Learning Objectives
In this section students will:
- 3.1.1 – Use the product rule of exponents
- 3.1.2 – Use the quotient rule of exponents
- 3.1.3 – Use the power rule of exponents
- 3.1.4 – Use the zero exponent rule of exponents
- 3.1.5 – Use the negative rule of exponents
- 3.1.6 – Find the power of a product and a quotient
- 3.1.7 – Simplify exponential expressions
Focus in on a square centimeter of your skin. Look closer. Closer still. If you could look closely enough, you would see hundreds of thousands of microscopic organisms. They are bacteria, and they are not only on your skin, but in your mouth, nose, and even your intestines. In fact, the bacterial cells in your body at any given moment outnumber your own cells. But that is no reason to feel bad about yourself. While some bacteria can cause illness, many are healthy and even essential to the body.
Bacteria commonly reproduce through a process called binary fission, during which one bacterial cell splits into two. When conditions are right, bacteria can reproduce very quickly. Unlike humans and other complex organisms, the time required to form a new generation of bacteria is often a matter of minutes or hours, as opposed to days or years.[1]
For simplicity’s sake, suppose we begin with a culture of one bacterial cell that can divide every hour. The table below shows the number of bacterial cells at the end of each subsequent hour. We see that the single bacterial cell leads to over one thousand bacterial cells in just ten hours! And if we were to extrapolate the table to twenty-four hours, we would have over 16 million!
| Hour | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Bacteria | 1 | 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 | 512 | 1024 |
In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. Before exploring exponential functions, we will review various properties of exponents.
3.1.1 – Using the Product Rule of Exponents
Consider the product [latex]\,{x}^{3}\cdot {x}^{4}.\,[/latex] Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.
[latex]$$\begin{array}{ccc}\hfill {x}^{3}\cdot {x}^{4}& =& \stackrel{3\text{ factors}}{\stackrel{}{x\cdot x\cdot x}}\cdot \stackrel{4\text{ factors}}{\stackrel{}{x\cdot x\cdot x\cdot x}}\hfill \\ & =& \stackrel{7\text{ factors}}{\stackrel{}{x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x}}\hfill \\ & =& {x}^{7}\hfill \end{array}$$[/latex]
The result is that [latex]\,{x}^{3}\cdot {x}^{4}={x}^{3+4}={x}^{7}.[/latex]
Notice that the exponent of the product is the sum of the exponents of the terms. In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents. This is the product rule of exponents.
[latex]$${a}^{m}\cdot {a}^{n}={a}^{m+n}$$[/latex]
Now consider an example with real numbers.
[latex]$${2}^{3}\cdot {2}^{4}={2}^{3+4}={2}^{7}$$[/latex]
We can always check that this is true by simplifying each exponential expression. We find that [latex]\,{2}^{3}\,[/latex] is 8, [latex]\,{2}^{4}\,[/latex] is 16, and [latex]\,{2}^{7}\,[/latex] is 128. The product [latex]\,8\cdot 16\,[/latex] equals 128, so the relationship is true. We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents.
The Product Rule of Exponents
For any real number [latex]\,a\,[/latex] and natural numbers [latex]\,m\,[/latex] and [latex]\,n,[/latex] the product rule of exponents states that
[latex]$${a}^{m}\cdot {a}^{n}={a}^{m+n}$$[/latex]
Example 1 – Using the Product Rule
Write each of the following products with a single base. Do not simplify further.
- [latex]{t}^{5}\cdot {t}^{3}[/latex]
- [latex]{\left(-3\right)}^{5}\cdot \left(-3\right)[/latex]
- [latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}[/latex]
Use the product rule to simplify each expression.
- [latex]{t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}[/latex]
- [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]
- [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.
Notice we get the same result by adding the three exponents in one step.
Try It
Write each of the following products with a single base. Do not simplify further.
- [latex]{k}^{6}\cdot {k}^{9}[/latex]
- [latex]{\left(\frac{2}{y}\right)}^{4}\cdot \left(\frac{2}{y}\right)[/latex]
- [latex]{t}^{3}\cdot {t}^{6}\cdot {t}^{5}[/latex]
Show answer
- [latex]{k}^{15}[/latex]
- [latex]{\left(\frac{2}{y}\right)}^{5}[/latex]
- [latex]{t}^{14}[/latex]
3.1.2 – Using the Quotient Rule of Exponents
The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as [latex]\,\frac{{y}^{m}}{{y}^{n}},[/latex] where [latex]\,m>n.\,[/latex] Consider the example [latex]\,\frac{{y}^{9}}{{y}^{5}}.\,[/latex] Perform the division by canceling common factors.
Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.
[latex]$$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$$[/latex]
In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.
[latex]$$\frac{{y}^{9}}{{y}^{5}}={y}^{9-5}={y}^{4}$$[/latex]
For the time being, we must be aware of the condition [latex]\,m>n.\,[/latex] Otherwise, the difference [latex]\,m-n\,[/latex] could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.
The Quotient Rule of Exponents
For any real number [latex]\,a\,[/latex] and natural numbers [latex]\,m\,[/latex] and [latex]\,n,[/latex] such that [latex]\,m>n,[/latex] the quotient rule of exponents states that
[latex]$$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$$[/latex]
Example 2 – Using the Quotient Rule
Write each of the following products with a single base. Do not simplify further.
- [latex]\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}[/latex]
- [latex]\frac{{t}^{23}}{{t}^{15}}[/latex]
- [latex]\frac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}[/latex]
Use the quotient rule to simplify each expression.
- [latex]\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}={\left(-2\right)}^{14-9}={\left(-2\right)}^{5}[/latex]
- [latex]\frac{{t}^{23}}{{t}^{15}}={t}^{23-15}={t}^{8}[/latex]
- [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]
Try It
Write each of the following products with a single base. Do not simplify further.
- [latex]\frac{{s}^{75}}{{s}^{68}}[/latex]
- [latex]\frac{{\left(-3\right)}^{6}}{-3}[/latex]
- [latex]\frac{{\left(e{f}^{2}\right)}^{5}}{{\left(e{f}^{2}\right)}^{3}}[/latex]
Show answer
- [latex]{s}^{7}[/latex]
- [latex]{\left(-3\right)}^{5}[/latex]
- [latex]{\left(e{f}^{2}\right)}^{2}[/latex]
3.1.3 – Using the Power Rule of Exponents
Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression [latex]\,{\left({x}^{2}\right)}^{3}.\,[/latex] The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3.
[latex]$$\begin{array}{ccc}\hfill {\left({x}^{2}\right)}^{3}& =& \stackrel{3\text{ factors}}{\stackrel{}{\left({x}^{2}\right)\cdot \left({x}^{2}\right)\cdot \left({x}^{2}\right)}}\hfill \\ & =& \hfill \stackrel{3\text{ factors}}{\stackrel{}{\left(\stackrel{2\text{ factors}}{\overbrace{x\cdot x}}\right)\cdot \left(\stackrel{2\text{ factors}}{\overbrace{x\cdot x}}\right)\cdot \left(\stackrel{2\text{ factors}}{\overbrace{x\cdot x}}\right)}}\\ & =& x\cdot x\cdot x\cdot x\cdot x\cdot x\hfill \\ & =& {x}^{6}\hfill \end{array}$$[/latex]
The exponent of the answer is the product of the exponents: [latex]\,{\left({x}^{2}\right)}^{3}={x}^{2\cdot 3}={x}^{6}.\,[/latex] In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents.
[latex]$${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$$[/latex]
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.
[latex]$$\begin{array}{cccccccccc}& & \text{Product Rule}\hfill & & & & & \text{ Power Rule}\hfill & & \\ \hfill {5}^{3}\cdot {5}^{4}& =& {5}^{3+4}\hfill & =& {5}^{7}\hfill & \phantom{\rule{1em}{0ex}}\text{but}\phantom{\rule{1em}{0ex}}& \hfill \text{ }{\left({5}^{3}\right)}^{4}& =& {5}^{3\cdot 4}\hfill & =& {5}^{12}\hfill \\ \hfill {x}^{5}\cdot {x}^{2}& =& {x}^{5+2}\hfill & =& {x}^{7}\hfill & \phantom{\rule{1em}{0ex}}\text{but}\phantom{\rule{1em}{0ex}}& \hfill {\left({x}^{5}\right)}^{2}& =& {x}^{5\cdot 2}\hfill & =\hfill & {x}^{10}\hfill \\ \hfill {\left(3a\right)}^{7}\cdot {\left(3a\right)}^{10}& =& {\left(3a\right)}^{7+10}\hfill & =& {\left(3a\right)}^{17}\hfill & \phantom{\rule{1em}{0ex}}\text{but}\phantom{\rule{1em}{0ex}}& \hfill {\left({\left(3a\right)}^{7}\right)}^{10}& =& {\left(3a\right)}^{7\cdot 10}\hfill & =& {\left(3a\right)}^{70}\hfill \end{array}$$[/latex]
The Power Rule of Exponents
For any real number [latex]\,a\,[/latex] and positive integers [latex]\,m\,[/latex] and [latex]\,n,[/latex] the power rule of exponents states that
[latex]$${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$$[/latex]
Example 3 – Using the Power Rule
Write each of the following products with a single base. Do not simplify further.
- [latex]{\left({x}^{2}\right)}^{7}[/latex]
- [latex]{\left({\left(2t\right)}^{5}\right)}^{3}[/latex]
- [latex]{\left({\left(-3\right)}^{5}\right)}^{11}[/latex]
Use the power rule to simplify each expression.
- [latex]{\left({x}^{2}\right)}^{7}={x}^{2\cdot 7}={x}^{14}[/latex]
- [latex]{\left({\left(2t\right)}^{5}\right)}^{3}={\left(2t\right)}^{5\cdot 3}={\left(2t\right)}^{15}[/latex]
- [latex]{\left({\left(-3\right)}^{5}\right)}^{11}={\left(-3\right)}^{5\cdot 11}={\left(-3\right)}^{55}[/latex]
Try It
Write each of the following products with a single base. Do not simplify further.
- [latex]{\left({\left(3y\right)}^{8}\right)}^{3}[/latex]
- [latex]{\left({t}^{5}\right)}^{7}[/latex]
- [latex]{\left({\left(-g\right)}^{4}\right)}^{4}[/latex]
Show answer
- [latex]{\left(3y\right)}^{24}[/latex]
- [latex]{t}^{35}[/latex]
- [latex]{\left(-g\right)}^{16}[/latex]
3.1.4 – Using the Zero Exponent Rule of Exponents
Return to the quotient rule. We made the condition that [latex]\,m>n\,[/latex] so that the difference [latex]\,m-n\,[/latex] would never be zero or negative. What would happen if [latex]\,m=n?[/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 4 – Using the Zero Exponent Rule
Simplify each expression using the zero exponent rule of exponents.
- [latex]\frac{{c}^{3}}{{c}^{3}}[/latex]
- [latex]\frac{-3{x}^{5}}{{x}^{5}}[/latex]
- [latex]\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}[/latex]
- [latex]\frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}[/latex]
Use the zero exponent and other rules to simplify each expression.
- [latex]\begin{array}{ccc}\hfill \frac{{c}^{3}}{{c}^{3}}& =& {c}^{3-3}\hfill \\ & =& {c}^{0}\hfill \\ & =& 1\hfill \end{array}[/latex]
- [latex]\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]
- [latex]\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 & \phantom{\rule{3em}{0ex}}\text{Use the product rule in the denominator}.\hfill \\ & =& \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}}\hfill & \phantom{\rule{3em}{0ex}}\text{Simplify}.\hfill \\ & =& {\left({j}^{2}k\right)}^{4-4}\hfill & \phantom{\rule{3em}{0ex}}\text{Use the quotient rule}.\hfill \\ & =& {\left({j}^{2}k\right)}^{0}\hfill & \phantom{\rule{3em}{0ex}}\text{Simplify}.\hfill \\ & =& 1\hfill & \end{array}[/latex]
- [latex]\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 & \phantom{\rule{5em}{0ex}}\text{Use the quotient rule}.\hfill \\ & =& 5{\left(r{s}^{2}\right)}^{0}\hfill & \phantom{\rule{5em}{0ex}}\text{Simplify}.\hfill \\ & =& 5\cdot 1\hfill & \phantom{\rule{5em}{0ex}}\text{Use the zero exponent rule}.\hfill \\ & =& 5\hfill & \phantom{\rule{5em}{0ex}}\text{Simplify}.\hfill \end{array}[/latex]
Try It
Simplify each expression using the zero exponent rule of exponents.
- [latex]\frac{{t}^{7}}{{t}^{7}}[/latex]
- [latex]\frac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}[/latex]
- [latex]\frac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}[/latex]
- [latex]\frac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}[/latex]
Show answer
- [latex]1[/latex]
- [latex]\frac{1}{2}[/latex]
- [latex]1[/latex]
- [latex]1[/latex]
3.1.5 – Using the Negative Rule of Exponents
Another useful result occurs if we relax the condition that [latex]\,m>n\,[/latex] in the quotient rule even further. For example, can we simplify [latex]\,\frac{{h}^{3}}{{h}^{5}}?\,[/latex] When [latex]\,m
Divide one exponential expression by another with a larger exponent. Use our example, [latex]\,\frac{{h}^{3}}{{h}^{5}}.[/latex]
[latex]$$\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h}\hfill \\ & =& \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h}\hfill \\ & =& \frac{1}{h\cdot h}\hfill \\ & =& \frac{1}{{h}^{2}}\hfill \end{array}$$[/latex]
If we were to simplify the original expression using the quotient rule, we would have
[latex]$$\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& {h}^{3-5}\hfill \\ & =& \text{ }{h}^{-2}\hfill \end{array}$$[/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 5 – Using the Negative Exponent Rule
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
- [latex]\frac{{\theta }^{3}}{{\theta }^{10}}[/latex]
- [latex]\frac{{z}^{2}\cdot z}{{z}^{4}}[/latex]
- [latex]\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}[/latex]
- [latex]\frac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3-10}={\theta }^{-7}=\frac{1}{{\theta }^{7}}[/latex]
- [latex]\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]
- [latex]\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]
Try It
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
- [latex]\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}[/latex]
- [latex]\frac{{f}^{47}}{{f}^{49}\cdot f}[/latex]
- [latex]\frac{2{k}^{4}}{5{k}^{7}}[/latex]
Show answer
- [latex]\frac{1}{{\left(-3t\right)}^{6}}[/latex]
- [latex]\frac{1}{{f}^{3}}[/latex]
- [latex]\frac{2}{5{k}^{3}}[/latex]
Example 6 – Using the Product and Quotient Rules
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
- [latex]{b}^{2}\cdot {b}^{-8}[/latex]
- [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}[/latex]
- [latex]\frac{-7z}{{\left(-7z\right)}^{5}}[/latex]
- [latex]{b}^{2}\cdot {b}^{-8}={b}^{2-8}={b}^{-6}=\frac{1}{{b}^{6}}[/latex]
- [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5-5}={\left(-x\right)}^{0}=1[/latex]
- [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]
Try It
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
- [latex]{t}^{-11}\cdot {t}^{6}[/latex]
- [latex]\frac{{25}^{12}}{{25}^{13}}[/latex]
Show answer
- [latex]{t}^{-5}=\frac{1}{{t}^{5}}[/latex]
- [latex]\frac{1}{25}[/latex]
3.1.6 – Finding the Power of a Product and a Quotient
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]$$\begin{array}{ccc}\hfill {\left(pq\right)}^{3}& =& \stackrel{3\text{ factors}}{\stackrel{}{\left(pq\right)\cdot \left(pq\right)\cdot \left(pq\right)}}\hfill \\ & =& p\cdot q\cdot p\cdot q\cdot p\cdot q\hfill \\ & =& \stackrel{3\text{ factors}}{\stackrel{}{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{\stackrel{}{q\cdot q\cdot q}}\hfill \\ & =& {p}^{3}\cdot {q}^{3}\hfill \end{array}$$[/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 – Using the Power of a Product Rule
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
- [latex]{\left(a{b}^{2}\right)}^{3}[/latex]
- [latex]{\left(2t\right)}^{15}[/latex]
- [latex]{\left(-2{w}^{3}\right)}^{3}[/latex]
- [latex]\frac{1}{{\left(-7z\right)}^{4}}[/latex]
- [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}[/latex]
Use the product and quotient rules and the new definitions to simplify each expression.
- [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]
- [latex]{\left(2t\right)}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}[/latex]
- [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]
- [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]
- [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]
Try It
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
- [latex]{\left({g}^{2}{h}^{3}\right)}^{5}[/latex]
- [latex]{\left(5t\right)}^{3}[/latex]
- [latex]{\left(-3{y}^{5}\right)}^{3}[/latex]
- [latex]\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}[/latex]
- [latex]{\left({r}^{3}{s}^{-2}\right)}^{4}[/latex]
Show answer
- [latex]{g}^{10}{h}^{15}[/latex]
- [latex]125{t}^{3}[/latex]
- [latex]-27{y}^{15}[/latex]
- [latex]\frac{1}{{a}^{18}{b}^{21}}[/latex]
- [latex]\frac{{r}^{12}}{{s}^{8}}[/latex]
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]$$\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}}\hfill \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}$$[/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 8 – Using the Power of a Quotient Rule
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
- [latex]{\left(\frac{4}{{z}^{11}}\right)}^{3}[/latex]
- [latex]{\left(\frac{p}{{q}^{3}}\right)}^{6}[/latex]
- [latex]{\left(\frac{-1}{{t}^{2}}\right)}^{27}[/latex]
- [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}[/latex]
- [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}[/latex]
- [latex]{\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]
- [latex]{\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]
- [latex]{\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]
- [latex]{\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]
- [latex]{\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]
Try It
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
- [latex]{\left(\frac{{b}^{5}}{c}\right)}^{3}[/latex]
- [latex]{\left(\frac{5}{{u}^{8}}\right)}^{4}[/latex]
- [latex]{\left(\frac{-1}{{w}^{3}}\right)}^{35}[/latex]
- [latex]{\left({p}^{-4}{q}^{3}\right)}^{8}[/latex]
- [latex]{\left({c}^{-5}{d}^{-3}\right)}^{4}[/latex]
Show answer
- [latex]\frac{{b}^{15}}{{c}^{3}}[/latex]
- [latex]\frac{625}{{u}^{32}}[/latex]
- [latex]\frac{-1}{{w}^{105}}[/latex]
- [latex]\frac{{q}^{24}}{{p}^{32}}[/latex]
- [latex]\frac{1}{{c}^{20}{d}^{12}}[/latex]
3.1.7 – Simplifying Exponential Expressions
Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.
Example 9 – Simplifying Exponential Expressions
Simplify each expression and write the answer with positive exponents only.
- [latex]{\left(6{m}^{2}{n}^{-1}\right)}^{3}[/latex]
- [latex]{17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}[/latex]
- [latex]{\left(\frac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}[/latex]
- [latex]\left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)[/latex]
- [latex]{\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}[/latex]
- [latex]\frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}[/latex]
- [latex]\begin{array}{cccc}\hfill {\left(6{m}^{2}{n}^{-1}\right)}^{3}& =& {\left(6\right)}^{3}{\left({m}^{2}\right)}^{3}{\left({n}^{-1}\right)}^{3}\hfill & \phantom{\rule{9em}{0ex}}\text{The power of a product rule}\hfill \\ & =& {6}^{3}{m}^{2\cdot 3}{n}^{-1\cdot 3}\hfill & \phantom{\rule{9em}{0ex}}\text{The power rule}\hfill \\ & =& \text{ }216{m}^{6}{n}^{-3}\hfill & \phantom{\rule{9em}{0ex}}\text{Simplify}.\hfill \\ & =& \frac{216{m}^{6}}{{n}^{3}}\hfill & \phantom{\rule{9em}{0ex}}\text{The negative exponent rule}\hfill \end{array}[/latex]
- [latex]\begin{array}{cccc}\hfill {17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}& =& {17}^{5-4-3}\hfill & \phantom{\rule{9em}{0ex}}\text{The product rule}\hfill \\ & =& {17}^{-2}\hfill & \phantom{\rule{9em}{0ex}}\text{Simplify}.\hfill \\ & =& \frac{1}{{17}^{2}}\text{ or }\frac{1}{289}\hfill & \phantom{\rule{9em}{0ex}}\text{The negative exponent rule}\hfill \end{array}[/latex]
- [latex]\begin{array}{cccc}\hfill {\left(\frac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}& =& \frac{{\left({u}^{-1}v\right)}^{2}}{{\left({v}^{-1}\right)}^{2}}\hfill & \phantom{\rule{12em}{0ex}}\text{The power of a quotient rule}\hfill \\ & =& \frac{{u}^{-2}{v}^{2}}{{v}^{-2}}\hfill & \phantom{\rule{12em}{0ex}}\text{The power of a product rule}\hfill \\ & =& {u}^{-2}{v}^{2-\left(-2\right)}& \phantom{\rule{12em}{0ex}}\text{The quotient rule}\hfill \\ & =& {u}^{-2}{v}^{4}\hfill & \phantom{\rule{12em}{0ex}}\text{Simplify}.\hfill \\ & =& \frac{{v}^{4}}{{u}^{2}}\hfill & \phantom{\rule{12em}{0ex}}\text{The negative exponent rule}\hfill \end{array}[/latex]
- [latex]\begin{array}{cccc}\hfill \left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)& =& -2\cdot 5\cdot {a}^{3}\cdot {a}^{-2}\cdot {b}^{-1}\cdot {b}^{2}\hfill & \phantom{\rule{3em}{0ex}}\text{Commutative and associative laws of multiplication}\hfill \\ & =& -10\cdot {a}^{3-2}\cdot {b}^{-1+2}\hfill & \phantom{\rule{3em}{0ex}}\text{The product rule}\hfill \\ & =& -10ab\hfill & \phantom{\rule{3em}{0ex}}\text{Simplify}.\hfill \end{array}[/latex]
- [latex]\begin{array}{cccc}\hfill {\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}& =& {\left({x}^{2}\sqrt{2}\right)}^{4-4}\hfill & \phantom{\rule{8em}{0ex}}\text{The product rule}\hfill \\ & =& \text{ }{\left({x}^{2}\sqrt{2}\right)}^{0}\hfill & \phantom{\rule{8em}{0ex}}\text{Simplify}.\hfill \\ & =& 1\hfill & \phantom{\rule{8em}{0ex}}\text{The zero exponent rule}\hfill \end{array}[/latex]
- [latex]\begin{array}{cccc}\hfill \frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}& =& \frac{{\left(3\right)}^{5}\cdot {\left({w}^{2}\right)}^{5}}{{\left(6\right)}^{2}\cdot {\left({w}^{-2}\right)}^{2}}\hfill & \phantom{\rule{13.5em}{0ex}}\text{The power of a product rule}\hfill \\ & =& \frac{{3}^{5}{w}^{2\cdot 5}}{{6}^{2}{w}^{-2\cdot 2}}\hfill & \phantom{\rule{13.5em}{0ex}}\text{The power rule}\hfill \\ & =& \frac{243{w}^{10}}{36{w}^{-4}}\hfill & \phantom{\rule{13.5em}{0ex}}\text{Simplify}.\hfill \\ & =& \frac{27{w}^{10-\left(-4\right)}}{4}\hfill & \phantom{\rule{13.5em}{0ex}}\text{The quotient rule and reduce fraction}\hfill \\ & =& \frac{27{w}^{14}}{4}\hfill & \phantom{\rule{13.5em}{0ex}}\text{Simplify}.\hfill \end{array}[/latex]
Try It
Simplify each expression and write the answer with positive exponents only.
- [latex]{\left(2u{v}^{-2}\right)}^{-3}[/latex]
- [latex]{x}^{8}\cdot {x}^{-12}\cdot x[/latex]
- [latex]{\left(\frac{{e}^{2}{f}^{-3}}{{f}^{-1}}\right)}^{2}[/latex]
- [latex]\left(9{r}^{-5}{s}^{3}\right)\left(3{r}^{6}{s}^{-4}\right)[/latex]
- [latex]{\left(\frac{4}{9}t{w}^{-2}\right)}^{-3}{\left(\frac{4}{9}t{w}^{-2}\right)}^{3}[/latex]
- [latex]\frac{{\left(2{h}^{2}k\right)}^{4}}{{\left(7{h}^{-1}{k}^{2}\right)}^{2}}[/latex]
Show answer
- [latex]\frac{{v}^{6}}{8{u}^{3}}[/latex]
- [latex]\frac{1}{{x}^{3}}[/latex]
- [latex]\frac{{e}^{4}}{{f}^{4}}[/latex]
- [latex]\frac{27r}{s}[/latex]
- [latex]1[/latex]
- [latex]\frac{16{h}^{10}}{49}[/latex]
Access these online resources for additional instruction and practice with exponents and scientific notation.
Key Equations
| Rules of Exponents< For nonzero real numbers [latex]\,a\,[/latex] and [latex]\,b\,[/latex] and integers [latex]\,m\,[/latex] and [latex]\,n\,[/latex] |
|
| Product rule | [latex]{a}^{m}\cdot {a}^{n}={a}^{m+n}[/latex] |
| Quotient rule | [latex]\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}[/latex] |
| Power rule | [latex]{\left({a}^{m}\right)}^{n}={a}^{m\cdot n}[/latex] |
| Zero exponent rule | [latex]{a}^{0}=1[/latex] |
| Negative rule | [latex]{a}^{-n}=\frac{1}{{a}^{n}}[/latex] |
| Power of a product rule | [latex]{\left(a\cdot b\right)}^{n}={a}^{n}\cdot {b}^{n}[/latex] |
| Power of a quotient rule | [latex]{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}[/latex] |
Key Concepts
- Products of exponential expressions with the same base can be simplified by adding exponents. See Example 1.
- Quotients of exponential expressions with the same base can be simplified by subtracting exponents. See Example 2.
- Powers of exponential expressions with the same base can be simplified by multiplying exponents. See Example 3.
- An expression with exponent zero is defined as 1. See Example 4.
- An expression with a negative exponent is defined as a reciprocal. See Example 5 and Example 6.
- The power of a product of factors is the same as the product of the powers of the same factors. See Example 7.
- The power of a quotient of factors is the same as the quotient of the powers of the same factors. See Example 8.
- The rules for exponential expressions can be combined to simplify more complicated expressions. See Example 9.
- Todar, PhD, Kenneth. Todar's Online Textbook of Bacteriology. http://textbookofbacteriology.net/growth_3.html. ↵