Introduction
In this lesson, you will learn how to use the quotient rule to divide with exponents. Just like with the product rule, in order to use the quotient rule, the bases must be the same. If the bases are the same, the quotient rule says you can subtract the power of the denominator from the power of the numerator.
This video illustrates the lesson material below. Watching the video is optional.
Quotient Rule of Exponents
Division is the opposite of multiplication because when \( 5\times3=15 \), then \( 15\div5=3 \). Similarly, the exponent rule of division, also known as the quotient rule, is the opposite of the product rule.
Review this math vocabulary word to help you understand this lesson better:
- Quotient: the answer when you divide a number by another number.
The following examples will explain how to divide terms with exponents.
Example 1
\( x^{4}\div x^{3} \)
If you write this out, this is the same as x multiplied together four times in the numerator, divided by x multiplied together three times in the denominator.
Remember, \( x^{4}\div x^{3}\) can also be written as \(\frac{x^4}{x^3}\). Using the quotient rule of exponents, this is the same as \(x^{4-3}\) which is equal to \(x^4\), or \(x\).
This demonstrates the quotient rule: when dividing terms with the same base, subtract the exponents.
\begin{align*}\color{black}\large\text{Quotient Rule of Exponents:} \;\frac{a^x} {a^y} =a^{x - y}\\\end{align*}
Example 2
\(\frac{m^{6}}{m^{2}}\)
\begin{align*}
&\frac{m^{6}}{m^{2}} &\color{red}\small\text{Simplify this expression}\\\\
&m^{6-2} &\color{red}\small\text{Quotient rule of exponents}\\\\
&m^{4} &\color{red}\small\text{Subtract the exponents}\\\\
\end{align*}
Example 3
\( \frac{x^{2}m^{3}x^{3}}{m^{2}x} \)
This equation has two different variables, or terms with different bases: \(x\) and \(m\).
\begin{align*}&\frac{x^{2}m^{3}x^{3}}{m^{2}x} &\color\red\small\text{Simplify this expression}\\\\&\frac{m^{3}x^{2}x^{3}}{m^{2}x} &\color\red\small\text{Rearrange into like bases}\\\\ &\frac{m^{3}x^{(2+3)}}{m^{2}x} &\color\red\small\text{Product rule of exponents}\\\\ &\frac{m^{3}x^{5}}{m^{2}x} &\color\red\small\text{Add the exponents}\\\\ &m^{(3-2)}x^{(5-1)} &\color\red\small\text{Quotient rule of exponents}\\\\ &mx^{4} &\color\red\small\text{Subtract exponents}\\\end{align*}
Remember, \(m^1=m\). The final answer is \(mx^{4}\) or \( x^{4}m\).
Things to Remember
- The quotient is the answer when you divide a number by another number.
- To use the quotient rule of exponents, the bases must be the same.
- To apply the quotient rule of exponents, subtract the power of the denominator from the power of the numerator.
Practice Problems
Simplify the following expressions:- \(\dfrac{{\text{m}}^{5}}{{\text{m}}^{2}}\) (
- \(\dfrac{{\text{x}}^{7}}{{\text{x}}^{5}}\) (
- \(\dfrac{{\text{m}}^{2}{\text{x}}^{2}}{\text{xm}}\) (
- \(\dfrac{{\text{x}}^{2}{\text{y}}^{4}{\text{x}}^{7}}{{\text{xy}}^{3}}\) (
Details:
Version 1: Solving using the quotient rule
It is helpful to first use the product rule in the numerator (top of the fraction) to simplify this equation.
The numerator \({\text{x}}^{2}{\text{y}}^{4}{\text{x}}^{7}\) can be rewritten using the product rule as:
\({\text{x}}^{\left ( 2+7 \right )}{\text{y}}^{4} = {\text{x}}^{9}{\text{y}}^{4}\)
Now that you have simplified the numerator, rewrite the original expression:
\(\displaystyle\frac{{\text{x}}^{2}{\text{y}}^{4}{\text{x}}^{7}}{{\text{x y}}^{3}}=\frac{{\text{x}}^{9}{\text{y}}^{4}}{{\text{x y}}^{3}}\)
Since there isn’t any addition or subtraction between any of the factors, you can use the quotient rule to further simplify this problem.
The quotient rule says that as long as factors have the same base, you can subtract the exponent of the factor in the denominator (bottom of the fraction) from the exponent of the factor in the numerator (top of the fraction).
This means you can do the following:
\(\displaystyle\frac{{\text{x}}^{9}{\text{y}}^{4}}{{\text{x y}}^{3}}=\frac{{\text{x}}^{\left ( 9-1 \right )}{\text{y}}^{\left ( 4-3 \right )}}{1}=\frac{{\text{x}}^{8}{\text{y}}^{1}}{1}={\text{x}}^{8}{\text{y}}\)
Remember that anything divided by 1 is just itself and anything raised to the exponent of 1 is just itself. You don’t have to place the 1 there. This is why \(\dfrac{{\text{x}}^{8}{\text{y}}^{1}}{1}={\text{x}}^{8}{\text{y}}\).
Version 2: Solving using the rules of multiplication and division
The following explanation demonstrates in more detail why the quotient rule works and will give you the same solution for this problem.
If you expand all the factors in the equation from their exponent form to their standard multiplication form you get the following:
\(\displaystyle\frac{{\text{x}}^{2}{\text{y}}^{4}{\text{x}}^{7}}{{\text{x y}}^{3}}=\frac{\text{x x y y y y x x x x x x x}}{\text{x y y y}}\)
Now you can “cancel out” factors in the numerator and denominator that are the same because anything divided by itself equals 1.
In this case, you can remove 1 x and 3 y's from both the numerator and denominator since they are being multiplied to other factors in both places.
It may appear now as though everything in the denominator is gone leaving nothing in the denominator, but this is not true. Remember, you haven’t actually gotten rid of these factors, instead, you have changed them into 1’s because \(\dfrac{\text{x}}{\text{x}}=\dfrac{1}{1}=1\) and \(\dfrac{\text{y}}{\text{y}}=\dfrac{1}{1}=1\). Also, \(1\cdot1=1\). So you still have a 1 in the denominator.
Rearranging the factors and putting them into exponent form you get: \(\dfrac{{\text{x}}^{8}{\text{y}}}{1}={\text{x}}^{8}{\text{y}}\).
The final answer is: \({\text{x}}^{8}{\text{y}}\). - \(\dfrac{{\text{x}}^{2}{\text{m}}^{3}{\text{x}}^{4}}{{\text{m}}^{2}{\text{x}}^{3}}\) (
- \(\dfrac{{\text{x}}^{5}{\text{y}}^{2}{\text{x}}}{{\text{x}}^{2}{\text{yz}}}\) (
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