Division is the opposite of multiplication because when \(5 \times 3 = 15 \), then \(15 \div 5 = 3 \). Similarly, the exponent rule of division (quotient rule) is the opposite of the product rule. Here is one math vocabulary word that will help you to understand this lesson video better:

• Quotient = what you get when you divide a number by another number

The following video will explain with some examples of how to divide with exponents:

Video Source (05:37 mins) | Transcript

Just like with the product rule, in order to use the quotient rule, our bases must be the same. Then, if the bases are the same, the division rule says we subtract the power of the denominator from the power of the numerator.
Examples:

\(\displaystyle \frac{{\text{m}}^6}{{\text{m}}^2}={\text{m}}^{6-2}={\text{m}}^4\)

\(\displaystyle \frac{{\text{x}}^2{\text{m}}^3{\text{x}}^3}{{\text{m}}^2{\text{x}}}=\frac{{\text{x}}^{2+3}{\text{m}}^3}{{\text{m}}^2{\text{x}}}={\text{x}}^{5-1}{\text{m}}^{3-2}={\text{x}}^4{\text{m}}\)

Additional Resources

• Khan Academy: Exponent Properties 2 (05:11 mins, Transcript)
• Khan Academy: Exponent Properties with Quotients (09:22 mins, Transcript)

Practice Problems

1. \(\dfrac{{\text{m}}^{5}}{{\text{m}}^{2}}\) (
   Solution

   x

   Solution:
   \({\text{m}}^{3}\)
   Details:
   
   Version 1: Using the Quotient Rule
   
   \(\dfrac{{\text{m}}^{5}}{{\text{m}}^{2}}\)
   
   This fraction has two factors: \({\text{m}}^{5}\) and \({\text{m}}^{2}\). The factor \({\text{m}}^{5}\) is in the numerator (top of the fraction). The factor \({\text{m}}^{2}\) is in the denominator (bottom of the fraction). Both of these factors have the same base m. The only operations being performed in this fraction are multiplication and division.
   
   According to the quotient rule for exponents, as long as the factors have the same base and everything is being multiplied or divided, then we can subtract the exponent of the factor in the denominator from the exponent of the factor in the numerator.
   
   Using the quotient rule:
   
   \(\dfrac{{\text{m}}^{5}}{{\text{m}}^{2}}={\text{m}}^{\left ( 5-2 \right )}={\text{m}}^{3}\)
   
   So our final answer is \({\text{m}}^{3}\).
   
   Version 2: Solving using the rules of multiplication and division
   
   This version shows what is happening behind the scenes in the quotient rule. The following is a more detailed explanation of why it works.
   
   We can multiply out the factors in both the top and bottom of the fraction.
   
   \(\dfrac{{\text{m}}^{5}}{{\text{m}}^{2}}=\dfrac{\text{m m m m m}}{\text{m m}}\)
   
   In this example we can separate the fraction into the following:
   
   \(\dfrac{\text{m m m m m}}{\text{m m}}=\dfrac{\text{m}}{\text{m}}\times\dfrac{\text{m}}{\text{m}}\times\dfrac{\text{m m m}}{1}\)
   
   (The denominator m m is the same as \(m \times m \times 1 = m m\). This is why the last fraction above has a 1 in the denominator.)
   
   Since
   
   \(\dfrac{\text{m}}{\text{m}}=1\) we can substitute 1 into our previous equation in two places.
   
   \(\dfrac{\text{m m m m m}}{\text{m m}}=1\times1\times\dfrac{\text{m m m}}{1}\)
   
   But 1 multiplied to anything is just itself and anything divided by 1 is just itself, so our final answer is:
   
   \(\dfrac{\text{m m m}}{1}={\text{m m m}}={\text{m}}^{3}\)
   )
2. \(\dfrac{{\text{x}}^{7}}{{\text{x}}^{5}}\) (
   Solution

   x

   Solution:
   \({\text{x}}^{2}\)
   )
3. \(\dfrac{{\text{m}}^{2}{\text{x}}^{2}}{\text{xm}}\) (
   Solution

   x

   Solution:
   \({\text{mx}}\)
   )
4. \(\dfrac{{\text{x}}^{2}{\text{y}}^{4}{\text{x}}^{7}}{{\text{xy}}^{3}}\) (
   Solution

   x

   Solution:
   \({\text{x}}^{8}{\text{y}}\)
   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 first 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 we have simplified the numerator, let’s rewrite our original expression:
   
   \(\displaystyle\frac{{\text{x}}^{2}{\text{y}}^{4}{\text{x}}^{7}}{{\text{x y}}^{3}}=\frac{{\text{x}}^{9}{\text{y}}^{4}}{{\text{x y}}^{3}}\)
   
   Now since there isn’t any addition or subtraction between any of the factors, we can use the quotient rule to further simplify this problem.
   
   The quotient rule says that as long as factors have the same base, we 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 we 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. We 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 us the same solution for this problem.
   
   If we expand all the factors in the equation from their exponent form to their standard multiplication form we 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 we can “cancel out” factors in the numerator and denominator that are the same because anything divided by itself equals 1.
   
   In this case, we can remove 1 x and 3 y 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, we haven’t actually gotten rid of these factors, instead, we 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 we still have a 1 in the denominator.
   
   Rearranging the factors and putting them into exponent form we get
   
   \(\dfrac{{\text{x}}^{8}{\text{y}}}{1}={\text{x}}^{8}{\text{y}}\)
   
   
   
   Our final answer is \({\text{x}}^{8}{\text{y}}\).
   )
5. \(\dfrac{{\text{x}}^{2}{\text{m}}^{3}{\text{x}}^{4}}{{\text{m}}^{2}{\text{x}}^{3}}\) (
   Video Solution

   x

   Solution:
   \({\text{mx}}^{3}\)
   Details:
   

   
   (Video Source | Transcript)
   )
6. \(\dfrac{{\text{x}}^{5}{\text{y}}^{2}{\text{x}}}{{\text{x}}^{2}{\text{yz}}}\) (
   Video Solution

   x

   Solution:
   \(\dfrac{{\text{x}}^{4}{\text{y}}}{\text{z}}\)
   Details:
   

   
   (Video Source | Transcript)
   )

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