Finding the Power of a Product and a Quotient

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

A General Note: 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 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.

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

Solution

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

${\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}$
$2{t}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}$
${\left(-2{w}^{3}\right)}^{3}={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=-8\cdot {w}^{3\cdot 3}=-8{w}^{9}$
$\frac{1}{{\left(-7z\right)}^{4}}=\frac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{1}{2,401{z}^{4}}$
${\left({e}^{-2}{f}^{2}\right)}^{7}={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}={e}^{-14}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}$

Try It 7

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

a. ${\left({g}^{2}{h}^{3}\right)}^{5}$
b. ${\left(5t\right)}^{3}$
c. ${\left(-3{y}^{5}\right)}^{3}$
d. $\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}$
e. ${\left({r}^{3}{s}^{-2}\right)}^{4}$

Solution

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, let’s look at the following example.

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

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

Let’s 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}

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

A General Note: 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 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.

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

Solution

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

Try It 8

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

a. ${\left(\frac{{b}^{5}}{c}\right)}^{3}$
b. ${\left(\frac{5}{{u}^{8}}\right)}^{4}$
c. ${\left(\frac{-1}{{w}^{3}}\right)}^{35}$
d. ${\left({p}^{-4}{q}^{3}\right)}^{8}$
e. ${\left({c}^{-5}{d}^{-3}\right)}^{4}$

Solution

Exponents and Scientific Notation