Product and Quotient Rules
Learning Outcomes
- Simplify exponential expressions with like bases using the product, quotient, and power rules
Use the Product Rule to Multiply Exponential Expressions
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 us 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 [latex]625[/latex]. Do not 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 [latex]7[/latex] is the sum of the original two exponents, [latex]3[/latex] and [latex]4[/latex]. 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
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]
Answer: 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]
[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]
Example
Multiply. [latex]x^{a+2}\cdot{x^{3a-9}}[/latex]Answer: We have two exponential terms with the same base. 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 cannot be simplified any further.
Use the Quotient Rule to Divide Exponential Expressions
Let us 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, [latex]3[/latex], is the difference between the two exponents in the original expression, [latex]5[/latex] and [latex]2[/latex]. 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
Write each of the following products with a single base. Do not simplify further.- [latex]\dfrac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}[/latex]
- [latex]\dfrac{{t}^{23}}{{t}^{15}}[/latex]
- [latex]\dfrac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}[/latex]
Answer: Use the quotient rule to simplify each expression.
- [latex]\dfrac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}={\left(-2\right)}^{14 - 9}={\left(-2\right)}^{5}[/latex]
- [latex]\dfrac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[/latex]
- [latex]\dfrac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}={\left(z\sqrt{2}\right)}^{5 - 1}={\left(z\sqrt{2}\right)}^{4}[/latex]
Example
Simplify. [latex]\dfrac{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]\dfrac{y^{x-3}}{y^{9-x}}=y^{(x-3)-(9-x)}=y^{2x-12}[/latex-display]
Licenses & Attributions
CC licensed content, Original
- Simplify Expressions Using the Quotient Rule of Exponents (Basic). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
CC licensed content, Shared previously
- 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. License terms: Download for free at http://cnx.org/contents/[email protected]:1/Preface.
- Ex: Expanding and Evaluating Exponential Notation . Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Ex: Simplify Exponential Expressions Using the Product Property of Exponents . Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.