ORCCA Introduction to Exponent Rules

Section 2.9 Introduction to Exponent Rules

Objectives PCC Course Content and Outcome Guide

A.1.1:2.i

In this section, we will look at some rules or properties we use when simplifying expressions that involve multiplication and exponents.

Figure 2.9.1 Alternative Video Lesson

Subsection 2.9.1 Exponent Basics

Before we discuss any exponent rules, we need to quickly remind ourselves of some important concepts and vocabulary.

When working with expressions with exponents, we have the following vocabulary:

\begin{equation*} \text{base}^{\text{exponent}} = \text{power} \end{equation*}

For example, when we calculate \(8^{2} = 64\text{,}\) the base is \(8\text{,}\) the exponent is \(2\text{,}\) and the expression \(8^{2}\) is called the 2nd power of 8.

The other foundational concept is that if an exponent is a positive integer, the power can be rewritten as repeated multiplication of the base. For example, the 4th power of \(3\) can be written as \(4\) factors of \(3\) like so:

\begin{equation*} 3^{4} = 3 \cdot 3 \cdot 3 \cdot 3 \end{equation*}

Subsection 2.9.2 Exponent Rules

Product Rule

If we write out \(3^5\cdot 3^2\) without using exponents, we'd have:

\begin{equation*} 3^5 \cdot 3^2 = \left(3 \cdot 3\cdot 3\cdot 3\cdot 3\right) \cdot \left(3 \cdot 3\right) \end{equation*}

If we then count how many \(3\)s are being multiplied together, we find we have \(5+2=7\text{,}\) a total of seven \(3\)s.

\begin{align*} 3^5\cdot 3^2 \amp= 3^{5+2}\\ \amp= 3^7 \end{align*}

Example 2.9.2

Simplify \(x^2\cdot x^3\text{.}\)

To simplify \(x^2\cdot x^3\text{,}\) we write this out in its expanded form, as a product of \(x\)'s, we have

\begin{align*} x^2\cdot x^3 \amp=(x\cdot x)(x \cdot x \cdot x)\\ \amp=x\cdot x\cdot x \cdot x \cdot x\\ \amp=x^5 \end{align*}

Note that we obtained the exponent of \(5\) by adding \(2\) and \(3\text{.}\)

This is our first rule, the Product Rule: when multiplying two expressions that have the same base, we can simplify the product by adding the exponents.

\begin{equation} x^m \cdot x^n = x^{m+n}\label{equation-product-rule}\tag{2.9.1} \end{equation}

Checkpoint 2.9.3

Power to a Power Rule

The second rule is an extension of the first rule. If we write out \(\left(3^5\right)^2\) without using exponents, we'd have \(3^5\) multiplied by itself:

\begin{align*} \left(3^5\right)^2 \amp= \left(3^5\right)\cdot \left(3^5\right)\\ \amp= \left(3\cdot 3\cdot 3\cdot 3 \cdot 3 \right) \cdot \left(3 \cdot 3\cdot 3\cdot 3\cdot 3\right) \end{align*}

If we again count how many \(3\)s are being multiplied, we have a total of two groups each with five \(3\)s. So we'd have \(2\cdot 5=10\) instances of a \(3\text{.}\)

\begin{align*} \left(3^5\right)^2 \amp= 3^{2\cdot 5}\\ \amp= 3^{10} \end{align*}

Example 2.9.4

Simplify \(\left(x^2\right)^3\text{.}\)

To simplify \(\left(x^2\right)^3\text{,}\) we write this out in its expanded form, as a product of \(x\)'s, we have

\begin{align*} \left(x^2\right)^3 \amp=\left(x^2\right) \cdot \left(x^2\right)\cdot\left(x^2\right)\\ \amp=(x \cdot x)\cdot (x \cdot x)\cdot (x \cdot x)\\ \amp=x^6 \end{align*}

Note that we obtained the exponent of \(6\) by multiplying \(2\) and \(3\text{.}\)

We have our second rule, the Power to a Power Rule: when a base is raised to an exponent and that expression is raised to another exponent, we multiply the exponents.

\begin{equation*} \left(x^m\right)^n = x^{m \cdot n} \end{equation*}

Checkpoint 2.9.5

Product to a Power Rule

The third exponent rule deals with having multiplication inside a set of parentheses and an exponent outside the parentheses. If we write out \(\left(3t\right)^5\) without using an exponent, we'd have \(3t\) multiplied by itself five times:

\begin{equation*} (3t)^5= (3t)(3t)(3t)(3t)(3t) \end{equation*}

Keeping in mind that there is multiplication between every \(3\) and \(t\) and multiplication between all of the parentheses, we can reorder and regroup the factors:

\begin{align*} \left(3t\right)^5 \amp= (3\cdot t)\cdot (3\cdot t)\cdot (3\cdot t)\cdot (3\cdot t)\cdot (3\cdot t)\\ \amp= \left(3\cdot 3\cdot 3\cdot 3\cdot 3 \right) \cdot \left(t \cdot t \cdot t \cdot t \cdot t\right) \\ \amp= 3^5 t^5 \end{align*}

We essentially applied the outer exponent to each factor inside the parentheses.

Example 2.9.6

Simplify \((xy)^5\text{.}\)

To simplify \((xy)^5\text{,}\) we write this out in its expanded form, as a product of \(x\)'s and \(y\)'s, we have

\begin{align*} (xy)^5 \amp=(x \cdot y) \cdot (x \cdot y) \cdot (x \cdot y) \cdot (x \cdot y) \cdot (x \cdot y)\\ \amp=(x \cdot x \cdot x \cdot x \cdot x) \cdot (y \cdot y \cdot y \cdot y \cdot y)\\ \amp=x^5 y^5 \end{align*}

Note that the exponent on \(xy\) can simply be applied to both \(x\) and \(y\text{.}\)

This is our third rule, the Product to a Power Rule: when a product is raised to an exponent, we can apply the exponent to each factor in the product.

\begin{equation*} \left(x\cdot y\right)^n = x^{n}\cdot y^{n} \end{equation*}

Checkpoint 2.9.7

If \(a\) and \(b\) are real numbers, and \(n\) and \(m\) are positive integers, then we have the following rules:

Product Rule: \(a^{n} \cdot a^{m} = a^{n+m}\)
Power to a Power Rule: \((a^{n})^{m} = a^{n\cdot m}\)
Product to a Power Rule: \((ab)^{n} = a^{n} \cdot b^{n}\)

List 2.9.8 Summary of the Rules of Exponents for Multiplication

Many examples we'll come across will make use of more than one exponent rule. In deciding which exponent rule to work with first, it's important to remember that the order of operations still applies.

Example 2.9.9

Simplify the following expressions.

\(\left(3^7r^5\right)^4\)
\(\left(t^3\right)^2\cdot \left(t^4\right)^5\)

Explanation

Since we cannot simplify anything inside the parentheses, we'll begin simplifying this expression using the Product to a Power Rule. We'll apply the outer exponent of 4 to each factor inside the parentheses. Then we'll use the Power to a Power Rule to finish out simplification process:

\begin{align*} \left(3^7r^5\right)^4 \amp= \left(3^7\right)^4 \cdot \left(r^5\right)^4\\ \amp= 3^{7\cdot4} \cdot r^{5\cdot 4} \\ \amp= 3^{28}r^{20} \end{align*}
According to the order of operations, we should first simplify any exponents before carrying out any multiplication. Therefore, we'll begin simplifying this by applying the Power to a Power Rule and then finish using the Product Rule:

\begin{align*} \left(t^3\right)^2\cdot \left(t^4\right)^5 \amp= t^{3\cdot2}\cdot t^{4\cdot5} \\ \amp= t^6 \cdot t^{20} \\ \amp= t^{6+20} \\ \amp= t^{26} \end{align*}

Remark 2.9.10

We cannot simplify an expression like \(x^2y^3\) using the Product Rule, as the factors \(x^2\) and \(y^3\) do not have the same base.

Subsection 2.9.3 Exercises

Review and Warmup

1

Evaluate the following.

\(\displaystyle{ 3^{2}= }\)
\(\displaystyle{ 4^{3}= }\)
\(\displaystyle{ (-4)^{2}= }\)
\(\displaystyle{ (-5)^{3}= }\)

2

Evaluate the following.

\(\displaystyle{ 3^{2}= }\)
\(\displaystyle{ 2^{3}= }\)
\(\displaystyle{ (-2)^{2}= }\)
\(\displaystyle{ (-3)^{3}= }\)

3

Evaluate the following.

\(\displaystyle{ 1^{7}= }\)
\(\displaystyle{ (-1)^{15}= }\)
\(\displaystyle{ (-1)^{14}= }\)
\(\displaystyle{ 0^{19}= }\)

4

Evaluate the following.

\(\displaystyle{ 1^{8}= }\)
\(\displaystyle{ (-1)^{13}= }\)
\(\displaystyle{ (-1)^{16}= }\)
\(\displaystyle{ 0^{17}= }\)

5

Evaluate the following.

\(\displaystyle{ (-3)^{2}= }\)
\(\displaystyle{ -10^{2}= }\)

6

Evaluate the following.

\(\displaystyle{ (-1)^{2}= }\)
\(\displaystyle{ -2^{2}= }\)

7

Evaluate the following.

\(\displaystyle{ (-1)^{3}= }\)
\(\displaystyle{ -2^{3}= }\)

8

Evaluate the following.

\(\displaystyle{ (-4)^{3}= }\)
\(\displaystyle{ -1^{3}= }\)