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Section 1.3 Exponents

Most fundamentally, exponents represent repeated multiplication in the way that multiplication represents repeated addition.
Table 1.3.1. Exponents and Multiplication as Repeated Operations
Multiplication as Repeated Addition Exponents as Repeated Multiplication
\(na = \underbrace{a+ a +\cdots + a}_\text{$n$ times}\) \(\longrightarrow\) \(a^n = \underbrace{a\cdot a \cdot\cdots \cdot a}_\text{$n$ times}\)
Based on this idea, we can see the following properties of exponents:
We can use these properties to re-write expressions with exponents in alternate forms. When we do this, it is helpful to do the following to solidify you understanding:
  • Take it one step at a time and write down which exponent property you’re using at each step.
  • If the exponents are positive whole numbers, try to write the expression out without exponents as an intermediate step, as in the "for example" lines in 1.3.2.

Example 1.3.3.

Suppose we are given the instruction to write the expression \(\frac{2^a4^a}{8^b}\) with a single exponent and base \(8\text{.}\) In other words, we want to find the expression for \(x\) in the equation
\begin{equation*} \frac{2^a4^a}{8^b} = 8^x\text{.} \end{equation*}
To do this, we’ll use the exponent properties.
\begin{align*} \frac{2^a4^a}{8^b} \amp = \frac{(2\cdot 4)^a}{8^b} \amp \text{ using property } 4 \\ \amp = \frac{8^a}{8^b} \amp \text{ multiply } \\ \amp = 8^{a-b} \amp \text{ using property } 2. \end{align*}
Thus, our expression for \(x\) is \(a-b\text{.}\)
To check whether this really makes sense, let \(a = 7\) and \(b=3\text{.}\) Then we can work it out without explicitly using any exponent properties to get the same answer.
\begin{align*} \frac{2^74^7}{8^3} \amp = \frac{2\cdot 2 \cdot 2 \cdot 2 \cdot 2\cdot 2\cdot 2\cdot 4 \cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4 }{8\cdot 8\cdot 8} \\ \amp = \frac{2\cdot 4 \cdot 2 \cdot 4 \cdot 2\cdot 4\cdot 2\cdot 4 \cdot 2\cdot 4\cdot 2\cdot 4\cdot 2\cdot 4}{8\cdot 8\cdot 8} \\ \amp = \frac{8\cdot 8\cdot 8\cdot 8\cdot \bcancel{8}\cdot \bcancel{8}\cdot \bcancel{8}}{\bcancel{8}\cdot \bcancel{8}\cdot \bcancel{8}} = 8^4 \end{align*}
Observing that \(4 = 7-3\text{,}\) we’re seeing the same answer as we got with exponent properties. Here we strictly used the definition of exponents.
Here are a some exercises for you to practice.

Checkpoint 1.3.4.

Rewrite the following using a single exponent.
\(4^{9} \cdot 4^n = 4^x\) for \(x\) =

Checkpoint 1.3.5.

Write the following expression in the form \(x^n\text{,}\) assuming \(x \not= 0\text{.}\) Simplify your answer as much as possible.
\(\displaystyle \frac{ \left( x \cdot x^{8} \right)^2 }{ \left( x^{5} \right)^3 }\) =

Checkpoint 1.3.6.

Rewrite the following as a perfect square.
\(25 b^{12} = \big(\) \(\big)^{2}\)
More subtlety arises when making sense out of zero, negative, and fractional exponents. The problem is that we don’t know what it means to multiply a number by itself zero times, a negative number of times, or a fractional number of times. Because of this, we use the properties listed above to define zero, negative, and fractional exponents.
First, let’s consider what \(a^0\) should equal, where \(a\) is a nonzero number. We know that if \(n\) is any number that \(0 = n-n\text{.}\) Using the second exponent property given above, we have
\begin{equation*} a^0 = a^{n-n} = \frac{a^n}{a^n}. \end{equation*}
Since \(\frac{a^n}{a^n} = 1\text{,}\) we define \(a^0 = 1\) whenever \(a\) is not zero (\(0^0\) is undefined).
Now consider a negative exponent, \(a^{-n}\text{.}\) Here we use the first two exponent properties to arrive at what \(a^{-n}\) must be. We have
\begin{equation*} a^{n}a^{-n} = a^{n+(-n)} = a^{n-n} = a^{0} = 1. \end{equation*}
Shortening this string of equalities, we have \(a^{n}a^{-n} = 1\text{.}\) Now we divide both sides by \(a^{n}\) to get \(a^{-n} = \frac{1}{a^{n}}\text{.}\) In words, a negative exponent indicates division by the same number with a positive exponent.

Checkpoint 1.3.7.

Simplify the following. Write numerical answers as fractions multiplied out and write expressions containing variables using positive exponents only.
(a) \(\left(4^{2}\right)^{-1} =\)
(b) \(\displaystyle \left(\frac{4}{5}\right)^{-2} =\)
(c) \(\displaystyle \left(y^{4}\right)^{-4} =\)
The next one looks daunting at first, but isn’t that bad if you take it slow.

Checkpoint 1.3.8.

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Finally, let’s consider what \(a^{\frac{1}{n}}\) should be. This time we’ll use the third exponent property that a power raised to a power reults in the exponents being multiplied. Thus we have,
\begin{equation*} (a^{\frac{1}{n}})^n = a^{\frac{n}{n}} = a^{1} = a. \end{equation*}
So \(a^{\frac{1}{n}}\) is the number that when raised to the \(n\)-th power results in \(a\text{.}\) This is exactly a description of the \(n\)-th root, \(\sqrt[n]{a}\text{.}\) Going slightly further with the third exponent property, we have \(a^\frac{m}{n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m\text{.}\)

Question 1.3.9.

Justify the last sentence of the last paragraph. Hint: First note that \(\frac{m}{n}\) can be written as \(m\cdot\frac{1}{n}\text{,}\) then apply property number three from 1.3.2
So in summary we have the following:

Checkpoint 1.3.11.

Evaluate the following without a calculator. Make sure you do it as efficiently as possible using properties of exponents.
  1. \(4^{\frac{3}{2}} =\)
  2. \(9^{-\frac{1}{2}} =\)
  3. \((\frac{1}{27})^{-\frac{2}{3}} =\)
  4. \(\frac{1}{7^{-2}} =\)
  5. \(\frac{10^9}{10^7} =\)
  6. \(16^{\frac{5}{4}} =\)

Question 1.3.12.

Use properties of exponents to write the following in a simpler way using only positive exponents.
  1. \(-4^0\)
  2. \(z^{-3} \cdot z^3\)
  3. \(w^{-4} \cdot(4x)^0 \cdot x^6\)
  4. \(\dfrac{5t^3}{t^{-6}}\)
  5. \(\left(\dfrac{a^2}{b^3} \right)^{-1}\)
  6. \(\left(\dfrac{x}{3} \right)^{-2}\)