Rules of Exponents:

Applying Them Together

In some of the exercises you will do, there will be multiple steps to simplifying the expression, much like in the Order of Operations. Each of these rules is a tool and all the tools can be used together to simplify expressions.

Video Source (03:43 mins) | Transcript

Just to review:

Product Rule: When the bases are the same, you add the powers
Quotient Rule: When the bases are the same, you subtract the powers
Negative Exponent Rule: Move it to the other part of the fraction (numerator or denominator) and the exponent becomes positive
Power Rule: Multiply the powers
Exponents of 0 and 1: anything raised to the 1 is itself, anything raised to the 0 is 1.
$(-1)$ Raised to an Exponent: if the exponent is even, the answer is positive, if the exponent is odd, the answer is negative.
Follow the Order of Operations (PEDMAS)

Additional Resources

Khan Academy: Exponent Properties 4 (03:06 mins, Transcript)
Khan Academy: Composite Problems (07:43 mins, Transcript)

Practice Problems

Simplify and Evaluate the following expressions:

${\text{a}}^{5}{\text{b}}^{3}\left ( {\text{a}}{\text{b}} \right )^{4}{\text{b}} =$

$\left ({\text{x}}{\text{y}} \right )^{3}{\text{x}}{\text{y}} =$

$\frac{{\text{x}}^{5}{\text{y}}^{3}{\text{x}}^{2}}{{\text{x}}^{6}{\text{y}}^{2}}=$

$\frac{{\text{m}}^{3}{\text{x}}^{7}}{{\text{m}}^{3}{\text{x}}^{2}}=$

$\left ( {\text{b}}^{4}{\text{x}}^{3}{\text{y}}{\text{b}} \right )^{2}{\text{x}} =$

$\left ( {-}{\text{m}} \right )^{3}{\text{b}}^{2}{\text{mx}}^{3} =$

$\left ( -3 \right )^{3}{\text{a}}^{2}{\text{b}}^{4}\left ( -2 \right )^{2} =$

Solutions

${\text{a}}^{9}{\text{b}}^{8}$ (Written Solution)

In this example, the first thing we 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{a}}{\text{b}} $ multiplied together 4 times.

This is the same as the variable ${\text{a}}$ multiplied together 4 times and the variable ${\text{b}}$ 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 we can substitute that back into our original expression.

Now we 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 ${\text{b}}$ on the end has an exponent of 1 and we need to include that in our solution.

${\text{a}}^{5}$ and ${\text{a}}^{4}$ become ${\text{a}}^{\left ( 5+4 \right )}={\text{a}}^{9}$

${\text{b}}^{3}$ and ${\text{b}}^{4}$ and ${\text{b}}$ become ${\text{b}}^{\left ( 3+4+1 \right )}={\text{b}}^{8}$

Our final solution is ${\text{a}}^{9}{\text{b}}^{8}$.
${\text{x}}^{4}{\text{y}}^{4}$
${\text{x}} {\text{y}}$ (Solution Video | Transcript)
${\text{x}}^{5}$ (Written Solution)

In this case, we are using the quotient rule on both variables ${\text{m}}$ and ${\text{x}}$.

$\frac{{\color{Blue} {\text{m}}}^{3}{\color{DarkOrange} {\text{x}}}^{7}}{{\color{Blue} {\text{m}}}^{3}{\color{DarkOrange} {\text{x}}}^{2}}$

First, look at the quotient rule for the ${\text{m}}$ variables.

$\frac{{\text{m}}^{3}}{{\text{m}}^{3}} = {\text{m}}^{\left ( 3-3 \right )} = {\text{m}}^{0} = 1$

$This image contains the equation: fraction m to the third power x to the seventh power over m to the third power x to the second power this equals m to the power of left parentheses 3 minus 3 right parentheses times the fraction x to the seventh power over x to the second power. The variable m is tinted blue and the variable x is tinted orange. The dot between the terms in the equation means multiplication$

Since ${\text{m}}^{\left ( 3-3 \right )} = {\text{m}}^{0} = 1$, and 1 multiplied to anything is itself, we only have the ${\text{x}}$ variables left.

${\color{Blue} {\text{m}}}^{\left ( 3-3 \right )}\cdot\frac{{\text{x}}^{7}}{{\text{x}}^{2}} = {\color{Blue} {\text{m}}}^{0} \cdot\frac{{\text{x}}^{7}}{{\text{x}}^{2}}={\color{Blue} 1}\cdot\frac{{\text{x}}^{7}}{{\text{x}}^{2}}$

Now we look at the $x$ variables and their exponents.

$\frac{{\text{x}}^{7}}{{\text{x}}^{2}} = {\text{x}}^{\left ( 7-2 \right )} = {\text{x}}^{5}$

${\color{Blue} 1}\cdot\frac{{\color{DarkOrange} {\text{x}}}^{7}}{{\color{DarkOrange} {\text{x}}}^{2}} = {\color{Blue} 1}\cdot {\color{DarkOrange} {\text{x}}}^{\left ( 7-2 \right )} = {\color{Blue} 1}\cdot {\color{DarkOrange} {\text{x}}}^{5} = {\color{DarkOrange} {\text{x}}}^{5}$

So we have $1 \cdot {\text{x}}^{5} = {\text{x}}^{5}$
${\text{b}}^{10}{\text{x}}^{7}{\text{y}}^{2}$ (Solution Video | Transcript)
$-{\text{m}}^{4}{\text{b}}^{2}{\text{x}}^{3}$
$-108{\text{a}}^{2}{\text{b}}^{4}$