Introduction

In this lesson, you will learn some of the rules of exponents and how to apply them together.

These videos illustrate the lesson material below. Watching the videos is optional.

• Rules of Exponents - Product Rule (07:19 mins) | Transcript
• Rules of Exponents: Dividing with Exponents (05:37 mins) | Transcript
• Rules of Exponents: Negative Exponents (07:39 mins) | Transcript
• Rules of Exponents: Power Rule (08:23 mins) | Transcript
• Rules of Exponents: Exponents of 0 and 1 (04:16 mins) | Transcript
• Rules of Exponents: Applying them Together (03:43 mins) | Transcript

Rules of Exponents Vocabulary

• Base: the number or variable that is being multiplied by itself.
• Power: the number of the exponent, meaning how many times the base is multiplied by itself.
• Product: the answer when you multiply a number/variable by another number/variable.
• Quotient: the answer when you divide a number/variable by another number/variable.

When working with exponents, it is helpful to learn the rules that can be applied to solve or simplify exponents.

The Product Rule of Exponents

$$\begin{array}{r}\text{Product Rule of Exponents:}\frac{a^x}{a^y}=a^{x-y}\end{array}$$

Remember that you can multiply in any order, so $(a)(b)=(b)(a)$. This means that if there are multiple bases, you can rearrange the order and add the exponents of any of the bases that are the same.

Example 1
Simplify $x^3{x}^4$.

This example involves variables, but it is performed in the same way as before.

\begin{align*}
&x^{3}x^{4}&\color{red}\small\text{Simplify this expression}\\&x^{3 + 4}&\color{red}\small\text{Product rule of exponents}\\&x^7 &\color{red}\small\text{Add the exponents with common base}\\ \end{align*

The Quotient Rule of Exponents

Remember, $x^4÷x^3$ 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 $x^1$, or $x$.

$$\begin{array}{r}\text{Quotient Rule of Exponents:}\frac{a^x}{a^y}=a^{x-y}\end{array}$$

Example 2
$\frac{m^6}{m^2}$

$$\begin{array}{rlr}& \frac{m^6}{m^2}& \text{Simplify this expression}\\ \\ & m^{6-2}& \text{Quotient rule of exponents}\\ \\ & m^4& \text{Subtract the exponents}\\ \end{array}$$

Negative Exponents

When using the quotient rule of exponents, you will sometimes have a negative exponent in your answer. $\frac{a^2}{a^5}=a^{(2-5)}=a^{-3}$. What does this mean? What does a to the power of -3 mean? Doing this by hand looks like this:

$$\begin{array}{r}\frac{a^2}{a^5}=\frac{\cancel{a}\cdot \cancel{a}}{\cancel{a}\cdot \cancel{a}\cdot a\cdot a\cdot a}=\frac{1}{a^3}\end{array}$$

In other words, $\frac{1}{a^3}$ is the simplified answer where there is no negative exponent.

So, any base raised to a negative exponent is actually equal to 1 over that number to the positive exponent, but in the denominator: $a^{-3}=\frac{1}{a^3}$.

This is also true when a base has a negative exponent in the denominator. In this case, the base is raised to the positive exponent and moved to the numerator, with 1 as the denominator.

$$\begin{array}{r}\text{Negative Exponents Rule: }{x}^{-a}=\frac{1}{x^a}\\ \\ \frac{1}{x^{-a}}=\frac{x^a}{1}\end{array}$$

Example 3

$\frac{x^2}{x^4}$

$$\begin{array}{rlr}& \frac{x^2}{x^4}& \text{Simplify this expression}\\ \\ & x^{(2-4)}& \text{Quotient rule of exponents}\\ \\ & x^{-2}& \text{Subtract exponents}\\ \\ & \frac{1}{x^2}& \text{Negative exponent rule}\end{array}$$

$x^{-2}$ can also be expressed as $\frac{1}{x^2}$.

The Power Rule of Exponents

$$\begin{array}{r}\text{Power Rule of Exponents: }(a^b{)}^c\text{nightarrow}{a}^{b\times c}\text{nightarrow}{a}^{bc}\end{array}$$

Example 4

$(3^4{)}^2$

$$\begin{array}{rlr}& (3^4{)}^2& \text{Simplify this expression}\\ \\ & 3^{4\times 2}& \text{Power rule of exponents}\\ \\ & 3^8& \text{Multiply exponents}\end{array}$$

Exponents of 1

Any base, like $a$, raised to the power of 1, $a^1$, is just equal to itself, a. So $a^1=a$.

Example 5

$$\begin{array}{r}\frac{a^4}{a^3}=a^{4-3}=a^1\end{array}$$

Because $a^1=a$, the answer is a.

Another way to simplify the expression $\frac{a^4}{a^3}$ is by canceling like bases from the numerator and denominator like this:

$$\begin{array}{r}\frac{a^4}{a^3}=\frac{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot a}{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}}=a\end{array}$$

So again, the answer is $a^1$, or a. Anything raised to the one power is just itself.

Exponents of 0

Any base, like $a$, raised to the power of 0, $a^0$, is equal to 1. So $a^0=1$.

Example 6

$$\begin{array}{r}\frac{a^2}{a^2}=a^{2-2}=a^0\end{array}$$

To understand what this means, simplify the same expression by canceling the common bases.

$$\begin{array}{r}\frac{a^2}{a^2}=\frac{\cancel{a}\cdot \cancel{a}}{\cancel{a}\cdot \cancel{a}}=1\cdot 1=1\end{array}$$

$\frac{a^2}{a^2}=a^0=1$. This is true for any value for $a$, except for zero, $0^0\ne 1$. Base 0, to the power of 0 is undefined.

Applying Exponents Together

Example 7
This example has several different variables. All the variables in this expression are being multiplied together and all are being raised to an exponent.
$$\begin{array}{r}(m^2{x}^{3}ymx{)}^2\end{array}$$

Use the order of operations and what you know about exponents to simplify this problem.

First, according to the order of operations you need to look at what is inside of the parentheses. Once you've simplified the variables inside the parentheses as much as possible, you can move on.

The first thing to notice is that, inside the parentheses, there are terms with the same base being multiplied together. You have $m^2$ and $m^1$, as well as $x^3$ and $x^1$.

According to the product rule, you should add the exponents of like terms together.

$$\begin{array}{rlr}& (m^2{x}^{3}ymx{)}^2& \text{Simplify the parentheses first}\\ \\ & (m^2{x}^3{y}^1{m}^1{x}^1{)}^2& \text{Note: Variables to the power of 1}\\ \\ & (m^2{m}^1{x}^3{x}^1{y}^1{)}^2& \text{Rearrange the like bases to be together}\\ \\ & (m^{(2+1)}{x}^{(3+1)}{y}^1{)}^2& \text{Product rule of exponents}\\ \\ & (m^3{x}^{4}y{)}^2& \text{Add the exponents of likes bases}\\ \\ & m^{3\cdot 2}{x}^{4\cdot 2}{y}^{1\cdot 2}& \text{Power rule of exponents}\\ \\ & m^6{x}^8{y}^2& \text{Multiply exponents and simplify}\end{array}$$

By following the order of operations and the rules of exponents, you were able to simplify this equation down to $m^6{x}^8{y}^2$.

Example 8
$\frac{x^2{m}^3{x}^3}{m^2{x}^7}$

This equation has two different variables, or terms with different bases: $x$ and $m$.

$$\begin{array}{rlr}& \frac{x^2{m}^3{x}^3}{m^2{x}^7}& \text{Simplify this expression}\\ \\ & \frac{m^3{x}^2{x}^3}{m^2{x}^7}& \text{Rearrange into like bases}\\ \\ & \frac{m^3{x}^{(2+3)}}{m^2{x}^7}& \text{Product rule of exponents}\\ \\ & \frac{m^3{x}^5}{m^2{x}^7}& \text{Add the exponents}\\ \\ & m^{(3-2)}{x}^{(5-7)}& \text{Quotient rule of exponents}\\ \\ & m{x}^{-2}& \text{Subtract exponents}\\ \\ & m\frac{1}{x^2}& \text{Negative exponent rule}\\ \\ & \frac{m}{x^2}& \text{Simplify}\end{array}$$

Remember, $m^1=m=\frac{m}{1}$. The final answer is $\frac{m}{x^2}$.

Things to Remember

• The Order of Operations is PEMDAS. 
• Product Rule: When multiplying two exponents and the bases are the same, add the powers.

$$\begin{array}{r}{x}^a{x}^b=x^{a+b}\end{array}$$

• Quotient Rule: When dividing exponents and the bases are the same, subtract the powers.

$$\begin{array}{r}\frac{a^x}{a^y}=a^{x-y}\end{array}$$

• Negative Exponent Rule: When an exponent has a negative power, move it to the other part of the fraction (numerator or denominator) and the power becomes positive.

$$\begin{array}{r}{x}^{-a}=\frac{1}{x^a}\end{array}$$

• Power Rule: When an exponent is linked to the outside of a parentheses, multiply the powers. 

$$\begin{array}{r}(x^a{)}^b=x^{(a)(b)}\end{array}$$

• Exponents of 0 and 1: Anything raised to the power of 0 is 1, anything raised to the power of 1 is itself.

$$\begin{array}{r}{a}^0=1\\ a^1=a\end{array}$$

Practice Problems

1. $\text{Extra left or missing right}$ (
   Solution

   x

   Solution: ${\text{a}}^9{\text{b}}^8$
   Details:
   In this example, the first thing you need to do is look at the ${\left(\text{a}\text{b}\right)}^4$ part.
   
   ${\left(\text{a}\text{b}\right)}^4$ is the same as $\text{ab}$ multiplied together 4 times.
   
    
   
   This is the same as the variable a multiplied together 4 times and the variable b multiplied together 4 times.
   
    
   
   Thus, the part ${\left(\text{a}\text{b}\right)}^4$ is the same as ${\text{a}}^4{\text{b}}^4$ and you can substitute that back into the original expression.
   
    
   
   Now you can add the exponents of factors with the same base. Remember that anything that doesn’t have an exponent actually has an invisible exponent of 1. So the variable b on the end has an exponent of 1 and you need to include that in the solution.
   
   ${\text{a}}^5$ and ${\text{a}}^4$ become ${\text{a}}^{(5+4)}={\text{a}}^9$
   
   ${\text{b}}^3$, ${\text{b}}^4$ and $\text{b}$ become ${\text{b}}^{(3+4+1)}={\text{b}}^8$
   
    
   
   The final solution is: ${\text{a}}^9{\text{b}}^8$.
   )
2. $\text{Extra left or missing right}$ (
   Solution

   x

   Solution: ${\text{x}}^4{\text{y}}^4$
   )
3. ${\displaystyle \frac{{\text{x}}^5{\text{y}}^3{\text{x}}^2}{{\text{x}}^6{\text{y}}^2}}=$ (
   Video Solution

   x

   Solution: $\text{x}\text{y}$
   Details:
   

   
   
   | Transcript)
4. ${\displaystyle \frac{{\text{m}}^3{\text{x}}^7}{{\text{m}}^3{\text{x}}^2}}=$ (
   Solution

   x

   Solution: ${\text{x}}^5$
   Details:
   In this case, you are using the quotient rule on both variables m and x.
   
   ${\displaystyle \frac{{\text{m}}^3{\text{x}}^7}{{\text{m}}^3{\text{x}}^2}}$
   
   First, look at the quotient rule for the m variables.
   
   ${\displaystyle \frac{{\text{m}}^3}{{\text{m}}^3}}={\text{m}}^{(3-3)}={\text{m}}^0=1$
   
    
   
   Note: The dot between the terms in the equation means multiplication.
   
   Since ${\text{m}}^{(3-3)}={\text{m}}^0=1$ and 1 multiplied to anything is itself, you only have the x variables left.
   
   ${\displaystyle {\text{m}}^{(3-3)}\cdot \frac{{\text{x}}^7}{{\text{x}}^2}={\text{m}}^0\cdot \frac{{\text{x}}^7}{{\text{x}}^2}=1\cdot \frac{{\text{x}}^7}{{\text{x}}^2}}$
   
   Now look at the x variables and their exponents.
   
   ${\displaystyle \frac{{\text{x}}^7}{{\text{x}}^2}}={\text{x}}^{(7-2)}={\text{x}}^5$
   
   ${\displaystyle 1\cdot \frac{{\text{x}}^7}{{\text{x}}^2}=1\cdot {\text{x}}^{(7-2)}=1\cdot {\text{x}}^5={\text{x}}^5}$
   
   So you have $1\cdot {\text{x}}^5={\text{x}}^5$.
   )
5. $\text{Extra left or missing right}$ (
   Video Solution

   x

   Solution: ${\text{b}}^{10}{\text{x}}^7{\text{y}}^2$
   Details:
   

   
   
   | Transcript)
6. $\text{Extra left or missing right}$ (
   Solution

   x

   Solution: $-{\text{m}}^4{\text{b}}^2{\text{x}}^3$
   )
7. $\text{Extra left or missing right}$ (
   Solution

   x

   Solution: $-108{\text{a}}^2{\text{b}}^4$
   )