponential Expressions Using the Product Property of Exponents . Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
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  • Simplify Expressions Using the Quotient and Zero Exponent Rules. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Simplify Expressions Using the Quotient and Negative Exponent Rules . Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Simplify Expressions Using Exponent Rules (Product, Quotient, Zero Exponent). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Simplify Expressions Using Exponent Rules (Power of a Product). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
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  • Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay, et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et. al. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.