10.1 Exponents

Natural number exponents

Using the Product Rule of Exponents

Consider the product \(x^3\times x^4\).  Both terms have the same base, \(x\), but they are raised to different exponents.  Expand each expression, and then rewrite the resulting expression.

\( \begin{align*} x^3 \times x^4 &= \overbrace{x \times x \times x}^{\text{3 factors}}\, \times \, \overbrace{ x \times x \times x\times x}^{\text{4 factors}} \\[4pt] &= \overbrace{x\times x\times x\times x\times x\times x\times x}^{\text{7 factors}} \\[4pt] &=x^7 \end{align*}\)

Notice that the exponent of a product is the sum of the exponents of the two factors.  In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents.  This is the product rule of exponents.

\[ x^3\times x^4=x^{3+4}=x^7 \nonumber\]

Now consider an example with real numbers.

\(2^3\times2^4=2^{3+4}=2^7\)

We can always check that this is true by simplifying each exponential expression.  We find that \(2^3\) is \(8\), \(2^4\) is \(16\), and \(2^7\) is \(128\).  The product \(8\times 16\) equals \(128\), so we see that the relationship is true.  We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents.

THE PRODUCT RULE OF EXPONENTS

For any real number \(a\) and natural numbers \(m\) and \(n\), the Product Rule of Exponents states that

\[a^m\times a^n=a^{m+n} \label{prod}\]

Example \(\PageIndex{1}\): Using the Product Rule

Write each of the following products with a single base.  Do not simplify further.

1. \(t^5\times t^3\)
2. \((−3)^5\times(−3)\)
3. \(x^2\times x^5\times x^3\)

Solution

Use the Product Rule (Equation \ref{prod}) to simplify each expression.

1. \(t^5\times t^3=t^{5+3}=t^8\)
2. \((−3)^5\times(−3)=(−3)^5\times(−3)^1=(−3)^{5+1}=(−3)^6\)
3. \(x^2\times x^5\times x^3\)

At first, it may appear that we cannot simplify a product of three factors.  However, using the associative property of multiplication, begin by simplifying the first two.

\[x^2\times x^5\times x^3=(x^2\times x^5) \times x^3=(x^{2+5})\times x^3=x^7\times x^3=x^{7+3}=x^{10} \nonumber\]

Notice we get the same result by adding the three exponents in one step.

\[x^2\times x^5\times x^3=x^{2+5+3}=x^{10} \nonumber\]

 

\(\PageIndex{1}\)

Write each of the following products with a single base.  Do not simplify further.

1. \(k^6\times k^9\)
2. \(\left(\dfrac{2}{y}\right)^4\times\left(\dfrac{2}{y}\right)\)
3. \(t^3\times t^6\times t^5\)

Answers

a. \(k^{15}\) \( \qquad \) b. \(\left(\dfrac{2}{y}\right)^5\) \( \qquad \) c. \(t^{14}\)

Using the Power Rule of Exponents

Suppose an exponential expression is raised to some power.  Can we simplify the result?  Yes.  To do this, we use the Power Rule of Exponents.  Consider the expression \((x^2)^3\) .  The variable \(x\) inside the parentheses is multiplied twice because it has an exponent of \(2\).  Then the result is multiplied three times because the entire expression has an exponent of \(3\).

\[\begin{align*} (x^2)^3 &= (x^2)\times(x^2)\times(x^2)\\ &= x\times x \, \times \, x\times x \, \times \, x\times x\\ &= x^6 \end{align*}\]

The exponent of the answer is the product of the exponents.  In other words, when raising an exponential expression to a power, we write the result with the common base and multiply the exponents.

\[ (x^2)^3=x^{2⋅3}=x^6 \nonumber\]

Be careful to distinguish between uses of the Product Rule and the Power Rule.  When using the Product Rule, factors with the same bases are raised to exponents and then multiplied together.  In this case, you add the exponents.  When using the Power Rule, a base raised to an exponent is raised to another exponent.  In this case, you multiply the exponents.

Product Rule	Power Rule
\(5^3\times5^4=5^{3+4}=5^7\)	\((5^3)^4=5^{3\times4}=5^{12}\)
\(x^5\times x^2=x^{5+2}=x^7\)	\((x^5)^2=x^{5\times2}=x^{10}\)
\((3a)^7\times(3a)^{10}=(3a)^{7+10}=(3a)^{17}\)	\(((3a)^7)^{10}=(3a)^{7\times10}=(3a)^{70}\)

THE POWER RULE OF EXPONENTS

For any real number \(a\) and natural numbers \(m\) and \(n\), the Power Rule of Exponents states that

\[(a^m)^n=a^{m⋅n} \label{power}\]

Example \(\PageIndex{2}\): Using the Power Rule

Write each of the following products with a single base.  Do not simplify further.

1. \((x^2)^7\)
2. \(((2t)^5)^3\)
3. \(((−3)^5)^{11}\)

Solution

Use the Power Rule (Equation \ref{power}) to simplify each expression.

1. \((x^2)^7=x^{2⋅7}=x^{14}\)
2. \(((2t)^5)^3=(2t)^{5⋅3}=(2t)^{15}\)
3. \(((−3)^5)^{11}=(−3)^{5⋅11}=(−3)^{55}\)

\(\PageIndex{2}\)

Write each of the following products with a single base.  Do not simplify further.

1. \(((3y)^8)^3\)
2. \((t^5)^7\)
3. \(((−g)^4)^4\)

Answers

a. \((3y)^{24}\) \( \qquad \) b. \(t^{35}\)  \( \qquad \) c. \((−g)^{16}\)

Using the Quotient Rule of Exponents

The Quotient Rule of Exponents allows us to simplify an expression that divides two numbers with the same base but different exponents.  In a similar way to the Product Rule, we can simplify an expression such as \(\dfrac{y^m}{y^n}\).  Consider the example \(\dfrac{y^9}{y^5}\) .  Perform the division by canceling common factors.

\[\begin{align*} \dfrac{y^9}{y^5} &= \dfrac{y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y}{y\cdot y\cdot y\cdot y\cdot y}\\ &= \dfrac{y\cdot y\cdot y\cdot y}{1}\\ &= y^4 \end{align*}\]

Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.  In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.

\(\dfrac{y^9}{y^5}=y^{9−5}=y^4\)

 

THE QUOTIENT RULE OF EXPONENTS

For any real number \(a \;(a \ne 0)\) and natural numbers \(m\) and \(n\), the Quotient Rule of Exponents states that

\[\dfrac{a^m}{a^n}=a^{m−n} \label{quot}\]

Instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.

Example \(\PageIndex{3}\): Using the Quotient Rule

Write each of the following products with a single base.  Do not simplify further.

1. \(\dfrac{(−2)^{14}}{(−2)^{9}}\)
2. \(\dfrac{t^{23}}{t^{15}}\)
3. \(\dfrac{(z\sqrt{2})^5}{z\sqrt{2}}\)

Solution

Use the Quotient Rule (Equation \ref{quot}) to simplify each expression.

1. \(\dfrac{(−2)^{14}}{(−2)^{9}}=(−2)^{14−9}=(−2)^5\)
2. \(\dfrac{t^{23}}{t^{15}}=t^{23−15}=t^8\)
3. \(\dfrac{(z\sqrt{2})^5}{z\sqrt{2}}=(z\sqrt{2})^{5−1}=(z\sqrt{2})^4\)

\(\PageIndex{3}\)

Write each of the following products with a single base.  Do not simplify further.

1. \(\dfrac{s^{75}}{s^{68}}\)
2. \(\dfrac{(−3)^6}{−3}\)
3. \(\dfrac{(ef^2)^5}{(ef^2)^3}\)

Answers

a. \(s^7\) \( \qquad \) b. \((−3)^5\) \( \qquad \) c. \((ef^2)^2\)

The Zero Exponent Definition

What would happen if the Quotient Rule were used and \(m=n\)?  Consider this example:

\[\dfrac{t^8}{t^8}=1 \qquad \text{because a nonzero number divided by itself is 1.}\nonumber\]

If we were to simplify the expression using the Quotient Rule instead, we would have

\[\dfrac{t^8}{t^8}=t^{8−8}=t^0. \nonumber\]

If we equate the two answers, the result is \(t^0=1\).  This leads to the natural definition of what it means to raise a quantity to a zero exponent.

THE ZERO EXPONENT DEFINITION

For any nonzero real number \(a\), the zero exponent definition states that

\[a^0=1 \label{zero}\]

Note: The sole exception is the expression \(0^0\), whose value is undefined (can you see the reason why?). 

Example \(\PageIndex{4}\): Using the Zero Exponent Definition

Simplify each expression using the zero exponent definition.

1. \(\dfrac{c^3}{c^3}\)
2. \(\dfrac{-3x^5}{x^5}\)
3. \(\dfrac{(j^2k)^4}{(j^2k)\times(j^2k)^3}\)
4. \(\dfrac{5(rs^2)^2}{(rs^2)^2}\)

Solution

Use the zero exponent definition and other rules to simplify each expression.

a. \[\begin{align*} \dfrac{c^3}{c^3} &= c^{3-3}\\ &= c^0\\ &= 1 \end{align*}\]

b. \[\begin{align*} \dfrac{-3x^5}{x^5} &= -3\times\dfrac{x^5}{x^5}\\[2mm] &= -3\times x^{5-5}\\ &= -3\times x^0\\ &= -3\times 1\\ &= -3 \end{align*}\]

c. \[\begin{align*} \dfrac{(j^2k)^4}{(j^2k)\times(j^2k)^3} &= \dfrac{(j^2k)^4}{(j^2k)^{1+3}} && \text{ Use the Product Rule in the denominator}\\[2mm] &= \dfrac{(j^2k)^4}{(j^2k)^4} && \text{ Simplify}\\[2mm] &= (j^2k)^{4-4} && \text{ Use the Quotient Rule}\\[2mm] &= (j^2k)^0 && \text{ Simplify}\\ &= 1 \end{align*}\]

d. \[\begin{align*} \dfrac{5(rs^2)^2}{(rs^2)^2} &= 5(rs^2)^{2-2} && \text{ Use the Quotient Rule}\\[2mm] &= 5(rs^2)^0 && \text{ Simplify}\\ &= 5\times1 && \text{ Use the zero exponent definition}\\ &= 5 && \text{ Simplify} \end{align*}\]

\(\PageIndex{4}\)

Simplify each expression using the zero exponent definition.

1. \(\dfrac{t^7}{t^7}\)
2. \(\dfrac{(de^2)^{11}}{2(de^2)^{11}}\)
3. \(\dfrac{w^4\times w^2}{w^6}\)
4. \(\dfrac{t^3\times t^4}{t^2\times t^5}\)

Answers

a. \(1\) \( \qquad \)  b. \(\dfrac{1}{2}\)  \( \qquad \)  c. \(1\)  \( \qquad \)  d. \(1\)

The Negative Exponent Definition

Consider the situation where one exponential expression is divided by another exponential expression that has a larger exponent.  For example, \(\dfrac{t^3}{t^5}\).

\[\begin{align*} \dfrac{t^3}{t^5} &= \dfrac{t\times t\times t}{t\times t\times t\times t\times t} \\ &= \dfrac{1}{t\times t}\\ &= \dfrac{1}{t^2} \end{align*}\]

If we were to simplify the original expression using the Quotient Rule, we would have

\[\begin{align*} \dfrac{t^3}{t^5} &= t^{3-5} \\ &= t^{-2} \end{align*}\]

Putting the answers together, we have \(t^{−2}=\dfrac{1}{t^2}\). This suggests a natural definition for negative integer exponents, assuming the base is a nonzero real number.

THE NEGATIVE EXPONENTS DEFINITION

For any nonzero real number \(a\) and integer \(n\), the negative exponent definition states that

\[a^{−n}=\dfrac{1}{a^n} \]

Using the definition of a negative exponent, a rule for working with negative exponents can be deduced: a factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa. 

THE NEGATIVE EXPONENT RULE

For any real number \(a \;(a \ne 0)\), the Negative Exponent Rule states that

\[a^{−n}=\dfrac{1}{a^n} \text{ and } \dfrac{1}{a^{−n}}=a^n \label{neg}\]

To see that \(\dfrac{1}{a^{−n}} = a^n \), we would need to simplify a complex fraction (see Section 10.2).

We have now defined the exponential expression \(a^n\) when \(n\) is a natural number, \(0\), or the negative of a natural number, in such a way that the product rule, power rule, and quotient rule holds true.  In other words, \(a^n\) is defined for any integer \(n\). 

Example \(\PageIndex{5}\): Using the Negative Exponent Definition

Write each of the following quotients with a single base.  Do not simplify further.  Write answers with positive exponents.

1. \(\dfrac{\theta^3}{\theta^{10}}\)
2. \(\dfrac{z^2\times z}{z^4}\)
3. \(\dfrac{(-5t^3)^4}{(-5t^3)^8}\)

Solution

1. \(\dfrac{\theta^3}{\theta^{10}}=\theta^{3-10}=\theta^{-7}=\dfrac{1}{\theta^7}\)
2. \(\dfrac{z^2\times z}{z^4}=\dfrac{z^{2+1}}{z^4}=\dfrac{z^3}{z^4}=z^{3-4}=z^{-1}=\dfrac{1}{z}\)
3. \(\dfrac{(-5t^3)^4}{(-5t^3)^8}=(-5t^3)^{4-8}=(-5t^3)^{-4}=\dfrac{1}{(-5t^3)^4}\)

\(\PageIndex{5}\)

Write each of the following quotients with a single base.  Do not simplify further.  Write answers with positive exponents.

1. \(\dfrac{(-3t)^2}{(-3t)^8}\)
2. \(\dfrac{f^{47}}{f^{49}\times f}\)
3. \(\dfrac{2k^4}{5k^7}\)

Answers

a. \(\dfrac{1}{(-3t)^6}\)  \( \qquad \)  b. \(\dfrac{1}{f^3}\) \( \qquad \)  c. \(\dfrac{2}{5k^3}\)

Let's look at some examples of the product, power and quotient rules that involve any integer \(n\). 

Example \(\PageIndex{6}\): Using the Product, Power and Quotient Rules

Write each of the following products with a single base.  Do not simplify further.  Write answers with positive exponents.

1. \(b^2\times b^{-8}\)
2. \((-x)^5\times(-x)^{-5}\)
3. \(\dfrac{-7z}{(-7z)^5}\)
4. \( ((3t)^{-2})^{-4} \)

Solution

1. \(b^2\times b^{-8}=b^{2\: + \: -8}=b^{-6}=\dfrac{1}{b^6}\)
2. \((-x)^5\times(-x)^{-5}=(-x)^{5\: + \:-5}=(-x)^0=1\)
3. \(\dfrac{-7z}{(-7z)^5}= \dfrac{(-7z)^1}{(-7z)^5}=(-7z)^{1\:-\:5}=(-7z)^{-4}=\dfrac{1}{(-7z)^4}\)
4. \( ((3t)^{-2})^{-4} = (3t)^{-2 \;\times \;-4} = (3t)^8 \)

\(\PageIndex{6}\)

Write each of the following products with a single base.  Do not simplify further.  Write answers with positive exponents.

1. \(t^{-11}\times t^6\)
2. \(\dfrac{25^{12}}{25^{13}}\)
3. \( ((-x)^{-5})^4 \)

Answers

a. \(t^{-5}=\dfrac{1}{t^5}\)  \( \qquad \) b. \(\dfrac{1}{25}\) \( \qquad \) c. \( \dfrac{1}{(-x)^{20}} \)

Finding the Power of a Product

To rewrite the power of a product of two exponential expressions, we can develop another rule of exponents.  Consider \((pq)^3\).  We begin by using the associative and commutative properties of multiplication to regroup the factors.

\[\begin{align*} (pq)^3 &= (pq)\times(pq)\times(pq)\\ &= p\times p\times p\times q\times q\times q\\ &= p^3\times q^3 \end{align*}\]

In other words, \((pq)^3=p^3\times q^3\). 

For an example using a negative exponent, consider \( (pq)^{-2}\).

\[ \begin{align*} (pq)^{-2} &=\dfrac{1}{(pq)^2}\\ &= \dfrac{1}{(pq)\times(pq)}\\ &= \dfrac{1}{p\times p\times q\times q}\\ &= \dfrac{1}{p^2}\times \dfrac{1}{q^2} \\&= p^{-2} \times q^{-2} \end{align*}\]

THE POWER OF A PRODUCT RULE OF EXPONENTS

For any real numbers \(a\) and \(b\) and any integer \(n\), the power of a product rule of exponents states that

\[(ab)^n=a^nb^n.\]​

Example \(\PageIndex{7}\): Using the Power of a Product Rule

Simplify each of the following products as much as possible using the power of a product rule.  Write answers with positive exponents.

1. \((ab^2)^3\)
2. \((2t)^{15}\)
3. \((-2w^3)^3\)
4. \(\dfrac{1}{(-7z)^4}\)
5. \((e^{-2}f^2)^7\)

Solution

Use the product and quotient rules and the definition of a negative exponent to simplify each expression.

a. \((ab^2)^3=(a)^3\times(b^2)^3=a^{1\times3}\times b^{2\times3}=a^3b^6\)

b. \((2t)^{15}=(2)^{15}\times(t)^{15}=2^{15}t^{15}=32,768t^{15}\)

c. \((−2w^3)^3=(−2)^3\times(w^3)^3=−8\times w^{3\times3}=−8w^9\)

d. \(\dfrac{1}{(-7z)^4}=\dfrac{1}{(-7)^4\times(z)^4}=\dfrac{1}{2401z^4}\)

e. \((e^{-2}f^2)^7=(e^{−2})^7\times(f^2)^7=e^{−2\times7}\times f^{2\times7}=e^{−14}f^{14}=\dfrac{f^{14}}{e^{14}}\)

\(\PageIndex{7}\)

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.

1. \((g^2h^3)^5\)
2. \((5t)^3\)
3. \((-3y^5)^3\)
4. \(\dfrac{1}{(a^6b^7)^3}\)
5. \((r^3s^{-2})^4\)

Answers

a. \(g^{10}h^{15}\)  \( \qquad \)  b. \(125t^3\)  \( \qquad \)  c. \(-27y^{15}\)  \( \qquad \)  d. \(\dfrac{1}{a^{18}b^{21}}\)  \( \qquad \)  e. \(\dfrac{r^{12}}{s^8}\)

Finding the Power of a Quotient

To simplify the power of a quotient of two expressions, consider the expression below.

\( \left( \dfrac{2}{x} \right)^3 = \dfrac{2}{x} \cdot \dfrac{2}{x} \cdot \dfrac{2}{x} = \dfrac{2 \cdot 2 \cdot 2}{x \cdot x \cdot x} = \dfrac{2^3}{x^3} \)

In general, the power of a quotient of expressions is the quotient of the powers of the expressions.

THE POWER OF A QUOTIENT RULE OF EXPONENTS

For any real numbers \(a\) and \(b\), \(b \ne 0 \), and any integer \(n\), the Power of a Quotient Rule of Exponents states that

\[\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}.\]

Example \(\PageIndex{8}\): Using the Power of a Quotient Rule

Simplify each of the following quotients as much as possible using the Power of a Quotient Rule. Write answers with positive exponents.

1. \(\left(\dfrac{4}{z^{11}}\right)^3\)
2. \(\left(\dfrac{p}{q^3}\right)^6\)
3. \(\left(\dfrac{-1}{t^2}\right)^{27}\)
4. \((j^3k^{-2})^4\)
5. \((m^{-2}n^{-2})^3\)

Solution

a. \(\left(\dfrac{4}{z^{11}}\right)^3=\dfrac{(4)^3}{(z^{11})^3}=\dfrac{64}{z^{11\times3}}=\dfrac{64}{z^{33}}\)

b. \(\left(\dfrac{p}{q^3}\right)^6=\dfrac{(p)^6}{(q^3)^6}=\dfrac{p^{1\times6}}{q^{3\times6}}=\dfrac{p^6}{q^{18}}\)

c. \(\left(\dfrac{-1}{t^2}\right)^{27}=\dfrac{(-1)^{27}}{(t^2)^{27}}=\dfrac{-1}{t^{2\times27}}=\dfrac{-1}{t^{54}}=-\dfrac{1}{t^{54}}\)

d. \((j^3k^{-2})^4=\left(\dfrac{j^3}{k^2}\right)^4=\dfrac{(j^3)^4}{(k^2)^4}=\dfrac{j^{3\times4}}{k^{2\times4}}=\dfrac{j^{12}}{k^8}\)

e. \((m^{-2}n^{-2})^3=\left(\dfrac{1}{m^2n^2}\right)^3=\dfrac{(1)^3}{(m^2n^2)^3}=\dfrac{1}{(m^2)^3(n^2)^3}=\dfrac{1}{m^{2\times3}n^{2\times3}}=\dfrac{1}{m^6n^6}\)

\(\PageIndex{8}\)

Simplify each of the following quotients as much as possible using the Power of a Quotient Rule. Write answers with positive exponents.

1. \(\left(\dfrac{b^5}{c}\right)^3\)
2. \(\left(\dfrac{5}{u^8}\right)^4\)
3. \(\left(\dfrac{-1}{w^3}\right)^{35}\)
4. \((p^{-4}q^3)^8\)
5. \((c^{-5}d^{-3})^4\)

Answers

a. \(\dfrac{b^{15}}{c^3}\)  \( \qquad \) b. \(\dfrac{625}{u^{32}}\)  \( \qquad \) c. \(\dfrac{-1}{w^{105}}\) \( \qquad \)  d. \(\dfrac{q^{24}}{p^{32}}\)  \( \qquad \) e. \(\dfrac{1}{c^{20}d^{12}}\)

Simplifying Exponential Expressions

Complicated-looking expressions that involve a lot of exponents can often be rewritten more simply by using the rules for exponents.

Example \(\PageIndex{9}\): Simplifying Exponential Expressions

Simplify each expression and write the answer with positive exponents only.

1. \((6m^2n^{-1})^3\)
2. \(17^5\times17^{-4}\times17^{-3}\)
3. \(\left(\dfrac{u^{-1}v}{v^{-1}}\right)^2\)
4. \((-2a^3b^{-1})(5a^{-2}b^2)\)
5. \((x^2\sqrt{2})^4(x^2\sqrt{2})^{-4}\)
6. \(\dfrac{(3w^2)^5}{(6w^{-2})^2}\)

Solution

a. \[\begin{align*} (6m^2n^{-1})^3 &= (6)^3(m^2)^3(n^{-1})^3 && \text{ Power of a Product Rule}\\ &= 6^3m^{2\times3}n^{-1\times3} && \text{ Power Rule}\\ &= 216m^6n^{-3} && \text{ Power Rule}\\ &= \dfrac{216m^6}{n^3} && \text{ Negative Exponent Rule} \end{align*}\]

b. \[\begin{align*} 17^5\times17^{-4}\times17^{-3} &= 17^{5+(-4)+(-3)} && \text{ Product Rule}\\ &= 17^{-2} && \text{ Simplify}\\ &= \dfrac{1}{17^2} \;\text{ or } \;\dfrac{1}{289} && \text{ Negative Exponent Rule} \end{align*}\]

The order in which you apply the rules of exponents generally does not matter, so you can try different orders to find what works best for you.  For instance, to avoid making mistakes in subtracting negative numbers, you might want to apply the Negative Exponent Rule before the Quotient Rule. Both approaches are illustrated below.  They both begin by using the Power of a Quotient Rule followed by the Power of a Product Rule.  Then, to the left, the Quotient Rule comes next; while to the right, the Negative Exponent Rule comes next.

c. \[\begin{align*} \left ( \dfrac{u^{-1}v}{v^{-1}} \right )^2
&= \dfrac{(u^{-1}v)^2}{(v^{-1})^2} && \text{ Power of a Quotient Rule}\\
&= \dfrac{u^{-2}v^2}{v^{-2}} && \text{ Power of a Product Rule}\\
\end{align*}\]

\( \begin{array}{ll|ll}
= u^{-2}v^{2-(-2)} & \text{Quotient Rule} & =\dfrac{v^2 {\color{Cerulean}{v^2}}}{  \color{Cerulean}{u^2}  } & \text{Negative Exponent Rule}\\
= u^{-2}v^4 & \text{Simplify} & =\dfrac{v^{2+2}}{u^2} & \text{Product Rule}\\
= \dfrac{v^4}{u^2} & \text{Negative Exponent Rule} & =\dfrac{v^4}{u^2} & \text{Simplify}\\
\end{array} \)

d. \[\begin{align*} \left (-2a^3b^{-1} \right ) \left(5a^{-2}b^2 \right )
&= -2 \cdot a^3 \cdot b^{-1} \cdot 5 \cdot a^{-2} \cdot b^2 && \text{ Associative law of multiplication}\\
&= -2 \cdot 5 \cdot a^3 \cdot a^{-2} \cdot b^{-1} \cdot b^2 && \text{ Commutative law of multiplication}\\
&= (-2 \cdot 5) \cdot (a^3 \cdot a^{-2}) \cdot (b^{-1} \cdot b^2) && \text{ Associative law of multiplication}\\
&= -10\times a^{3+(-2)}\times b^{-1+2} && \text{ The product rule}\\
&= -10ab && \text{ Simplify} \end{align*}\]

e. \[\begin{align*} \left (x^2\sqrt{2})^4(x^2\sqrt{2} \right )^{-4} &= \left (x^2\sqrt{2} \right )^{4-4} && \text{ Product Rule}\\ &= \left (x^2\sqrt{2} \right )^0 && \text{ Simplify}\\ &= 1 && \text{ Zero Exponent Definition} \end{align*}\]

f. \[\begin{align*} \dfrac{(3w^2)^5}{(6w^{-2})^2}
&= \dfrac{(3)^5\times(w^2)^5}{(6)^2\times(w^{-2})^2} && \text{ Power of a Product Rule}\\
&= \dfrac{3^5w^{2\times5}}{6^2w^{-2\times2}} && \text{ Power Rule}\\
&= \dfrac{243w^{10}}{36w^{-4}} && \text{ Simplify}\\
&= \dfrac{243w^{10}w^4}{36} && \text{Negative Exponent Rule}\\
&= \dfrac{243w^{10+4}}{36} && \text{ Product Rule}\\
&= \dfrac{27w^{14}}{4} && \text{ Reduce fraction} \end{align*}\]

\(\PageIndex{9}\)

Simplify each of the following exponential expressions. Write answers with positive exponents.

1. \(  (2uv^{-2})^{-3}   \)
2. \(  x^8 \cdot c^{-12} \cdot x \)
3. \( \Big(  \dfrac{e^2f^{-3}}{f^{-1}} \Big)^2 \)
4. \(  (9r^{-5}s^3)(3r^6s^{-4})  \)
5. \(  ( \frac{4}{9}tw^{-2} )^{-3} ( \frac{4}{9}tw^{-2} )^3  \)
6. \( \dfrac{ (2h^2k)^4 }{ (7h^{-1}k^2)^2 } \)

Answers

a. \(\dfrac{v^6}{8u^3}\)  \( \qquad \) b. \(\dfrac{x^9}{c^12}\)  \( \qquad \) c. \(\dfrac{e^4}{f^4}\) \( \qquad \)  d. \(\dfrac{27r}{s}\)  \( \qquad \) e. \( 1 \) \( \qquad \)  f. \( \dfrac{8h^{16}}{49}\)

Rational Exponents

n^th Roots

Rational (or fractional) exponents are another way of writing expressions with radicals. When we use rational exponents, we can apply the properties of exponents to simplify expressions.  You can study radicals in detail in Section 10.2, but for a brief reminder, we have the following.

Definition: Principal \(n\)th root of \(a\)

For any whole number \(n\) greater than 1, the principal \(n\)th root of \(a\) is written \( \sqrt[n]{a} \) and has the same sign as \(a\).
When \(a\) is negative and \(n\) is even, the principal \(n\)th root is undefined as a real number.

\[ \text{The notation } \sqrt[n]{a} = b  \text{ means that } a=b^n. \]

The Definition of the Fractional Exponent \(\frac{1}{n}\)

Suppose we wish to define fractional (or, rational) exponents in such a way that the rules of exponents continue to hold true.  For example, consider what the meaning of \( a^{1/2}\) should be. Using the Power Rule, \((a^{1/2})^2=a^{(1/2)⋅2}=a^1=a  \). Combining these two different forms for \(a\), it can be concluded that \( a^{1/2} = \sqrt{a}\), which exists as long as \( a \ge 0 \).

Let's extend this idea to find an appropriate meaning for \(a^{1/3}\). Again using the Power Rule, \((a^{1/3})^3=a^{(1/3)⋅3}=a^1=a  \). Combining these two different forms for \(a\), it can be concluded that \( a^{1/3} = \sqrt[3]{a}\). However, unlike square roots, the radicand \(a\) can be any real number (not only non-negative numbers) and its cube root will always be defined. This is because negative numbers have cube roots; for example, the cube root of \(-8\) is \(-2\). 

These ideas can be extended in a similar fashion to any \(n\)th root, with \(a^{1/n}\) defined for non-negative values of \(a\) if \(n\) is even, and for all values of \(a\) if \(n\) is odd.

Definition:  \(a^{1/n}\)

\[\text{For any whole number } n \text{ greater than 1, } a^{\frac{1}{n}}=\sqrt[n]{a}. \label{nth}\]

For example,

\[ \begin{align*} (25)^{1/2} &= \sqrt{25} = 5 \\ (-27)^{1/3} &= \sqrt[3]{-27} = -3 . \end{align*} \]

There will be times when working with expressions will be easier if you use rational exponents and times when it will be easier if you use radicals. In these next examples, you'll practice converting expressions between the two notations.

Example \(\PageIndex{10}\)

Write as a radical expression:

1. \(x^{\frac{1}{2}}\)
2. \(y^{\frac{1}{3}}\)
3. \(z^{\frac{1}{4}}\)

Solution:

We want to write each expression in the form \(\sqrt[n]{a}\).

a.  \(x^{\frac{1}{2}}\)

The denominator of the rational exponent is \(2\), so the root is a second root, known as a square root. We do not show the index when it is \(2\).

\(\sqrt{x}\)

b.  \(y^{\frac{1}{3}}\)

The denominator of the exponent is \(3\), so the root is a third root, known as a cube root.

\(\sqrt[3]{y}\)

c.  \(z^{\frac{1}{4}}\)

The denominator of the exponent is \(4\), so the root is a fourth root.

\(\sqrt[4]{z}\)

Exercise \(\PageIndex{10}\)

Write as a radical expression:

1. \(b^{\frac{1}{6}}\)
2. \(z^{\frac{1}{5}}\)
3. \(p^{\frac{1}{16}}\)

Answer

1. \(\sqrt[6]{b}\)
2. \(\sqrt[5]{z}\)
3. \(\sqrt[16]{p}\)

In the next exercise, write each radical using a rational exponent.

Exercise \(\PageIndex{11}\)

Write as a base with a rational exponent:

1. \(\sqrt[3]{a}\)
2. \(\sqrt[5]{t}\)
3. \(\sqrt[11]{x}\)

Answer

1. \(a^{\frac{1}{3}}\)
2. \(t^{\frac{1}{5}}\)
3. \(x^{\frac{1}{11}}\)

Rational Exponential Expressions \(a^{m/n}\)

An expression with a rational exponent is equivalent to a radical where the denominator indicates the root and the numerator indicates the exponent. Any radical expression can be written with a rational exponent, which we call exponential form.

Let \(m\) and \(n\) be positive integers with no common factor other than 1.  Using the Power Rule, we have

\( (a^{1/n})^m = a^{(1/n)⋅m} = a^{m/n}. \) 

Thus, using the definition of \(a^{1/n}\), we have the following rule.

Rule for Rewriting Rational Exponents

\[a^{m/n}=(\sqrt[n]{a})^m=\sqrt[n]{a^m} \label{rat}\]

If \(a\) is negative and \(n\) is even, no real number can be assigned to this expression.

Note that, as long as the numerator and denominator of the exponent have no common factor, it does not matter if we apply the power first or the root first. For example, we can apply the power before the \(n\)th root:

\((-8) ^ { 2 / 3 } = ((-8)^2)^{1/3} = (64)^{1/3} = \sqrt [ 3 ] {64 } = 4.\)

Or we can apply the \(n\)th root before the power:

\((-8) ^ { 2 / 3 } = ((-8)^{1/3})^2 = ( \sqrt [ 3 ] { -8 } ) ^ { 2 } = (-2) ^ { 2 } = 4.\)

The results are the same, although when doing calculations by hand it is often easier to find the \(n\)th root first because the numbers are smaller.

However, if the numerator and denominator do share a common factor and the base is a negative number, there can be problems. For example, suppose we try to calculate the value of \((-2)^{2/2}\).  We would expect to get \(-2\) as our answer, since \(\frac{2}{2}=1,\) and \( (-2)^1=-2\).  Here's what happens if we apply the power before the root:

\[(-2)^{2/2} = ((-2)^2)^{1/2} = 4^{1/2} = \sqrt{4} = 2.  \quad\text{ Not what we expected!} \nonumber\]

Here's what happens if we apply the root before the power:

\[ (-2)^{2/2} = ((-2)^{1/2})^2 = (\sqrt{-2})^2, \text{ undefined as a real number, since there is no real square root of } -2.  \quad\text{Again, not what we expected!}\nonumber \]

Neither calculation gives us the expected answer, and the results are different if we change the order!  Since the goal is to have a set of rules that are well-behaved in all circumstances, this is the reason rational exponents with an even denominator are defined only for a non-negative base; and why we require the fraction to be reduced.

Exercise \(\PageIndex{10}\)

Try to calculate \((-1)^{4/10}\) in these two different orders:

1. \( ((-1)^{\frac{1}{10}})^4 \)
2. \( ((-1)^4)^{\frac{1}{10}} \)

Answer

1. Undefined
2. 1

 

Given an expression with a rational exponent, write the expression as a radical.

1. Determine the power by looking at the numerator of the exponent.
2. Determine the root by looking at the denominator of the exponent.

Key Equations

Rules of Exponents For nonzero real numbers a and b and integers m and n
Product rule	\(a^m⋅a^n=a^{m+n}\)
Quotient rule	\(\dfrac{a^m}{a^n}=a^{m−n}\)
Power rule	\((a^m)^n=a^{m⋅n}\)
Zero exponent definition	\(a^0=1\)
Negative exponent definition	\(a^{−n}=\dfrac{1}{a^n}\)
Power of a product rule	\((a⋅b)^n=a^n⋅b^n\)
Power of a quotient rule	\(\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\)
\(n\)th root definition (for \(n>1\))	\(a^{\frac{1}{n}}=\sqrt[n]{a}\)
Rational exponent rule (where \(m\) and \(n\) have no common factor)	\(a^{m/n}=(\sqrt[n]{a})^m=\sqrt[n]{a^m}\)

 

Key Concepts

• Products of exponential expressions with the same base can be simplified by adding exponents. 
• Quotients of exponential expressions with the same base can be simplified by subtracting exponents. 
• Powers of exponential expressions with the same base can be simplified by multiplying exponents. 
• An expression with exponent zero is defined as 1.
• An expression with a negative exponent is defined as a reciprocal. 
• The power of a product of factors is the same as the product of the powers of the same factors.
• The power of a quotient of factors is the same as the quotient of the powers of the same factors.
• An expression with a fractional exponent can be rewritten as a radical expression.
• The rules for exponential expressions can be combined to simplify more complicated expressions.

Contributors

• Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.

• Margaret Dean