Find the Power of a Product and a Quotient

Learning Outcomes

Simplify compound expressions using the exponent rules

Finding the Power of a Product

To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider ${\left(pq\right)}^{3}$ . We begin by using the associative and commutative properties of multiplication to regroup the factors.

\begin{array}{ccc} {(p q)}^{3} & = & \overset{3 factors}{(p q) \cdot (p q) \cdot (p q)} \\ = & p \cdot q \cdot p \cdot q \cdot p \cdot q \\ = & \overset{3 factors}{p \cdot p \cdot p} \cdot \overset{3 factors}{q \cdot q \cdot q} \\ = & p^{3} \cdot q^{3} \end{array}

$\begin{array}{ccc}\hfill {\left(pq\right)}^{3}& =& \stackrel{3\text{ factors}}{{\left(pq\right)\cdot \left(pq\right)\cdot \left(pq\right)}}\hfill \\ & =& p\cdot q\cdot p\cdot q\cdot p\cdot q\hfill \\ & =& \stackrel{3\text{ factors}}{{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{{q\cdot q\cdot q}}\hfill \\ & =& {p}^{3}\cdot {q}^{3}\hfill \end{array}$

In other words, ${\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}$ .

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

{(a b)}^{n} = a^{n} b^{n}

${\left(ab\right)}^{n}={a}^{n}{b}^{n}$

Example

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

${\left(a{b}^{2}\right)}^{3}$
${\left(2^a{t}\right)}^{15}$
${\left(-2{w}^{3}\right)}^{3}$
$\frac{1}{{\left(-7z\right)}^{4}}$
${\left({e}^{-2}{f}^{2}\right)}^{7}$

Show Solution

Use the product and quotient rules and the new definitions to simplify each expression.

${\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}$
${\left(2^a{t}\right)}^{15}={\left(2^a\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{a\cdot15}\cdot{t}^{15}=2^{15a}\cdot{t}^{15}$
${\left(-2{w}^{3}\right)}^{3}={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=-8\cdot {w}^{3\cdot 3}=-8{w}^{9}$
$\frac{1}{{\left(-7z\right)}^{4}}=\frac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{1}{2,401{z}^{4}}$
${\left({e}^{-2}{f}^{2}\right)}^{7}={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}={e}^{-14}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}$

Caution! Do not try to apply this rule to sums.

Think about the expression $\left(2+3\right)^{2}$

Does $\left(2+3\right)^{2}$ equal $2^{2}+3^{2}$ ?

No, it does not because of the order of operations!

$\left(2+3\right)^{2}=5^{2}=25$

and

$2^{2}+3^{2}=4+9=13$

Therefore, you can only use this rule when the numbers inside the parentheses are being multiplied or divided.

In the following video, we provide more examples of how to find the power of a product.

Finding the Power of a Quotient

To simplify the power of a quotient of two expressions, we can use the power of a quotient rule which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, look at the following example:

{(e^{- 2} f^{2})}^{7} = \frac{f^{14}}{e^{14}}

${\left({e}^{-2}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}$

Rewrite the original problem differently and look at the result:

\begin{array}{ccc} {(e^{- 2} f^{2})}^{7} & = & {(\frac{f^{2}}{e^{2}})}^{7} \\ = & \frac{f^{14}}{e^{14}} \end{array}

$\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}$

It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.

\begin{array}{ccc} {(e^{- 2} f^{2})}^{7} & = & {(\frac{f^{2}}{e^{2}})}^{7} \\ = & \frac{{(f^{2})}^{7}}{{(e^{2})}^{7}} \\ = & \frac{f^{2 \cdot 7}}{e^{2 \cdot 7}} \\ = & \frac{f^{14}}{e^{14}} \end{array}

$\normalsize\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}}\hfill \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}$

The Power of a Quotient Rule of Exponents

For any real numbers $a$ and $b$ and any integer $n$ , the power of a quotient rule of exponents states that

{(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}

${\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}$

Example

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

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

Show Solution

$\Large{\left(\frac{4}{{z}^{11}}\right)}^{3}=\frac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\frac{64}{{z}^{11\cdot 3}}=\frac{64}{{z}^{33}}$
$\Large{\left(\frac{p}{{q}^{3}}\right)}^{6}=\frac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\frac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\frac{{p}^{6}}{{q}^{18}}$
$\Large{\left(\frac{-1}{{t}^{2}}\right)}^{27}=\frac{{\left(-1\right)}^{27}}{{\left({t}^{2}\right)}^{27}}=\frac{-1}{{t}^{2\cdot 27}}=\frac{-1}{{t}^{54}}=-\frac{1}{{t}^{54}}$
$\Large{\left({j}^{3}{k}^{-2}\right)}^{4}={\left(\frac{{j}^{3}}{{k}^{2}}\right)}^{4}=\frac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\frac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\frac{{j}^{12}}{{k}^{8}}$
$\Large{\left({m}^{-2}{n}^{-2}\right)}^{3}={\left(\frac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\frac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\frac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\frac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\frac{1}{{m}^{6}{n}^{6}}$

The following video provides more examples of simplifying expressions using the power of a quotient and other exponent rules.

Summary

Evaluating expressions containing exponents is the same as evaluating any expression. You substitute the value of the variable into the expression and simplify.
The product rule for exponents: For any number x and any integers a and b, $\left(x^{a}\right)\left(x^{b}\right) = x^{a+b}$ .
The quotient rule for exponents: For any non-zero number x and any integers a and b, $\displaystyle \frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}$
The power rule for exponents:
1. For any nonzero numbers a and b and any integer n, $\left(ab\right)^{n}=a^{n}\cdot{b^{n}}$ .
2. For any number a, any non-zero number b, and any integer n, $\displaystyle {\left(\frac{a}{b}\right)}^{n}=\frac{a^{n}}{b^{n}}$

Module 7: Exponents