4.2: Laws of Exponents

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• Review of the Laws of Exponents (not including negative or rational exponents)

The Laws of Exponents: Product Law

Remember that an exponent indicates repeated multiplication of the same quantity. For example, in the expression \(a^m\), the exponent \(m\) tells us how many times we use the base \(a\) as a factor.\[a^{m}= \underbrace{a \cdot a \cdot a \cdot \ldots \cdot a}_{\color{blue}{m \text{ factors}}} \nonumber\]For example\[(-9)^{5}= \underbrace{ (-9)\cdot (-9)\cdot (-9)\cdot (-9) \cdot (-9)}_{\color{blue}{5 \text{ factors}}} \nonumber\]Let’s review the vocabulary for expressions with exponents.

Definition: Exponential Notation

\[ a^m = \underbrace{a \cdot a \cdot a \cdot \ldots \cdot a}_{\color{blue}{m \text{ factors}}} \nonumber \]This is read \(a\) to the \(m^{\text{th}}\) power. The exponent \(m\) tells us how many times we use the base \(a\) as a factor.

When we combine like terms by adding and subtracting, we need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different, too.

First, we will look at an example that leads to the Product Law.\[ x^2 \cdot x^3 = \underbrace{\underbrace{x \cdot x}_{\color{blue}{2 \text{ factors}}} \cdot \underbrace{x \cdot x \cdot x}_{\color{blue}{3 \text{ factors}}}}_{\color{purple}{5 \text{ factors}}} = x^5. \nonumber \]Notice that 5 is the sum of the exponents, 2 and 3. We see \(x^2 \cdot x^3\) is \(x^{2+3}\) or \(x^5\). The base stayed the same and we added the exponents. This leads to the Product Law for Exponents.

Theorem: Product Law For Exponents

If \( a \) is a real number, and \( m \) and \( n \) are integers, then\[a^m \cdot a^n = a^{m+n}. \nonumber\]That is, to multiply with like bases, add the exponents.

Example \(\PageIndex{1}\)

Simplify each expression:

1. \(y^5 \cdot y^6\)
2. \(2^x \cdot 2^{3x}\)
3. \(2a^7 \cdot 3a\)
4. \(d^4 \cdot d^5 \cdot d^2\)

Solutions

1.    \[ \begin{array}{rclcl}
   y^5 \cdot y^6 & = & y^{5 + 6} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & y^{11} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
2.    \[ \begin{array}{rclcl}
   2^x \cdot 2^{3x} & = & 2^{x + 3x} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & 2^{4x} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
3.    \[ \begin{array}{rclcl}
   2a^7 \cdot 3a & = & 2a^7 \cdot 3a^1 & \quad & \left( \text{rewriting }a\text{ as }a^1 \right) \\[6pt]
   & = & 2 \cdot 3 \cdot a^7 \cdot a^1 & \quad & \left( \text{Commutative Property of Multiplication} \right) \\[6pt]
   & = & 6 \cdot a^7 \cdot a^1 & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & 6 \cdot a^{7 + 1} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & 6 a^{8} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
4.    \[ \begin{array}{rclcl}
   d^4 \cdot d^5 \cdot d^2 & = & d^{4 + 5 + 2} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & d^{11} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]

Checkpoint \(\PageIndex{1}\)

Simplify each expression:

1. \(b^9 \cdot b^8\)
2. \(4^{2x} \cdot 4^x\)
3. \(3p^5 \cdot 4p\)
4. \(x^6 \cdot x^4 \cdot x^8\).

Answers

1. \(b^{17}\)
2. \(4^{3x}\)
3. \(12p^6\)
4. \(x^{18}\)

The theorem for the Product Law stated in this Corequisite Codex is technically not complete. In truth, the exponents can be any real numbers as long as the base is positive. If the base is negative, the exponents cannot include rational numbers whose simplified forms have even numbers in their denominators. This is due to the relationship between rational exponents and radicals (a topic to be reviewed in a different section). For now, it is enough to know that, in general, \( a^m \cdot a^n = a^{m + n} \).

The Laws of Exponents: Quotient Law

Now we will look at an exponent property for division. As before, we’ll try to discover a property by looking at some examples.\[ \begin{array}{cccc}
\textbf{Expression} & \textbf{Meaning} & \textbf{Road to Simplification} & \textbf{Simplified Expression} \\[6pt]
\hline \dfrac{x^5}{x^2} & \dfrac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x} & \dfrac{\cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot x}{\cancel{x} \cdot \cancel{x}} & x^3 \\[6pt]
\hline \dfrac{x^2}{x^3} & \dfrac{x \cdot x}{x \cdot x \cdot x} & \dfrac{\cancel{x} \cdot \cancel{x} \cdot 1}{\cancel{x} \cdot \cancel{x} \cdot x} & \dfrac{1}{x}\\[6pt]
\end{array} \nonumber \]Notice, in each case the bases were the same and we subtracted exponents. We see \(\frac{x^5}{x^2}\) is \(x^{5−2}\), or \(x^3\). We also see that \(\frac{x^2}{x^3}\) is \(\frac{1}{x}\). When the larger exponent was in the numerator, we were left with factors in the numerator. When the larger exponent was in the denominator, we were left with factors in the denominator - notice the numerator of 1. When all the factors in the numerator have been removed, remember this is really dividing the factors to one, and so we need a 1 in the numerator. This leads to the Quotient Law for Exponents.

Theorem: Quotient Law for Exponents

If \(a\) is a real number, where \( a \neq 0 \), and \( m \) and \( n \) are integers, then\[ \begin{array} {lllll} {\dfrac{a^m}{a^n}=a^{m−n},} &{m>n} &{\text{and}} &{\dfrac{a^m}{a^n}=\dfrac{1}{a^{n−m}},} &{n>m} \\[6pt] \nonumber \end{array} \]

Example \(\PageIndex{2}\)

Simplify each expression:

1. \(\frac{x^9}{x^7}\)
2. \(\frac{3^{10}}{3^2}\)
3. \(\frac{b^8}{b^{12}}\)
4. \(\frac{7^3}{7^5}\).

Solutions

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

1. Since \(9>7\), there are more factors of \(x\) in the numerator.\[ \begin{array}{rclcl}
   \dfrac{x^9}{x^7} & = & x^{9 - 7} & \quad & \left( \text{Laws of Exponents: Quotient Law} \right) \\[6pt]
   & = & x^{2} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
2. Since \(10>2\), there are more factors of \(3\) in the numerator.\[ \begin{array}{rclcl}
   \dfrac{3^{10}}{3^2} & = & 3^{10 - 2} & \quad & \left( \text{Laws of Exponents: Quotient Law} \right) \\[6pt]
   & = & 3^{8} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.
3. Since \(12>8\), there are more factors of \( b \) in the denominator.\[ \begin{array}{rclcl}
   \dfrac{b^{8}}{b^{12}} & = & \dfrac{1}{b^{12 - 8}} & \quad & \left( \text{Laws of Exponents: Quotient Law} \right) \\[6pt]
   & = & \dfrac{1}{b^4} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
4. Since \(5>3\), there are more factors of \(3\) in the denominator.\[ \begin{array}{rclcl}
   \dfrac{7^{3}}{7^{5}} & = & \dfrac{1}{7^{5 - 3}} & \quad & \left( \text{Laws of Exponents: Quotient Law} \right) \\[6pt]
   & = & \dfrac{1}{7^2} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{1}{49} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]Notice that when the larger exponent is in the denominator, we are left with factors in the denominator.

Checkpoint \(\PageIndex{2}\)

Simplify each expression:

1. \(\dfrac{x^{15}}{x^{10}}\)
2. \(\dfrac{6^{14}}{6^5}\)
3. \(\dfrac{x^{18}}{x^{22}}\)
4. \(\dfrac{12^{15}}{12^{30}}\).

Answers

1. \(x^5\)
2. \(6^9\)
3. \(\frac{1}{x^4}\)
4. \(\frac{1}{12^{15}}\)

As with the Product Law (and for the same reasons), the Quotient Law stated in this Corequisite Codex is technically not complete. For now, it is enough to know that the Quotient Law holds for most cases.

The Laws of Exponents: Zero Exponent Law

The Quotient Law for Exponents shows us how to simplify \(\frac{a^m}{a^m}\) when \(m>n\) and when \(n<m\) by subtracting exponents. What if \(m=n\)? We will simplify \(\frac{a^m}{a^m}\) in two ways to lead us to the definition of the Zero Exponent Law. In general, for \(a \neq 0\):\[ \begin{array}{rccccl}
\textbf{Approach #1: } & \dfrac{a^m}{a^m} & = & a^{m - m} & = & a^0 \\[6pt]
\textbf{Approach #2: } & \dfrac{a^m}{a^m} & = & \dfrac{\overbrace{\cancel{a} \cdot \cancel{a} \cdot \cancel{a} \cdot \ldots \cdot \cancel{a}}^{\color{blue}{m \text{ factors}}}}{\underbrace{\cancel{a} \cdot \cancel{a} \cdot \cancel{a} \cdot \ldots \cdot \cancel{a}}_{\color{blue}{m \text{ factors}}}} & = & 1 \\[6pt]
\end{array} \nonumber \]We see \(\frac{a^m}{a^m}\) simplifies to \(a^0\) and to \( 1 \). So \(a^0=1\). Any non-zero base raised to the power of zero equals \(1\).

Definition: Zero Exponent Law

If \(a\) is a non-zero number, then \(a^0=1\).

In most textbooks, it is assumed any variable raised to the zero power is not zero.

Example \(\PageIndex{3}\)

Simplify each expression:

1. \(9^0\)
2. \(n^0\)

Solutions

The definition says any non-zero number raised to the zero power is \(1\).

1. Use the definition of the zero exponent. \(9^0 = 1\)
2. Use the definition of the zero exponent. \(n^0 = 1\)

Checkpoint \(\PageIndex{3}\)

Simplify each expression:

1. \(11^0\)
2. \(q^0\)

Answers

1. 1
2. 1

Negative Exponents

We saw that the Quotient Law for Exponents has two forms depending on whether the exponent is larger in the numerator or the denominator. What if we just subtract exponents regardless of which is larger?

Let's consider \(\frac{x^2}{x^5}\). If we subtract the exponent in the denominator from the exponent in the numerator, we find that \(\frac{x^2}{x^5}\) is \(x^{2−5}\) or \(x^{−3}\).

We can also simplify \(\frac{x^2}{x^5}\) by dividing out common factors:\[ \dfrac{x^2}{x^5} = \dfrac{\cancel{x} \cdot \cancel{x}}{\cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot x} = \dfrac{1}{x^3}. \nonumber \]This implies that \(x^{−3}=\frac{1}{x^3}\) and it leads us to the definition of a negative exponent. If \( n \) is a real number and \(a \neq 0\), then \(a^{−n} = \frac{1}{a^n}\).

Let's now look at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.\[ \dfrac{1}{a^{-n}} = \dfrac{1}{\frac{1}{a^n}} = \dfrac{a^n}{a^n} \cdot \dfrac{1}{\frac{1}{a^n}} = \dfrac{a^n}{1} = a^n. \nonumber \]This implies \(\dfrac{1}{a^{−n}}=a^n\) and is another form of the Properties of Negative Exponents.

Theorem: Properties of Negative Exponents

If \(n\) is an integer and \(a \neq 0\), then \(a^{−n}=\frac{1}{a^n}\) and \(\frac{1}{a^{−n}}=a^n\).

The negative exponent tells us we can rewrite the expression by taking the reciprocal of the base and then changing the sign of the exponent.

A Debatable Statement

Some would say that any expression having negative exponents is not considered to be in simplest form. While this is generally true for all levels of Mathematics below Calculus, once you reach Calculus it is better to leave expressions with negative exponents. For now, check with your instructor as to their preference. For consistency, we will use the definition of a negative exponent and other properties of exponents to write the expression with only positive exponents in this Corequisite Codex (and other textbooks below the Calculus level).

Example \(\PageIndex{4}\)

Simplify each expression:

1. \(x^{−5}\)
2. \(10^{−3}\)
3. \(\frac{1}{y^{−4}}\)
4. \(\frac{1}{3^{−2}}\)

Solutions

1.    \[ \begin{array}{rclcl}
   x^{-5} & = & \dfrac{1}{x^5} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   \end{array} \nonumber \]
2.    \[ \begin{array}{rclcl}
   10^{-3} & = & \dfrac{1}{10^3} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   & = & \dfrac{1}{1000} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
3.    \[ \begin{array}{rclcl}
   \dfrac{1}{y^{-4}} & = & y^4 & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   \end{array} \nonumber \]
4.    \[ \begin{array}{rclcl}
   \dfrac{1}{3^{-2}} & = & 3^2 & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   & = & 9 & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]

Checkpoint \(\PageIndex{4}\)

Simplify each expression:

1. \(z^{−3}\)
2. \(10^{−7}\)
3. \(\frac{1}{p^{−8}}\)
4. \(\frac{1}{4^{−3}}\)

Answers

1. \(\frac{1}{z^3}\)
2. \(\frac{1}{10^7}\)
3. \(p^8\)
4. \(64\)

Suppose now we have a fraction raised to a negative exponent. Let's use our definition of negative exponents to lead us to a new property. Consider \( \left( \frac{3}{4} \right)^{-2} \).\[ \begin{array}{rclcl}
\left( \dfrac{3}{4} \right)^{-2} & = & \dfrac{1}{\left( \frac{3}{4} \right)^2} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
& = & \dfrac{1}{\left( \frac{3}{4} \right) \cdot \left( \frac{3}{4} \right)} & & \\[6pt]
& = & \dfrac{1}{\frac{9}{16}} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{16}{16} \cdot \dfrac{1}{\frac{9}{16}} & \quad & \left( \text{simplifying the compound fraction} \right) \\[6pt]
& = & \dfrac{16}{9} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{4}{3} \cdot \dfrac{4}{3} & & \\[6pt]
& = & \left(\dfrac{4}{3}\right)^2 & & \\[6pt]
\end{array} \nonumber \]This tells us that\[ \left( \dfrac{3}{4} \right)^{-2} = \left( \dfrac{4}{3} \right)^{2}. \nonumber \]To get from the original fraction raised to a negative exponent to the final result, we took the reciprocal of the base - the fraction - and changed the sign of the exponent. This leads us to the Quotient to a Negative Power Law.

Theorem: Quotient to a Negative Power Law

If \(a\) and \(b\) are real numbers, \(a \neq 0\), \(b \neq 0\), and \(n\) is an integer, then\[\left(\dfrac{a}{b}\right)^{−n}=\left(\dfrac{b}{a}\right)^n. \nonumber \]

Example \(\PageIndex{5}\)

Simplify each expression:

1. \(\left( \frac{5}{7} \right)^{−2}\)
2. \(\left( −\frac{x}{y} \right)^{−3}\).

Solutions

1.    \[ \begin{array}{rclcl}
   \left( \dfrac{5}{7} \right)^{-2} & = & \left( \dfrac{7}{5} \right)^2 & \quad & \left( \text{Quotient to a Negative Power Law} \right) \\[6pt]
   & = & \dfrac{7}{5} \cdot \dfrac{7}{5}  & & \\[6pt]
   & = & \dfrac{49}{25} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
2.    \[ \begin{array}{rclcl}
   \left( -\dfrac{x}{y} \right)^{-3} & = & \left( -\dfrac{y}{x} \right)^3 & \quad & \left( \text{Quotient to a Negative Power Law} \right) \\[6pt]
   & = & \left(-\dfrac{y}{x}\right)\left(-\dfrac{y}{x}\right)\left(-\dfrac{y}{x}\right) & & \\[6pt]
   & = & -\dfrac{y^3}{x^3} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]

Checkpoint \(\PageIndex{5}\)

Simplify each expression:

1. \(\left(\frac{2}{3}\right)^{−4}\)
2. \(\left(−\dfrac{m}{n}\right)^{−2}\)

Answers

1. \(\frac{81}{16}\)
2. \(\frac{n^2}{m^2}\)

Now that we have negative exponents, we will use the Product Law with expressions that have negative exponents.

Example \(\PageIndex{6}\)

Simplify each expression:

1. \(z^{−5} \cdot z^{−3}\)
2. \((m^4n^{−3})(m^{−5}n^{−2})\)
3. \((2x^{−6}y^8)(−5x^5y^{−3})\)

Solutions

1.    \[ \begin{array}{rclcl}
   z^{−5} \cdot z^{−3} & = & z^{-5 + (-3)} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & z^{-8} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{1}{z^{8}} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   \end{array} \nonumber \]
2.    \[ \begin{array}{rclcl}
   (m^4n^{−3})(m^{−5}n^{−2}) & = & m^4 m^{-5} \cdot n^{-3} n^{-2} & \quad & \left( \text{Commutative Property of Multiplication} \right) \\[6pt]
   & = & m^{4 + (-5)} n^{-3 + (-2)} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & m^{-1} n^{-5} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{1}{m^1} \cdot \dfrac{1}{n^5} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   & = & \dfrac{1}{m n^5} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
3.    \[ \begin{array}{rclcl}
   (2x^{−6}y^8)(−5x^5y^{−3}) & = & (2)(-5) \cdot x^{-6} x^{5} \cdot y^{8} y^{-3} & \quad & \left( \text{Commutative Property of Multiplication} \right) \\[6pt]
   & = & -10 x^{-6 + 5} y^{8 + (-3)} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & -10 x^{-1} y^{5} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & -10 \cdot \dfrac{1}{x^1} \cdot y^5 & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   & = & -\dfrac{10y^5}{x} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]

Checkpoint \(\PageIndex{6}\)

Simplify each expression:

1. \(z^{−4} \cdot z^{−5}\)
2. \((p^6q^{−2})(p^{−9}q^{−1})\)
3. \((3u^{−5}v^7)(−4u^4v^{−2})\)

Answers

1. \(\frac{1}{z^9}\)
2. \(\frac{1}{p^3q^3}\)
3. \(−\frac{12v^5}{u}\)

The Laws of Exponents: Power Law

Now let's look at an exponential expression that contains a power raised to a power. See if you can discover a general property.\[ \left( x^2 \right)^3 = x^2 \cdot x^2 \cdot x^2 = \underbrace{\overbrace{x \cdot x}^{\color{blue}{2 \text{ factors}}} \quad \overbrace{x \cdot x}^{\color{blue}{2 \text{ factors}}} \quad \overbrace{x \cdot x}^{\color{blue}{2 \text{ factors}}}}_{\color{blue}{\text{for }6\text{ total factors}}} = x^6 \nonumber \]Notice the 6 is the product of the exponents, 2 and 3. We see that \((x^2)^3\) is \(x^{2 \cdot 3}\) or \(x^6\).

We multiplied the exponents. This leads to the Power Law for Exponents.

Theorem: Power Law for Exponents

If \(a\) is a real number and \(m\) and \(n\) are integers, then\[(a^m)^n=a^{m \cdot n}. \nonumber \]That is, powers raised to powers multiply.

Example \(\PageIndex{7}\)

Simplify each expression:

1. \((y^5)^9\)
2. \((4^4)^7\)
3. \((y^3)^6(y^5)^4\)

Solutions

1.    \[ \begin{array}{rclcl}
   \left( y^5 \right)^9 & = & y^{5 \cdot 9} & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
   & = & y^{45} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
2.    \[ \begin{array}{rclcl}
   \left( 4^4 \right)^7 & = & 4^{4 \cdot 7} & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
   & = & 4^{28} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
3.    \[ \begin{array}{rclcl}
   \left( y^3 \right)^6 \left( y^5 \right)^4 & = & y^{3 \cdot 6} y^{5 \cdot 4} & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
   & = & y^{18} y^{20} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & y^{18 + 20} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & y^{38} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]

Checkpoint \(\PageIndex{7}\)

Simplify each expression:

1. \((b^7)^5\)
2. \((5^4)^3\)
3. \((a^4)^5(a^7)^4\)

Answers

1. \(b^{35}\)
2. \(5^{12}\)
3. \(a^{48}\)

The Laws of Exponents: Product to a Power Law

We will now look at an expression containing a product that is raised to a power. Can you find this pattern?\[ \left( 2x \right)^3 = 2x \cdot 2x \cdot 2x = 2 \cdot 2 \cdot 2 \cdot x \cdot x \cdot x = 2^3 x^3. \nonumber \]Notice that each factor was raised to the power and \((2x)^3\) is \(2^3 \cdot x^3\). The exponent applies to each of the factors! This leads to the Product to a Power Law for Exponents.

Theorem: Product to a Power Law for Exponents

If \(a\) and \(b\) are real numbers and \(m\) is a whole number, then\[(ab)^m=a^mb^m. \nonumber \]That is, to raise a product to a power, raise each factor to that power.

Example \(\PageIndex{8}\)

Simplify each expression:

1. \((−3mn)^3\)
2. \((6k^3)^{−2}\)
3. \((5x^{−3})^2\)

Solutions

1.    \[ \begin{array}{rclcl}
   \left( -3mn \right)^3 & = & (-3)^3 m^3 n^3 & \quad & \left( \text{Laws of Exponents: Product to a Power Law} \right) \\[6pt]
   & = & -27 m^3 n^3 & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
2.    \[ \begin{array}{rclcl}
   \left(6k^3\right)^{−2} & = & (6)^{-2} \left( k^3 \right)^{-2} & \quad & \left( \text{Laws of Exponents: Product to a Power Law} \right) \\[6pt]
   & = & 6^{-2} k^{-6} & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
   & = & \dfrac{1}{6^{2}} \cdot \dfrac{1}{k^{6}} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   & = & \dfrac{1}{36} \cdot \dfrac{1}{k^{6}} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{1}{36k^6} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
3.    \[ \begin{array}{rclcl}
   \left(5x^{−3}\right)^2 & = & (5)^{2} \left( x^{-3} \right)^{2} & \quad & \left( \text{Laws of Exponents: Product to a Power Law} \right) \\[6pt]
   & = & 5^{2} x^{-6} & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
   & = & 5^2 \cdot \dfrac{1}{x^{6}} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   & = & 25 \cdot \dfrac{1}{x^{6}} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{25}{x^6} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]

Checkpoint \(\PageIndex{8}\)

Simplify each expression:

1. \((2wx)^5\)
2. \((−11pq^3)^0\)
3. \((2b^3)^{−4}\)
4. \((8a^{−4})^2\)

Answers

1. \(32w^5x^5\)
2. 1
3. \(\frac{1}{16b^{12}}\)
4. \(\frac{64}{a^8}\)

The Laws of Exponents: Quotient to a Power Law

Now we will look at an example that will lead us to the Quotient to a Power Law.\[ \left( \dfrac{x}{y} \right)^3 = \dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y} = \dfrac{x \cdot x \cdot x}{y \cdot y \cdot y} = \dfrac{x^3}{y^3}. \nonumber \]Notice that the exponent applies to both the numerator and the denominator. We see that \(\left(\frac{x}{y}\right)^3\) is \(\frac{x^3}{y^3}\). This leads to the Quotient to a Power Law for Exponents.

Theorem: Quotient to a Power Law for Exponents

If \(a\) and \(b\) are real numbers, \(b \neq 0\), and \(m\) is an integer, then\[\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}. \nonumber \]That is, to raise a fraction to a power, raise the numerator and denominator to that power.

Example \(\PageIndex{9}\)

Simplify each expression:

1. \(\left(\frac{b}{3}\right)^4\)
2. \(\left(\frac{k}{j}\right)^{−3}\)
3. \(\left(\frac{2xy^2}{z}\right)^3\)
4. \(\left(\frac{4p^{−3}}{q^2}\right)^2\)

Solutions

1.    \[ \begin{array}{rclcl}
   \left(\dfrac{b}{3}\right)^4 & = & \dfrac{b^4}{3^4} & \quad & \left( \text{Laws of Exponents: Quotient to a Power Law} \right) \\[6pt]
   & = & \dfrac{b^4}{81} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
2.    \[ \begin{array}{rclcl}
   \left(\dfrac{k}{j}\right)^{−3} & = & \dfrac{k^{-3}}{j^{-3}} & \quad & \left( \text{Laws of Exponents: Quotient to a Power Law} \right) \\[6pt]
   & = & \dfrac{k^{-3}}{1} \cdot \dfrac{1}{j^{-3}} & & \\[6pt]
   & = & \dfrac{1}{k^{3}} \cdot \dfrac{j^{3}}{1} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   & = & \dfrac{j^3}{k^{3}} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
3.    \[ \begin{array}{rclcl}
   \left(\dfrac{2xy^2}{z}\right)^3 & = & \dfrac{\left( 2xy^2 \right)^3}{z^3} & \quad & \left( \text{Laws of Exponents: Quotient to a Power Law} \right) \\[6pt]
   & = & \dfrac{2^3 x^3 \left( y^2 \right)^3}{z^3} & \quad & \left( \text{Laws of Exponents: Product to a Power Law} \right) \\[6pt]
   & = & \dfrac{2^3 x^3 y^{2 \cdot 3}}{z^3} & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
   & = & \dfrac{8 x^3 y^{6}}{z^3} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
4.    \[ \begin{array}{rclcl}
   \left(\dfrac{4p^{−3}}{q^2}\right)^2 & = & \dfrac{\left( 4p^{-3} \right)^2}{\left( q^2 \right)^2} & \quad & \left( \text{Laws of Exponents: Quotient to a Power Law} \right) \\[6pt]
   & = & \dfrac{4^2 \left( p^{-3} \right)^2}{q^4} & \quad & \left( \text{Laws of Exponents: Product to a Power Law and Power Law} \right) \\[6pt]
   & = & \dfrac{4^2 p^{-3 \cdot 2}}{q^4} & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
   & = & \dfrac{16 p^{-6}}{q^4} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{16}{q^4} \cdot \dfrac{p^{-6}}{1} & & \\[6pt]
   & = & \dfrac{16}{q^4} \cdot \dfrac{1}{p^{6}} & \quad & \left( \text{Laws of Exponents: Negative Exponent Property} \right) \\[6pt]
   & = & \dfrac{16}{p^6 q^4} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]

Checkpoint \(\PageIndex{9}\)

Simplify each expression:

1. \(\left(\frac{p}{10}\right)^4\)
2. \(\left(\frac{m}{n}\right)^{−7}\)
3. \(\left(\frac{3ab^3}{c^2}\right)^4\)
4. \(\left(\frac{3x^{−2}}{y^3}\right)^3\).

Answers

1. \(\frac{p^4}{10000}\)
2. \(\frac{n^7}{m^7}\)
3. \(\frac{81a^4b^{12}}{c^8}\)
4. \(\frac{27}{x^6y^9}\)

A Summary

We now have several properties for exponents. Let's summarize them and then we'll do some more examples that use more than one of the properties.

Theorem: Laws of Exponents

If \(a\) and \(b\) are real numbers, and \(m\) and \(n\) are integers, then\[ \begin{array}{rrcl}
\textbf{Product Law: } & a^m \cdot a^n & = & a^{m+n} \\[6pt]
\textbf{Quotient Law: } & \dfrac{a^m}{a^n} & = & a^{m−n}, \text{ where }a\neq 0 \\[6pt]
\textbf{Power Law: } & \left(a^m\right)^n & = & a^{m \cdot n} \\[6pt]
\textbf{Product to a Power Law: } & (ab)^n & = & a^nb^n \\[6pt]
\textbf{Quotient to a Power Law: } & \left(\dfrac{a}{b}\right)^m & = & \dfrac{a^m}{b^m}, \text{ where }b \neq 0 \\[6pt]
\textbf{Zero Exponent Law: } & a^0 & = & 1, \text{ where }a \neq 0 \\[6pt]
\textbf{Properties of Negative Exponents: } & a^{−n} & = & \dfrac{1}{a^n} \\[6pt]
 & \dfrac{1}{a^{−n}} & = & a^n \\[6pt]
 & \left(\dfrac{a}{b}\right)^{−n} & = & \left(\dfrac{b}{a}\right)^n \\[6pt]
\end{array} \nonumber \]

Example \(\PageIndex{10}\)

Simplify each expression by applying several properties:

1. \((3x^2y)^4(2xy^2)^3\)
2. \(\frac{(x^3)^4(x^{−2})^5}{(x^6)^5}\)
3. \(\left(\frac{2xy^2}{x^3y^{−2}}\right)^2 \left(\frac{12xy^3}{x^3y^{−1}}\right)^{−1}\)

Solutions

1.    \[ \begin{array}{rclcl}
   (3x^2y)^4(2xy^2)^3 & = & (3)^4 \left( x^2 \right)^4 y^4 \cdot (2)^3 x^3 \left( y^2 \right)^3 & \quad & \left( \text{Laws of Exponents: Product to a Power Law} \right) \\[6pt]
   & = & 81 x^{2 \cdot 4} y^4 \cdot 8 x^3 y^{2 \cdot 3} & \quad & \left( \text{simplifying and Laws of Exponents: Power Law} \right) \\[6pt]
   & = & (81 \cdot 8) x^{8} x^3 y^4 y^{6} & \quad & \left( \text{simplifying and Commutative Property of Multiplication} \right) \\[6pt]
   & = & 648 x^{8 + 3} y^{4 + 6} & \quad & \left( \text{simplifying and Laws of Exponents: Product Law} \right) \\[6pt]
   & = & 648 x^{11} y^{10} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]
2.    \[ \begin{array}{rclcl}
   \dfrac{(x^3)^4(x^{−2})^5}{(x^6)^5} & = & \dfrac{x^{3 \cdot 4} \cdot x^{-2 \cdot 5}}{x^{6 \cdot 5}} & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
   & = & \dfrac{x^{12} \cdot x^{-10}}{x^{30}} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{x^{12 + (-10)}}{x^{30}} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & \dfrac{x^{2}}{x^{30}} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & x^{2 - 30} & \quad & \left( \text{Laws of Exponents: Quotient Law} \right) \\[6pt]
   & = & x^{-28} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{1}{x^{28}} & \quad & \left( \text{Laws of Exponents: Properties of Negative Exponents} \right) \\[6pt]
   \end{array} \nonumber \]
3.    \[ \begin{array}{rclcl}
   \left(\dfrac{2xy^2}{x^3y^{−2}}\right)^2 \left(\dfrac{12xy^3}{x^3y^{−1}}\right)^{−1} & = & \left(2x^{1 - 3} y^{2 - (-2)}\right)^2 \left(12x^{1 - 3}y^{3 - (-1)}\right)^{−1} & \quad & \left( \text{Laws of Exponents: Quotient Law} \right) \\[6pt]
   & = & \left(2x^{-2} y^{4}\right)^2 \left(12x^{-2}y^{4}\right)^{−1} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & 2^2 \left(x^{-2}\right)^2 \left(y^{4}\right)^2 \cdot 12^{-1} \left(x^{-2}\right)^{-1} \left(y^{4}\right)^{−1} & \quad & \left( \text{Laws of Exponents: Product to a Power Law} \right) \\[6pt]
   & = & 4 x^{-2 \cdot 2} y^{4 \cdot 2} \cdot \dfrac{1}{12} x^{-2 \cdot -1} y^{4 \cdot -1} & \quad & \left( \text{simplifying and Laws of Exponents: Power Law} \right) \\[6pt]
   & = & \dfrac{4}{12} x^{-4} y^{8} x^{2} y^{-4} & \quad & \left( \text{simplifying and Commutative Property of Multiplication} \right) \\[6pt]
   & = & \dfrac{1}{3} x^{-4} x^{2} y^{8} y^{-4} & \quad & \left( \text{simplifying and Commutative Property of Multiplication} \right) \\[6pt]
   & = & \dfrac{1}{3} x^{-4 + 2} y^{8 + (-4)} & \quad & \left( \text{Laws of Exponents: Product Law} \right) \\[6pt]
   & = & \dfrac{1}{3} x^{-2} y^{4} & \quad & \left( \text{simplifying} \right) \\[6pt]
   & = & \dfrac{1}{3} \cdot \dfrac{1}{x^{2}} \cdot y^{4} & \quad & \left( \text{Laws of Exponents: Properties of Negative Exponents} \right) \\[6pt]
   & = & \dfrac{y^4}{3x^2} & \quad & \left( \text{simplifying} \right) \\[6pt]
   \end{array} \nonumber \]

Checkpoint \(\PageIndex{10}\)

Simplify each expression:

1. \((c^4d^2)^5(3cd^5)^4\)
2. \(\frac{(a^{−2})^3(a^2)^4}{(a^4)^5}\)
3. \(\left(\frac{3xy^2}{x^2y^{−3}}\right)^2\)

Answers

1. \(81c^{24}d^{30}\)
2. \(\frac{1}{a^{18}}\)
3. \(\frac{9y^{10}}{x^2}\)