5.3.1: Properties of Exponents: Product, Quotient and Power Rules

Learning Outcomes

Apply the product rule for exponents.
Apply the quotient rule for exponents.
Apply the power rule for exponents.

The Product Rule for Exponents

Consider the product ${x}^{3}\cdot {x}^{4}$ . Both terms have the same base, $x$ , but they are raised to different exponents. Let’s expand each expression, and then rewrite the resulting expression using exponents:

\begin{aligned} x^{3} \cdot x^{4} & = \overset{3 factors}{(x \cdot x \cdot x)} \overset{4 factors}{(x \cdot x \cdot x \cdot x)} \\ = \overset{7 factors}{x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x} \\ = x^{7} \end{aligned}

$\begin{align}x^{3}\cdot x^{4}&=\stackrel{\text{3 factors }}{(x\cdot x\cdot x)} \stackrel{\text{ 4 factors}}{(x\cdot x\cdot x\cdot x)} \\ & =\stackrel{7 \text{ factors}}{x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x} \\ & =x^{7}\end{align}$

The result is that ${x}^{3}\cdot {x}^{4}={x}^{3+4}={x}^{7}$ .

Notice that the exponent of the product is the sum of the exponents of the terms. 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 for exponents.

a^{m} \cdot a^{n} = a^{m + n}

${a}^{m}\cdot {a}^{n}={a}^{m+n}$

The Product Rule for Exponents

For any real numbers $a, m$ and $n$ , the product rule of exponents states that ${a}^{m}\cdot {a}^{n}={a}^{m+n}$

Example 1

Simplify the expressions.

${t}^{5}\cdot {t}^{3}$
$\left(-3\right)^{5}\cdot \left(-3\right)$
${x}^{2}\cdot {x}^{5}\cdot {x}^{3}$

Solution

${t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}$
${\left(-3\right)}^{5}\cdot \left(-3\right)={\left(-3\right)}^{5}\cdot {\left(-3\right)}^{1}={\left(-3\right)}^{5+1}={\left(-3\right)}^{6}$
${x}^{2}\cdot {x}^{5}\cdot {x}^{3}={x}^{2+5+3}={x}^{10}$

Try It 1

Simplify each expression.

${k}^{6}\cdot {k}^{9}$
${\left(\dfrac{2}{y}\right)}^{4}\cdot \left(\dfrac{2}{y}\right)$
${t}^{3}\cdot {t}^{6}\cdot {t}^{5}$

Show Answer

${k}^{15}$
${\left(\dfrac{2}{y}\right)}^{5}$
${t}^{14}$

TRY IT 2

The following video shows more examples of how to use the product rule to simplify an expression with exponents.

The Quotient Rule for Exponents

The quotient rule for 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{aligned} \frac{y^{9}}{y^{5}} & = \frac{y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y}{y \cdot y \cdot y \cdot y \cdot y} \\ = \frac{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} \\ = \frac{y \cdot y \cdot y \cdot y}{1} \\ = y^{4} \end{aligned}

$\begin{align}\frac{y^{9}}{y^{5}} &=\frac{y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y}{y\cdot y\cdot y\cdot y\cdot y} \\[1mm] &=\frac{\cancel{y}\cdot\cancel{y}\cdot\cancel{y}\cdot\cancel{y}\cdot\cancel{y}\cdot y\cdot y\cdot y\cdot y}{\cancel{y}\cdot\cancel{y}\cdot\cancel{y}\cdot\cancel{y}\cdot\cancel{y}} \\[1mm] & =\frac{y\cdot y\cdot y\cdot y}{1} \\[1mm] & =y^{4}\\ \text{ }\end{align}$

Remember that when we cancel

\frac{y}{y}

$\dfrac{y}{y}$ , we are simplifying by division and

\frac{y}{y} = 1

$\dfrac {y}{y}=1$ as long as

y \neq 0

$y\neq0$ .

Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.

\frac{a^{m}}{a^{n}} = a^{m - n}

$\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}$

In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.

The Quotient Rule for Exponents

For any real numbers $a, m$ and $n$ , provided $a\neq0$ , the quotient rule of exponents states that $\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}$

Example 2

Simplify the expressions.

$\dfrac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}$
$\dfrac{{t}^{23}}{{t}^{15}}$
$\dfrac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}$

Solution

Use the quotient rule to simplify each expression.

$\dfrac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}={\left(-2\right)}^{14 - 9}={\left(-2\right)}^{5}$
$\dfrac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}$
$\dfrac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}={\left(z\sqrt{2}\right)}^{5 - 1}={\left(z\sqrt{2}\right)}^{4}$

Try It 3

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

$\dfrac{{s}^{75}}{{s}^{68}}$
$\dfrac{{\left(-3\right)}^{6}}{-3}$
$\dfrac{{\left(e{f}^{2}\right)}^{5}}{{\left(e{f}^{2}\right)}^{3}}$

Show Answer

${s}^{7}$
${\left(-3\right)}^{5}$
${\left(e{f}^{2}\right)}^{2}$

TRY IT 4

TRY IT 5

Watch this video to see more examples of how to use the quotient rule for exponents.

The Power Rule for Exponents

Suppose an exponential expression is raised to some power. To simplify such an expression, we use the power rule of exponents. Consider the expression ${\left({x}^{2}\right)}^{3}$ . The expression 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{aligned} {(x^{2})}^{3} & = \overset{3 factors}{(x^{2}) \cdot (x^{2}) \cdot (x^{2})} \\ = \overset{3 factors}{\overset{⏞}{(\overset{2 factors}{\overset{⏞}{x \cdot x}}) \cdot (\overset{2 factors}{\overset{⏞}{x \cdot x}}) \cdot (\overset{2 factors}{\overset{⏞}{x \cdot x}})}} \\ = x \cdot x \cdot x \cdot x \cdot x \cdot x \\ = x^{6} \end{aligned}

$\begin{align} {\left({x}^{2}\right)}^{3}& = \stackrel{{3\text{ factors}}}{{{\left({x}^{2}\right)\cdot \left({x}^{2}\right)\cdot \left({x}^{2}\right)}}} \\ & = \stackrel{{3\text{ factors}}}{\overbrace{{\left(\stackrel{{2\text{ factors}}}{{\overbrace{x\cdot x}}}\right)\cdot \left(\stackrel{{2\text{ factors}}}{{\overbrace{x\cdot x}}}\right)\cdot \left(\stackrel{{2\text{ factors}}}{{\overbrace{x\cdot x}}}\right)}}}\\ & = x\cdot x\cdot x\cdot x\cdot x\cdot x\hfill \\ & = {x}^{6} \end{align}$

The exponent of the answer is the product of the exponents: ${\left({x}^{2}\right)}^{3}={x}^{2\cdot 3}={x}^{6}$ because there are three groups of two $x$ s multiplied together. Therefore, the total number of $x$ s multiplied together is $3 \times 2 = 6$ . In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents.

{(a^{m})}^{n} = a^{m \cdot n}

${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$

The Power Rule of Exponents

For any real number $a$ and positive integers $m$ and $n$ , the power rule of exponents states that

{(a^{m})}^{n} = a^{m \cdot n}

${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$

Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, we multiply the exponents.

Product Rule					Power Rule
$5^{3}\cdot5^{4}$	=	$5^{3+4}$	=	$5^{7}$	but	$\left(5^{3}\right)^{4}$	=	$5^{3\cdot4}$	=	$5^{12}$
$x^{5}\cdot x^{2}$	=	$x^{5+2}$	=	$x^{7}$	but	$\left(x^{5}\right)^{2}$	=	$x^{5\cdot2}$	=	$x^{10}$
$\left(3a\right)^{7}\cdot\left(3a\right)^{10}$	=	$\left(3a\right)^{7+10}$	=	$\left(3a\right)^{17}$	but	$\left(\left(3a\right)^{7}\right)^{10}$	=	$\left(3a\right)^{7\cdot10}$	=	$\left(3a\right)^{70}$

Example 3

Simplify the expressions.

${\left({x}^{2}\right)}^{7}$
${\left({\left(2t\right)}^{5}\right)}^{3}$
${\left({\left(-3\right)}^{5}\right)}^{11}$

Solution

We can apply the power rule to simplify each expression.

${\left({x}^{2}\right)}^{7}={x}^{2\cdot 7}={x}^{14}$
${\left({\left(2t\right)}^{5}\right)}^{3}={\left(2t\right)}^{5\cdot 3}={\left(2t\right)}^{15}$
${\left({\left(-3\right)}^{5}\right)}^{11}={\left(-3\right)}^{5\cdot 11}={\left(-3\right)}^{55}$

Try It 6

Simplify the expressions.

${\left({\left(3y\right)}^{8}\right)}^{3}$
${\left({t}^{5}\right)}^{7}$
${\left({\left(-g\right)}^{4}\right)}^{4}$

Show Answer

${\left(3y\right)}^{24}$
${t}^{35}$
${\left(-g\right)}^{16}$

TRY IT 7

TRY IT 8

The following video shows more examples of using the power rule to simplify expressions with exponents.

Chapter 5: Exponential Functions