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Palladium Algebra

Section 4.2 More Exponent Rules

There are a few more properties for working with exponents that we have not yet covered.
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Figure 4.2.1. Alternative Video Lesson
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Subsection 4.2.1 Review of Exponent Rules for Products and Exponents

In Section 1, we introduced three important properties of products with exponents. We begin this section with a recap of these exponent rules.
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List 4.2.2. Summary of Exponent Rules (Thus Far)
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Product Rule
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When multiplying two expressions that have the same base, simplify the product by adding the exponents.
\begin{equation*} x^m \cdot x^n = x^{m+n} \end{equation*}
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Power to a Power Rule
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When a base is raised to an exponent and that expression is raised to another exponent, multiply the exponents.
\begin{equation*} \left(x^m\right)^n = x^{m \cdot n} \end{equation*}
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Product to a Power Rule
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When a product is raised to an exponent, apply the exponent to each factor in the product.
\begin{equation*} \left(x\cdot y\right)^n = x^{n}\cdot y^{n} \end{equation*}
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Quotient of Powers Rule
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When dividing two expressions that have the same base, simplify the quotient by subtracting the exponents. We require that the base \(x\) here not be zero.
\begin{equation*} \frac{x^m}{x^n} = x^{m-n} \end{equation*}
For now, we only know this rule when \(m\gt n\text{.}\)
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Checkpoint 4.2.3.

  1. Simplify \(r^{16}\cdot r^{5}\text{.}\)
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  2. Simplify \(\left(x^{11}\right)^{10}\text{.}\)
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  3. Simplify \((3r)^{4}\text{.}\)
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  4. Simplify \(\frac{3y^7}{y^3}\text{.}\)
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Explanation.
  1. We add the exponents because this is a product of powers with the same base:
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    \begin{equation*} \begin{aligned} r^{16}\cdot r^{5}\amp=r^{16+5}\\ \amp=r^{21} \end{aligned} \end{equation*}
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  2. We multiply the exponents because this is a power being raised to a power:
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    \begin{equation*} \begin{aligned} \left(x^{11}\right)^{10}\amp=x^{11\cdot10}\\ \amp=x^{110} \end{aligned} \end{equation*}
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  3. We apply the power to each factor in the product:
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    \begin{equation*} \begin{aligned} (3r)^{4}\amp=3^4r^4\\ \amp=81r^4 \end{aligned} \end{equation*}
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  4. We subtract the exponents because this expression is dividing powers with the same base:
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    \begin{equation*} \begin{aligned} \frac{3y^7}{y^3}\amp=\frac{3}{1}\frac{y^7}{y^3}\\ \amp=3y^{7-3}\\ \amp=3y^4 \end{aligned} \end{equation*}
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Subsection 4.2.2 Quotient to a Power Rule

One property of exponent we have learned is the Product to a Power Rule, as in \((2x)^{3}=2^{3}x^{3}\text{.}\) When two factors are multiplied and the product is raised to a power, we may apply the exponent to each of those factors individually. We can use the rules of fractions to extend this property to a quotient raised to a power.
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Example 4.2.4.

Let \(y\) be a real number, where \(y \neq 0\text{.}\) Find another way to write \(\left(\frac{5}{y}\right)^4\text{.}\)
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Explanation.
Writing the expression without an exponent and then simplifying, we have:
\begin{align*} \left( \frac{5}{y} \right)^4 \amp= \left( \frac{5}{y} \right) \left( \frac{5}{y} \right) \left( \frac{5}{y} \right) \left( \frac{5}{y} \right)\\ \amp= \frac{5 \cdot 5 \cdot 5 \cdot 5}{y \cdot y \cdot y \cdot y}\\ \amp= \frac{5^4}{y^4}\\ \amp= \frac{625}{y^4} \end{align*}
Similar to the Product to a Power Rule, we essentially applied the outer exponent to the “factors” inside the parentheses—to factors of the numerator and factors of the denominator. The general pattern is:
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This property says that when you raise a fraction to a power, you may separately raise the numerator and denominator to that power. In Example 4, this means that we can directly calculate \(\left( \frac{5}{y} \right)^4\text{:}\)
\begin{align*} \left( \frac{5}{y} \right)^4 \amp= \frac{5^4}{y^4}\\ \amp=\frac{625}{y^4} \end{align*}
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Checkpoint 4.2.6.

  1. Simplify \(\left(\dfrac{p}{2}\right)^6\text{.}\)
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  2. Simplify \(\left(\dfrac{5^6w^7}{5^2w^4}\right)^9\text{.}\) If you end up with a large power of a specific number, leave it written that way.
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  3. Simplify \(\dfrac{\left(2r^5\right)^7}{\left(2^2 r^8\right)^3}\text{.}\) If you end up with a large power of a specific number, leave it written that way.
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Explanation.
  1. We can use the quotient to a power rule:
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    \begin{equation*} \begin{aligned} \left(\frac{p}{2}\right)^6\amp=\frac{p^6}{2^6}\\ \amp=\frac{p^6}{64} \end{aligned} \end{equation*}
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  2. If we stick closely to the order of operations, we should first simplify inside the parentheses and then work with the outer exponent. Going this route, we will first use the quotient rule:
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    \begin{equation*} \begin{aligned} \left(\frac{5^6w^7}{5^2w^4}\right)^9 \amp= \left(5^{6-2}w^{7-4}\right)^9\\ \amp= \left(5^{4}w^{3}\right)^9 \end{aligned} \end{equation*}
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    Now we can apply the outer exponent to each factor inside the parentheses using the product to a power rule.
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    \begin{equation*} \begin{aligned} \amp= \left(5^{4}\right)^9\cdot \left(w^{3}\right)^9 \end{aligned} \end{equation*}
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    To finish, we need to use the power to a power rule.
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    \begin{equation*} \begin{aligned} \amp= 5^{4\cdot 9}\cdot w^{3 \cdot 9}\\ \amp= 5^{36}\cdot w^{27} \end{aligned} \end{equation*}
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  3. According to the order of operations, we should simplify inside parentheses first, then apply exponents, then divide. Since we cannot simplify inside the parentheses, we must apply the outer exponents to each factor inside the respective set of parentheses first:
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    \begin{equation*} \begin{aligned} \frac{\left(2r^5\right)^7}{\left(2^2 r^8\right)^3} \amp= \frac{2^7 \left(r^5\right)^7}{\left(2^2\right)^3 \left(r^8\right)^3} \end{aligned} \end{equation*}
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    At this point, we need to use the power-to-a-power rule:
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    \begin{equation*} \begin{aligned} \phantom{\frac{\left(2r^5\right)^7}{\left(2^2 r^8\right)^3}} \amp= \frac{2^{7} r^{5\cdot 7}}{ 2^{2\cdot 3} r^{8 \cdot 3}}\\ \amp= \frac{2^{7} r^{35}}{ 2^{6} r^{24}} \end{aligned} \end{equation*}
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    To finish simplifying, we’ll conclude with the quotient rule:
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    \begin{equation*} \begin{aligned} \phantom{\frac{\left(2r^5\right)^7}{\left(2^2 r^8\right)^3}} \amp= 2^{7-6} r^{35-24}\\ \amp= 2^{1} r^{11}\\ \amp= 2 r^{11} \end{aligned} \end{equation*}
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Subsection 4.2.3 Zero as an Exponent

So far, we have been working with exponents that are natural numbers (\(1, 2, 3, \ldots\)). By the end of this section, we will expand our understanding to include exponents that are any integer, as with \(5^{0}\) and \(12^{-2}\text{.}\) As a first step, let’s explore how \(0\) should behave as an exponent by considering the pattern of decreasing powers of \(2\) in Figure 7.
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Power Product Result
\(2^4\) \(=\) \(2 \cdot 2 \cdot 2 \cdot 2\) \(=\) \(16\)
\(2^3\) \(=\) \(2 \cdot 2 \cdot 2\) \(=\) \(8\) (divide by \(2\))
\(2^2\) \(=\) \(2 \cdot 2\) \(=\) \(4\) (divide by \(2\))
\(2^1\) \(=\) \(2\) \(=\) \(2\) (divide by \(2\))
\(2^0\) \(=\) \(\mathord{?}\) \(=\) \(\mathord{?}\)
Figure 4.2.7. Descending Powers of \(2\)
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As we move down from one row to the row below it, we reduce the exponent in the power by \(1\) and we remove a factor of \(2\) from the product. The result in one row is half of the result of the previous row. The question is, what happens when the exponent gets down to \(0\) and you remove the last remaining factor of \(2\text{?}\) Following that pattern with the final results, moving from \(2^1\) to \(2^0\) should mean the result of \(2\) is divided by \(2\text{,}\) leaving \(1\text{.}\) So we have:
\begin{equation*} 2^0 = 1 \end{equation*}
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Definition 4.2.8. Zero as an Exponent.

For any real number \(a\text{,}\) we define \(a^0\) to mean \(1\text{.}\) That is,
\begin{equation*} a^0 = 1 \end{equation*}
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It is worth noting that this is the definition for what it means to use \(0\) as an exponent. Before this, we only had an official meaning for using “natural numbers” (\(1,2,3,\ldots\)) as exponents.
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Aside: Zero to the Zero.

Checkpoint 4.2.9.

Simplify the following expressions. Assume all variables represent nonzero real numbers.
  1. \(\displaystyle \left(173 x^4 y^{251}\right)^0\)
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  2. \(\displaystyle (-8)^0\)
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  3. \(\displaystyle -8^0\)
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  4. \(\displaystyle 3x^0\)
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Explanation.
To simplify any of these expressions, it is critical that we remember an exponent only applies to what it is touching or immediately next to.
  1. In the expression \(\left(173 x^4 y^{251}\right)^0\text{,}\) the exponent \(0\) applies to everything inside the parentheses.
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    \begin{equation*} \left(173 x^4 y^{251}\right)^0 = 1 \end{equation*}
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  2. In the expression \((-8)^0\) the exponent applies to everything inside the parentheses, \(-8\text{.}\)
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    \begin{equation*} (-8)^0 = 1 \end{equation*}
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  3. In contrast to the previous example, the exponent only applies to the \(8\text{.}\) The exponent has a higher priority than negation in the order of operations. We should consider that \(-8^0 = -\mathopen{}\left(8^0\right)\mathclose{}\text{,}\) and so:
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    \begin{equation*} \begin{aligned} -8^0 \amp= -\mathopen{}\left(8^0\right)\mathclose{}\\ \amp= -1 \end{aligned} \end{equation*}
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  4. In the expression \(3x^0\text{,}\) the exponent \(0\) only applies to the \(x\text{:}\)
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    \begin{equation*} \begin{aligned} 3x^0 \amp= 3\cdot x^0\\ \amp= 3\cdot 1\\ \amp= 3 \end{aligned} \end{equation*}
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Subsection 4.2.4 Negative Exponents

We understand what it means for a variable to have a natural number exponent. For example, \(x^5\) means \(\overbrace{x\cdot x\cdot x\cdot x\cdot x}^{\text{five times}}\text{.}\) Now we will try to give meaning to an exponent that is a negative integer, like in \(x^{-5}\text{.}\)
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To consider what it could possibly mean to have a negative integer exponent, let’s extend the pattern we saw in Figure 7. In that table, each time we move down a row, we reduce the power by \(1\) and we divide the value by \(2\text{.}\) We can continue this pattern in the power and value columns, going all the way down into when the exponent is negative.
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Power Result
\(2^3\) \(8\)
\(2^2\) \(4\) (divide by \(2\))
\(2^1\) \(2\) (divide by \(2\))
\(2^0\) \(1\) (divide by \(2\))
\(2^{-1}\) \(\sfrac{1}{2}=\sfrac{1}{2^1}\) (divide by \(2\))
\(2^{-2}\) \(\sfrac{1}{4}=\sfrac{1}{2^2}\) (divide by \(2\))
\(2^{-3}\) \(\sfrac{1}{8}=\sfrac{1}{2^3}\) (divide by \(2\))
Figure 4.2.10. Negative Powers of \(2\)
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We see a pattern where \(2^{\text{negative number}}\) is equal to \(\frac{1}{2^{\text{positive number}}}\text{.}\) Note that the choice of base \(2\) was arbitrary, and this pattern works for all bases except \(0\text{,}\) since we cannot divide by \(0\) in moving from one row to the next.
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Definition 4.2.11. Negative Integers as Exponents.

For any nonzero real number \(a\) and any natural number \(n\text{,}\) we define \(a^{-n}\) to mean the reciprocal of \(a^n\text{.}\) That is,
\begin{equation*} a^{-n} = \frac{1}{a^n} \end{equation*}
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As with Definition 8, we note that this is a definition. Prior to this point in this textbook, we only had an official meaning for when an exponent was \(0\) or a positive integer.
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Taking reciprocals of both sides leads to:
\begin{equation*} \frac{1}{a^{-n}}=a^n\text{.} \end{equation*}
Taken together, these things tell us that when you have an expression and there is a negative exponent in the numerator, that power “belongs” in the denominator (with a positive exponent). And similarly, when you have an expression and there is a negative exponent in the denominator, that power “belongs” in the numerator (with a positive exponent). In other words, you can view a negative exponent as telling you to move something to/from the numerator/denominator of an expression, changing the sign of the exponent at the same time.
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You may be expected to simplify expressions so that they do not have any negative exponents. This can always be accomplished using the above understanding. Try it with these exercises.
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Checkpoint 4.2.12.

  1. Write \(4y^{-6}\) without using negative exponents.
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  2. Write \(\dfrac{ 3 x^{-4} }{ yz^{-2} }\) without using negative exponents.
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  3. Simplify \(\left(-5x^{-5}\right)\left(-8x^4\right)\) and write it without using negative exponents.
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Explanation.
  1. An exponent only applies to whatever it is “touching”. In the expression \(4y^{-6}\text{,}\) only the \(y\) is affected by the exponent.
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    \begin{equation*} \begin{aligned} 4y^{-6} \amp= 4\cdot \frac{1}{y^{6}}\\ \amp= \frac{4}{y^6} \end{aligned} \end{equation*}
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  2. Negative exponents tell us to move some variables between the numerator and denominator to make the exponents positive. The \(x^{-4}\) in the numerator should become \(x^4\) in the denominator. The \(z^{-2}\) in the denominator should become \(z^2\) in the numerator.
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    \begin{equation*} \frac{ 3 \highlight{x^{-4}} }{ y\highlight{z^{-2}} }=\frac{ 3 \highlight{z^{2}} }{ y\highlight{x^{4}} } \end{equation*}
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    Notice that the factors of \(3\) and \(y\) did not move, as both of those factors had positive exponents.
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  3. The product of powers rule still applies, and we can add exponents even when one or both are negative:
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    \begin{equation*} \begin{aligned} \left(-5x^{-5}\right)\left(-8x^4\right)\amp=(-5)(-8)x^{-5}x^4\\ \amp=40x^{-1}\\ \amp=\frac{40}{x} \end{aligned} \end{equation*}
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Subsection 4.2.5 Summary of Exponent Rules

Now that we have some new exponent rules beyond those from Section 1, let’s summarize.
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List 4.2.13. Summary of the Rules of Exponents for Multiplication and Division
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If \(a\) and \(b\) are real numbers, and \(m\) and \(n\) are integers, then we have the following rules:
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Product Rule
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\(\displaystyle a^{m} \cdot a^{n} = a^{m+n}\)
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Power to a Power Rule
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\(\displaystyle (a^{m})^{n} = a^{m\cdot n}\)
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Product to a Power Rule
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\(\displaystyle (ab)^{m} = a^{m} \cdot b^{m}\)
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Quotient Rule
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\(\dfrac{a^{m}}{a^{n}} = a^{m-n}\text{,}\) as long as \(a \neq 0\)
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Quotient to a Power Rule
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\(\left( \dfrac{a}{b} \right)^{m} = \dfrac{a^{m}}{b^{m}}\text{,}\) as long as \(b \neq 0\)
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Zero Exponent Definition
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\(\displaystyle a^{0} = 1\)
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Negative Exponent Definition
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\(\displaystyle a^{-m} = \frac{1}{a^m}\)
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Negative Exponent Reciprocal Rule
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\(\displaystyle \frac{1}{a^{-m}} = a^m\)
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Remark 4.2.14. Why we have “\(a \neq 0\)” and “\(b \neq 0\)” for some rules.

We have to be careful to make sure the rules we state don’t suggest that it would ever be OK to divide by zero. Dividing by zero leads us to expressions that have no meaning. For example, both \(\frac{9}{0}\) and \(\frac{0}{0}\) are undefined, meaning no one has defined what it means to divide a number by \(0\text{.}\)
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Warning 4.2.15. A Common Mistake.

It may be tempting to apply the rules of exponents to expressions containing addition or subtraction. However, none of the Summary of the Rules of Exponents for Multiplication and Division involve addition or subtraction in the initial expression. Because whole number exponents mean repeated multiplication, not repeated addition or subtraction, trying to apply exponent rules in situations that do not use multiplication simply doesn’t work.
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Can we say something like \(a^m + a^n = a^{m+n}\text{?}\) How would that work out when \(a=2\text{,}\) \(m=3\text{,}\) and \(n=4\text{?}\)
\begin{align*} 2^3 + 2^4 \amp\wonder{=} 2^{3+4}\\ 8 + 16 \amp\wonder{=} 2^{7}\\ 24 \amp\reject{=} 128 \end{align*}
As we can see, that’s not even close. This attempt at a “sum rule” falls apart. In fact, without knowing values for \(a\text{,}\) \(n\text{,}\) and \(m\text{,}\) there’s no way to simplify the expression \(a^n + a^m\text{.}\)
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Checkpoint 4.2.16.

Decide whether each statement is true or false.
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  1. \(\left(7+8\right)^3=7^3+8^3\qquad\)
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  2. \((xy)^3=x^3 y^3\qquad\)
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  3. \(2x^3 \cdot 4x^2 \cdot 5x^6=(2 \cdot 4 \cdot 5)x^{3+2+6}\qquad\)
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  4. \(\left(x^3y^5\right)^4=x^{3+4}y^{5+4}\qquad\)
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  5. \(2\left(x^2y^5\right)^3=8x^6y^{15}\qquad\)
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  6. \(x^2+x^3=x^5\qquad\)
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  7. \(x^3+x^3=2x^3\qquad\)
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  8. \(x^3\cdot x^3=2x^6\qquad\)
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  9. \(3^2 \cdot 2^3=6^5\qquad\)
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  10. \(3^{-2} =-\frac{1}{9}\qquad\)
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Explanation.
  1. False, \(\left(7+8\right)^3\neq 7^3+8^3\text{.}\) Following the order of operations, on the left \(\left(7+8\right)^3\) would simplify as \(15^3\text{,}\) which is \(3375\text{.}\) However, on the right side, we have
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    \begin{equation*} \begin{aligned} 7^3+8^3\amp=343+512\\ \amp=855 \end{aligned} \end{equation*}
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    Since \(3375\neq 855\text{,}\) the equation is false.
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  2. True. As the cube applies to the product of \(x\) and \(y\text{,}\) \((xy)^3=x^3 y^3\text{.}\)
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  3. True. The coefficients do get multiplied together and the exponents added when the expressions are multiplied, so \(2x^3 \cdot 4x^2 \cdot 5x^6=(2 \cdot 4 \cdot 5)x^{3+2+6}\text{.}\)
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  4. False, \(\left(x^3y^5\right)^4\neq x^{3+4}y^{5+4}\text{.}\) When we have a power to a power, we multiply the exponents rather than adding them. So
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    \begin{equation*} \left(x^3y^5\right)^4=x^{3 \cdot 4}y^{5 \cdot 4} \end{equation*}
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  5. False, \(2\left(x^2y^5\right)^3\neq 8x^6y^{15}\text{.}\) The exponent of \(3\) applies to \(x^2\) and \(y^{5}\text{,}\) but does not apply to the \(2\text{.}\) So
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    \begin{equation*} \begin{aligned} 2\left(x^2y^5\right)^3\amp=2x^{2\cdot 3}6y^{5\cdot 3}\\ \amp=2x^6y^{15} \end{aligned} \end{equation*}
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  6. False, \(x^2+x^3\neq x^5\text{.}\) The two terms on the left hand side are not like terms and there is no way to combine them.
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  7. True. The terms \(x^3\) and \(x^3\) are like terms, so \(x^3+x^3=2x^3\text{.}\)
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  8. False, \(x^3\cdot x^3 \neq 2x^6\text{.}\) When \(x^3\) and \(x^3\) are multiplied, their coefficients are each \(1\text{.}\) So the coefficient of their product is still \(1\text{,}\) and we have \(x^3 \cdot x^3 = x^6\text{.}\)
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  9. False, \(3^2 \cdot 2^3\neq 6^5\text{.}\) Note that neither the bases nor the exponents are the same. Following the order of operations, on the left \(3^2 \cdot 2^3\) would simplify as \(9 \cdot 8\text{,}\) which is \(72\text{.}\) However, on the right side, we have \(6^5=7776\text{.}\) Since \(72\neq 7776\text{,}\) the equation is false.
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  10. False, \(3^{-2}\neq -\frac{1}{9}\text{.}\) The exponent of \(-2\) on the number \(3\) does not result in a negative number. Instead, \(3^{-2}= \frac{1}{3^2}\text{,}\) which is \(\frac{1}{9}\text{.}\)
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As we mentioned before, many situations we’ll come across will require us to use more than one exponent rule. In these situations, we’ll have to decide which rule to use first. There are often different, correct approaches we could take. But if we rely on order of operations, we will have a straightforward approach to simplify the expression correctly. To bring it all together, try these exercises.
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Checkpoint 4.2.17.

(a)
Simplify \(\dfrac{6x^{3}}{2x^{7}}\) and write it without using negative exponents.
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Explanation.
In the expression \(\frac{6x^{3}}{2x^{7}}\text{,}\) the coefficients reduce using the properties of fractions. One way to simplify the variable powers is:
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\begin{equation*} \begin{aligned} \frac{6x^{3}}{2x^{7}} \amp= \frac{6}{2} \cdot \frac{x^{3}}{x^{7}}\\ \amp= 3 \cdot x^{3-7}\\ \amp= 3 \cdot x^{-4}\\ \amp= 3 \cdot \frac{1}{x^4}\\ \amp= \frac{3}{x^4} \end{aligned} \end{equation*}
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(b)
Simplify \(4\left(\frac{1}{5}tv^{-4}\right)^{2}\) and write it without using negative exponents.
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Explanation.
In the expression \(4\left(\frac{1}{5}tv^{-4}\right)^{2}\text{,}\) the exponent \(2\) applies to each factor inside the parentheses.
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\begin{equation*} \begin{aligned} 4\left(\frac{1}{5}tv^{-4}\right)^{2} \amp= 4\left(\frac{1}{5}\right)^{2}\left(t\right)^2\left(v^{-4}\right)^2\\ \amp= 4\left(\frac{1}{25}\right)\left(t^2\right)\left(v^{-4\cdot 2}\right)\\ \amp= 4\left(\frac{1}{25}\right)\left(t^2\right)\left(v^{-8}\right)\\ \amp= 4\left(\frac{1}{25}\right)\left(t^2\right)\left(\frac{1}{v^{8}}\right)\\ \amp= \frac{4t^2}{25v^{8}} \end{aligned} \end{equation*}
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(c)
Simplify \(\left(\dfrac{3^{0}y^{4}\cdot y^{5}}{6y^2}\right)^{3}\) and write it without using negative exponents.
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Explanation.
To follow the order of operations in the expression \(\left(\frac{3^{0}y^{4}\cdot y^{5}}{6y^2}\right)^{3}\text{,}\) the numerator inside the parentheses should be dealt with first. After that, we’ll simplify the quotient inside the parentheses. As a final step, we’ll apply the exponent to that simplified expression:
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\begin{equation*} \begin{aligned} \left(\frac{3^0 y^4 \cdot y^5}{6y^2}\right)^3 \amp= \left(\frac{1\cdot y^{4+5}}{6y^2}\right)^3\\ \amp= \left(\frac{y^{9}}{6y^2}\right)^{3}\\ \amp= \left(\frac{y^{9-2}}{6}\right)^{3}\\ \amp= \left(\frac{y^7}{6}\right)^3\\ \amp= \frac{\left(y^7\right)^3}{6^3}\\ \amp= \frac{y^{7\cdot 3}}{216}\\ \amp= \frac{y^{21}}{216} \end{aligned} \end{equation*}
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(d)
Simplify \(\left( 7^{4} x^{-6} t^{2} \right)^{-5} \left( 7 x^{-2} t^{-7} \right)^{4}\) and write it without using negative exponents. Leave larger numbers (such as \(7^{10}\)) in exponent form.
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Explanation.
We’ll again rely on the order of operations, and look to simplify anything inside parentheses first and then apply exponents. In this example, we will begin by applying the product to a power rule, followed by the power to a power rule.
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\begin{equation*} \begin{aligned} \left( 7^{4} x^{-6} t^{2} \right)^{-5} \left( 7 x^{-2} t^{-7} \right)^{4} \amp= \left( 7^{4} \right)^{-5} \left( x^{-6} \right)^{-5} \left( t^{2} \right)^{-5} \cdot \left( 7 \right)^{4} \left(x^{-2}\right)^{4} \left(t^{-7} \right)^{4}\\ \amp= 7^{-20} x^{30} t^{-10} \cdot 7^{4} x^{-8} t^{-28}\\ \amp= 7^{-20+4} x^{30-8} t^{-10-28}\\ \amp= 7^{-16} x^{22} t^{-38}\\ \amp= \frac{x^{22}}{7^{16}t^{38}} \end{aligned} \end{equation*}
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Reading Questions 4.2.6 Reading Questions

1.

When you are considering using the exponent rule \(a^m\cdot a^n=a^{m+n}\text{,}\) are \(m\) and \(n\) allowed to be negative integers?
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2.

What are the differences between these three expressions?
\begin{equation*} x+0\qquad0x\qquad x^0 \end{equation*}
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3.

If you rearrange \(\frac{xy^{-3}}{a^2b^8c}\) so that it is written without negative exponents, how many factors will you have “moved?”
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Exercises 4.2.7 Exercises

Review and Warmup

Exercise Group.
Evaluate each power expression.
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3.
(a)
\(\left({{\frac{4}{5}}}\right)^3\)
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(b)
\(\left({-{\frac{1}{4}}}\right)^4\)
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(c)
\(\left({-{\frac{5}{2}}}\right)^5\)
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(d)
\(-\left({{\frac{4}{5}}}\right)^2\)
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4.
(a)
\(\left({{\frac{5}{3}}}\right)^2\)
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(b)
\(\left({-{\frac{3}{5}}}\right)^4\)
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(c)
\(\left({-{\frac{3}{4}}}\right)^3\)
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(d)
\(-\left({{\frac{5}{3}}}\right)^2\)
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Skills Practice

Simplifying Expressions with Exponents.
Use the properties of exponents to simplify the expression. If there are any exponents in your answer, they should be positive.
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7.
\(\displaystyle{\left(x^{4}\right)^{10}}\)
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8.
\(\displaystyle{\left(y^{5}\right)^{6}}\)
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11.
\(\displaystyle{({-7t^{14}})\cdot({7t^{9}})}\)
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12.
\(\displaystyle{({3x^{16}})\cdot({-6x^{3}})}\)
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13.
\(\displaystyle{\left({\frac{x^{18}}{7}}\right) \cdot \left({\frac{x^{15}}{6}}\right)}\)
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14.
\(\displaystyle{\left({-\frac{x^{20}}{9}}\right) \cdot \left({-\frac{x^{8}}{5}}\right)}\)
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15.
\(-2\left(-7y^{2}\right)^2\)
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16.
\(-3\left(-4y^{4}\right)^3\)
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21.
\(38^0+\left(-38\right)^0\)
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22.
\(44^0+\left(-44\right)^0\)
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27.
\(\left(\displaystyle\frac{x^{9}}{5}\right)^{2}\)
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28.
\(\left(\displaystyle\frac{x^{6}}{6}\right)^{3}\)
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29.
\(\left(\displaystyle\frac{-7}{2x^{3}}\right)^{2}\)
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30.
\(\left(\displaystyle\frac{-7}{10x^{7}}\right)^{3}\)
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31.
\(\left(\displaystyle\frac{3x^{9}}{4}\right)^{2}\)
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32.
\(\left(\displaystyle\frac{5x^{10}}{4}\right)^{2}\)
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33.
\(\left(\displaystyle\frac{x^{8}}{2y^{9}z^{10}}\right)^{2}\)
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34.
\(\left(\displaystyle\frac{x^{5}}{2y^{4}z^{8}}\right)^{2}\)
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35.
\(\left(\displaystyle\frac{-3x^{4}}{8y^{2}}\right)^{3}\)
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36.
\(\left(\displaystyle\frac{-3x^{5}}{4y^{8}}\right)^{2}\)
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37.
\(\left(\frac{1}{6}\right)^{-2}\)
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38.
\(\left(\frac{1}{7}\right)^{-2}\)
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39.
\(\displaystyle\frac{7^{-3}}{6^{-2}}\)
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40.
\(\displaystyle\frac{7^{-2}}{4^{-3}}\)
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43.
\(\displaystyle {9x^{-4}}\)
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44.
\(\displaystyle {19x^{-5}}\)
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45.
\(\displaystyle {\frac{14}{x^{-7}}}\)
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46.
\(\displaystyle {\frac{8}{x^{-8}}}\)
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47.
\(\displaystyle {\frac{19x^{-9}}{x}}\)
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48.
\(\displaystyle {\frac{13x^{-10}}{x}}\)
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49.
\(\displaystyle {\frac{13x^{-18}}{x^{-9}}}\)
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50.
\(\displaystyle {\frac{19x^{-20}}{x^{-35}}}\)
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51.
\(\displaystyle {\frac{10x^{-4}}{11x^{-13}}}\)
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52.
\(\displaystyle {\frac{17x^{-8}}{18x^{-6}}}\)
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53.
\(\displaystyle\frac{y^{-5}}{t^{-10}}\)
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54.
\(\displaystyle\frac{y^{-4}}{r^{-20}}\)
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55.
\(\displaystyle\frac{r^{-2}}{x^{18}}\)
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56.
\(\displaystyle\frac{r^{-10}}{t^{13}}\)
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57.
\(\displaystyle\frac{1}{22r^{-18}}\)
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58.
\(\displaystyle\frac{1}{5t^{-7}}\)
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59.
\(\displaystyle\frac{t^{26}}{t^{37}}\)
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60.
\(\displaystyle\frac{x^{3}}{x^{20}}\)
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61.
\(\displaystyle\frac{27x^{3}}{3x^{4}}\)
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62.
\(\displaystyle\frac{-90y^{33}}{10y^{35}}\)
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63.
\(\displaystyle\frac{11y^{7}}{9y^{18}}\)
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64.
\(\displaystyle\frac{7r^{39}}{5r^{50}}\)
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65.
\(\displaystyle\frac{r^{3}}{\left(r^{9}\right)^{2}}\)
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66.
\(\displaystyle\frac{r^{3}}{\left(r^{5}\right)^{9}}\)
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67.
\(\displaystyle\frac{t^{-2}}{\left(t^{12}\right)^{6}}\)
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68.
\(\displaystyle\frac{t^{-4}}{\left(t^{8}\right)^{4}}\)
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71.
\((5y^{-12})\cdot (-7y^{8})\)
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72.
\((3y^{-6})\cdot (-3y^{2})\)
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73.
\(\displaystyle\left(\frac{9}{8}\right)^{-2}\)
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74.
\(\displaystyle\left(\frac{9}{7}\right)^{-2}\)
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77.
\(\displaystyle\frac{1}{(-10)^{-2}}\)
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78.
\(\displaystyle\frac{1}{(-2)^{-2}}\)
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79.
\(\displaystyle\frac{-8}{(-2)^{-2}}\)
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80.
\(\displaystyle\frac{-5}{(-2)^{-3}}\)
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85.
\(\displaystyle\frac{1}{9^{-3}}\)
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86.
\(\displaystyle\frac{1}{10^{-2}}\)
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89.
\(\displaystyle\frac{\left(4y^{5}\right)^{2}}{y^{24}}\)
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90.
\(\displaystyle\frac{\left(4y^{12}\right)^{3}}{y^{37}}\)
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91.
\(\displaystyle\frac{\left(4y^{8}\right)^{2}}{y^{-14}}\)
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92.
\(\displaystyle\frac{\left(4r^{5}\right)^{2}}{r^{-10}}\)
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93.
\(\displaystyle\left(\frac{r^{19}}{r^{16}}\right)^{-3}\)
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94.
\(\displaystyle\left(\frac{t^{12}}{t^{4}}\right)^{-2}\)
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95.
\(\displaystyle\left(\frac{16t^{6}}{4t^{4}}\right)^{-5}\)
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96.
\(\displaystyle\left(\frac{8x^{18}}{4x^{3}}\right)^{-4}\)
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97.
\(\left(-5x^{-12}\right)^{-3}\)
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98.
\(\left(-2y^{-5}\right)^{-3}\)
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99.
\(\left(4y^{-17}\right)^{-2}\)
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100.
\(\left(3y^{-11}\right)^{-3}\)
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101.
\(\displaystyle\frac{9r^{3}\cdot9r^{3}}{8r^{2}}\)
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102.
\(\displaystyle\frac{5r^{8}\cdot10r^{7}}{7r^{13}}\)
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103.
\(\left(t^{8}\right)^{2}\cdot t^{-9}\)
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104.
\(\left(t^{3}\right)^{3}\cdot t^{-4}\)
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105.
\(\left(2x^{12}\right)^{4}\cdot x^{-2}\)
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106.
\(\left(2x^{7}\right)^{2}\cdot x^{-10}\)
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107.
\(\displaystyle\frac{\left(y^{3}\right)^{4}}{\left(y^{8}\right)^{3}}\)
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108.
\(\displaystyle\frac{\left(y^{8}\right)^{2}}{\left(y^{12}\right)^{5}}\)
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109.
\(\left(y^{9}\right)^{-4}\)
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110.
\(\left(r^{20}\right)^{-5}\)
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111.
\(\left(r^{7}y^{5}\right)^{-4}\)
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112.
\(\left(t^{13}y^{3}\right)^{-2}\)
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113.
\(\left(t^{-13}x^{5}\right)^{-4}\)
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114.
\(\left(x^{-11}r^{9}\right)^{-2}\)
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115.
\(\displaystyle\left(\frac{x^{5}}{2}\right)^{-2}\)
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116.
\(\displaystyle\left(\frac{y^{14}}{4}\right)^{-3}\)
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117.
\(\displaystyle\left(\frac{y^{14}}{r^{11}}\right)^{-2}\)
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118.
\(\displaystyle\left(\frac{y^{6}}{x^{8}}\right)^{-2}\)
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119.
\(\displaystyle\frac{\left(r^{7}t^{-3}\right)^{-2}}{\left(r^{-6}t^{6}\right)^{-2}}\)
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120.
\(\displaystyle\frac{\left(r^{8}x^{-3}\right)^{-2}}{\left(r^{-6}x^{6}\right)^{-3}}\)
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121.
\(5x^{-6}y^{2}z^{-5}\left(3x^{6}\right)^{-3}\)
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122.
\(7x^{-3}y^{7}z^{-4}\left(3x^{3}\right)^{-4}\)
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123.
\(\displaystyle \left( \frac{x^{7}y^{5}z^{6}}{x^{-3}y^{-4}z^{-6}}\right)^{-3}\)
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124.
\(\displaystyle \left( \frac{x^{6}y^{5}z^{3}}{x^{-3}y^{-3}z^{-2}}\right)^{-2}\)
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Challenge

125.
Consider the expression \(\dfrac{x^a\cdot x^b}{x^c}\) where \(a\gt0, b\lt0\text{,}\) and \(c \gt 0\text{.}\)
  1. Are there values for \(a\text{,}\) \(b\text{,}\) and \(c\) so that the expression equals \(x^{8}\text{?}\) If so, fill in the blanks below with possible values for \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) If not, fill in the blanks below with the word none.
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    \(a\) = , \(b\) = , and \(c\) =
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  2. Are there values for \(a\text{,}\) \(b\text{,}\) and \(c\) so that the exponential expression equals \(\frac{1}{x^{9}}\text{?}\) If so, fill in the blanks below with possible values for \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) If not, fill in the blanks below with the word none.
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    \(a\) = , \(b\) = , and \(c\) =
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126.
Consider the expression \(\dfrac{x^a\cdot x^b}{x^c}\) where \(a\lt0, b\lt0\text{,}\) and \(c \gt0\text{.}\)
  1. Are there values for \(a\text{,}\) \(b\text{,}\) and \(c\) so that the expression equals \(x^{8}\text{?}\) If so, fill in the blanks below with possible values for \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) If not, fill in the blanks below with the word none.
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    \(a\) = , \(b\) = , and \(c\) =
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  2. Are there values for \(a\text{,}\) \(b\text{,}\) and \(c\) so that the expression equals \(\frac{1}{x^{10}}\text{?}\) If so, fill in the blanks below with possible values for \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) If not, fill in the blanks below with the word none.
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    \(a\) = , \(b\) = , and \(c\) =
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127.
Consider the exponential expression \(\dfrac{x^a\cdot x^b}{x^c}\) where \(a\gt0, b\gt0\text{,}\) and \(c \lt 0\text{.}\)
  1. Are there values for \(a\text{,}\) \(b\text{,}\) and \(c\) so that the expression equals \(x^{8}\text{?}\) If so, fill in the blanks below with possible values for \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) If not, fill in the blanks below with the word none.
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    \(a\) = , \(b\) = , and \(c\) =
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  2. Are there values for \(a\text{,}\) \(b\text{,}\) and \(c\) so that the expression equals \(\frac{1}{x^{13}}\text{?}\) If so, fill in the blanks below with possible values for \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) If not, fill in the blanks below with the word none.
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    \(a\) = , \(b\) = , and \(c\) =
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