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}}}{{{\left({x}^{2}\right)\cdot \left({x}^{2}\right)\cdot \left({x}^{2}\right)}}}\hfill \\ & =& \hfill \stackrel{{3\text{ factors}}}{{{\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.
| 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+1-} [/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] |
A General Note: 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]
Solution
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 3
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]
Licenses & Attributions
CC licensed content, Specific attribution
- College Algebra. Provided by: OpenStax Authored by: OpenStax College Algebra. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
