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Section 1.1 Exponents

Learning Objectives.

  1. Recall the rules of exponents to be able to simplify complicated expressions.

In this sections, we review the rules of exponents and look at ways to simplify expressions using positive and negative exponents and rational exponents. We will use these skills throughout the course. This is a new section unique to this custom edition. It was written by Katie Hall.

Subsection 1.1.1 Exponent Overview

We have seen that \(x^3=x\cdot x \cdot x.\) Using this rule, we can see several properties of exponents. First, let's explore what we get if we have an exponent of 0 or a negative number. We will start by looking at the values of \(2^x \) for several values of \(x \text{.}\)

Table 1.2. Values of \(2^x\)
\(x\) \(2^x\)
\(5\) \(32\)
\(4\) \(16\)
\(3\) \(8\)
\(2\) \(4\)
\(1\) \(2\)

Notice that if we start in the middle of the table, we have to multiply by 2 to move up a row and divide by 2 to move down. Using this division principle, we can fill in several more rows of the table.

Table 1.3. Values of \(2^x\)
\(x\) \(2^x\)
\(5\) \(32\)
\(4\) \(16\)
\(3\) \(8\)
\(2\) \(4\)
\(1\) \(2\)
\(0\) \(1\)
\(-1\) \(1/2\)
\(-2\) \(1/4\)

This general pattern holds turn for whatever base we have. From this, we get the following rules:

Note 1.4. Exponent Rules.

For any constants \(a, b\)

  1. \(\displaystyle a^0=1\)

  2. \(a^{-b}=\frac{1}{a^b}\) when \(a \ne 0 \)

Example 1.5. Using Exponent Rules.

Using the basic exponent rules above, evaluate the following

  1. \(\displaystyle 5^{-2}\)

  2. \(\displaystyle 4^{0}\)

  3. \(\displaystyle 2^{3}\cdot3^{-2}\)

Solution.

  1. \(\displaystyle 5^{-2}=\frac{1}{5^2}=\frac{1}{25}\)

  2. \(\displaystyle 4^{0}=1\)

  3. \(\displaystyle 2^{3}\cdot 3^{-2}=8 \cdot \frac{1}{3^2} = 8 \cdot \frac{1}{9}=8/9\)

Checkpoint 1.6.

Using the basic exponent rules above, evaluate the following

  1. \(\displaystyle 3^{-3}\)

  2. \(\displaystyle 2^{0}\)

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

Hint.

Use each of the rules above. For the third one, find each part separately, then multiply them.

Solution.

  1. \(\displaystyle 3^{-3}=\frac{1}{27}\)

  2. \(\displaystyle 2^0=1\)

  3. \(\displaystyle 4^{2}\cdot 2^{-5}= 1/2\)

Subsection 1.1.2 More Exponent Rules

By considering the definition of an expressions like \(x^2 \) or \(x^3\text{,}\) we can start to derive more rules of exponents. For example, if we have \(x^2 \cdot x^3 \) this is the same as \((x \cdot x) \cdot (x \cdot x \cdot x) = x^5 \) since we end up with five \(x\)s all multiplied together. This sort of analysis leads to the first five rules below.

You may have seen in an algebra class that \(x^{\frac{1}{n}}=\sqrt[n]{x} \text{.}\) From the third rule, we know that \((x^{1/n})^n=x^{(1/n) \cdot n } = x \text{.}\) This allows us to conclude that \(x^{1/n}\) is the same as the number which when raised to the \(n\)th power becomes \(x\text{,}\) which is the defintion of \(\sqrt[n]{x} \text{.}\) One way to think about fraction powers, like \(x^{m/n} \) is the the numerator gives the power on \(x \) and the denominator tells which root it is. When dealing with roots, it is important to remember that we can't take an even root of a negative number.

Note 1.7. More Exponent Rules.

For any constants \(m, n\) and for all \(x\) and \(y\text{,}\)

  1. \(\displaystyle x^m\cdot x^n=x^{m+n}\)

  2. \(\frac{x^m}{x^n}=x^{m−n}\) when \(x \ne 0 \)

  3. \(\displaystyle (x^m)^n=x^{mn}\)

  4. \(\displaystyle (xy)^n=x^ny^n\)

  5. \(\displaystyle \left(\frac{x}{y}\right)^n=\frac{x^n}{y^n} \)

  6. \(\sqrt[n]{x}=x^{1/n}\) when defined

  7. \(\sqrt[n]{x^m}=x^{m/n}\) when defined

We can use these rules to simplify complicated expressions. When we simplify (for now) we want to have positive and negative powers but no radicals and division. For example, to simplify

\begin{equation*} \frac{x^3\sqrt{z}}{y^2z^5}, \end{equation*}

we can first split it to focus on the \(x\text{,}\) \(y\text{.}\) and \(z\) separately. We get

\begin{equation*} x^3 \cdot \frac{1}{y^2}\cdot\frac{\sqrt{z}}{z^5}. \end{equation*}

Then since \(\frac{1}{y^2}=y^{-2}\) and \(\frac{\sqrt{z}}{z^5}=\frac{z^{1/2}}{z^5}=z^{1/2-5}=z^{-9/2}\text{,}\) we get

\begin{equation*} x^3y^{-2}z^{-9/2}. \end{equation*}

Example 1.8. Simplifying Expressions with Exponents.

Using rules above, simplify the expression so tht we have positive and negative rational powers and not radicals or division.

  1. \(\displaystyle \frac{x^2\sqrt[4]{z^8}}{(xy)^2z^{-4}}\)

  2. \(\displaystyle \frac{x^3\sqrt[2]{z^7}}{(xz)^4y^{-5}}\)

Solution.

  1. \(\displaystyle \frac{x^2\sqrt[4]{z^8}}{(xy)^2z^{-4}} = \frac{x^2z^{8/4}}{x^2y^2z^{-4}}=\frac{x^2}{x^2}\frac{1}{y^2}\frac{z^2}{z^{-4}}=y^{-2}z^6\)

  2. \(\displaystyle \frac{x^3\sqrt[2]{z^7}}{(xz)^4y^{-5}} = \frac{x^3z^{7/2}}{x^4z^4y^{-5}}=\frac{x^3}{x^4}\frac{1}{y^{-5}}\frac{z^{7/2}}{z^4}=x^{-1}y^5z^{-1/2} \)

Checkpoint 1.9.

Using rules above, simplify the expression so tht we have positive and negative rational powers and not radicals or division.

  1. \(\displaystyle \frac{\sqrt[3]{x^2y^6}}{(xy)^3z^{4}}\)

  2. \(\displaystyle \frac{x^5\sqrt[3]{z^6}}{(xy)^5\sqrt{z}}\)

Hint.

Use each of the rules above. Break the expression into a product of a term with just \(x\)'s, a term with just \(y\)'s and a term with just \(z\)'s.

Solution.

  1. \(\displaystyle x^{-7/3}y^{-1}z^{-4} \)

  2. \(\displaystyle y^{-5}z^{3/2}\)

This book is a custom edition based on OpenStax Calculus Volume 1. You can download the original for free at https://openstax.org/details/books/calculus-volume-1. 1 

https://openstax.org/details/books/calculus-volume-1