# Lesson 1

Properties of Exponents

- Let’s use integer exponents.

### 1.1: Which One Doesn’t Belong: Exponents and Equations

A. \(2^3 = 9\)

B. \(9 = 3^2\)

C. \(2 \boldcdot 2 \boldcdot 2 \boldcdot 2 = 16\)

D. \(a \boldcdot 2^0 = a\)

### 1.2: Name That Power

Find the value of each variable that makes the equation true. Be prepared to explain your reasoning.

- \(2^3 \boldcdot 2^5 = 2^a\)
- \(3^b \boldcdot 3^7 = 3^{11}\)
- \(\frac{4^3}{4^2} = 4^c\)
- \(\frac{5^8}{5^d} = 5^2\)
- \(6^m \boldcdot 6^m \boldcdot 6^m = 6^{21}\)
- \((7^n)^4 = 7^{20}\)
- \(2^4 \boldcdot 3^4 = 6^s\)
- \(5^3 \boldcdot t^3 = 50^3\)

### 1.3: The Power of Zero

- Use exponent rules to write each expression as a single power of 2. Find the value of the expression. Record these in the table. The first row is done for you.
expression power of 2 value \(\frac{2^5}{2^1}\) \(2^4\) 16 \(\frac{2^5}{2^2}\) \(\frac{2^5}{2^3}\) \(\frac{2^5}{2^4}\) \(\frac{2^5}{2^5}\) \(\frac{2^5}{2^6}\) \(\frac{2^5}{2^7}\) - What is the value of \(5^0\)?
- What is the value of \(3^{\text{-}1}\)?
- What is the value of \(7^{\text{-}3}\)?

Explain why the argument used to assign a value to the expression \(2^0\) does not apply to make sense of the expression \(0^0\).

### 1.4: Matching Exponent Expressions

Sort expressions that are equal into groups. Some expressions may not have a match, and some may have more than one match. Be prepared to explain your reasoning.

- \(2^{\text{-}4}\)
- \(\frac{1}{2^4}\)
- \(\text{-}2^4\)
- \(\text{-}\frac{1}{2^4}\)
- \(4^2\)
- \(4^{\text{-}2}\)
- \(\text{-}4^2\)
- \(\text{-}4^{\text{-}2}\)
- \(2^7 \boldcdot 2^{\text{-}3}\)
- \(\frac{2^7}{2^{\text{-}3}}\)
- \(2^{\text{-}7} \boldcdot 2^{3}\)
- \(\frac{2^{\text{-}7}}{2^{\text{-}3}}\)
- \((\text-4)^2\)

### Summary

Exponent rules help us keep track of a base’s repeated factors. Negative exponents help us keep track of repeated factors that are the *reciprocal* of the base. We can define a number to the power of 0 to have a value of 1. These rules can be written symbolically as:

\(\begin{align}b^m \boldcdot b^n &= b^{m+n} \\ \left(b^m\right)^n &= b^{m \boldcdot n} \\ \frac{b^m}{b^n} &= b^{m-n} \\ b^{\text-n} &= \frac{1}{b^n} \\ b^0 &= 1 \\ a^n \boldcdot b^n &= (a \boldcdot b)^n \end{align}\)

Here, the base \(b\) can be any positive number, and the exponents \(n\) and \(m\) can be any integer.