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
(pq)3{\left(pq\right)}^{3}
. We begin by using the associative and commutative properties of multiplication to regroup the factors.

(pq)3=(pq)(pq)(pq)3 factors=pqpqpq=ppp3 factorsqqq3 factors=p3q3\begin{array}{ccc}\qquad {\left(pq\right)}^{3}& =& \stackrel{3\text{ factors}}{{\left(pq\right)\cdot \left(pq\right)\cdot \left(pq\right)}}\qquad \\ & =& p\cdot q\cdot p\cdot q\cdot p\cdot q\qquad \\ & =& \stackrel{3\text{ factors}}{{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{{q\cdot q\cdot q}}\qquad \\ & =& {p}^{3}\cdot {q}^{3}\qquad \end{array}
In other words,
(pq)3=p3q3{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}
.

A General Note: The Power of a Product Rule of Exponents

For any real numbers
aa
and
bb
and any integer
nn
, the power of a product rule of exponents states that

(ab)n=anbn{\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.

  1. (ab2)3{\left(a{b}^{2}\right)}^{3}
  2. (2t)15{\left(2t\right)}^{15}
  3. (2w3)3{\left(-2{w}^{3}\right)}^{3}
  4. 1(7z)4\frac{1}{{\left(-7z\right)}^{4}}
  5. (e2f2)7{\left({e}^{-2}{f}^{2}\right)}^{7}


Solution

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

  1. (ab2)3=(a)3(b2)3=a13b23=a3b6{\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. 2t15=(2)15(t)15=215t15=32,768t152{t}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}
  3. (2w3)3=(2)3(w3)3=8w33=8w9{\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}
  4. 1(7z)4=1(7)4(z)4=12,401z4\frac{1}{{\left(-7z\right)}^{4}}=\frac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{1}{2,401{z}^{4}}
  5. (e2f2)7=(e2)7(f2)7=e27f27=e14f14=f14e14{\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.

(g2h3)5{\left({g}^{2}{h}^{3}\right)}^{5}


b.
(5t)3{\left(5t\right)}^{3}


c.
(3y5)3{\left(-3{y}^{5}\right)}^{3}


d.
1(a6b7)3\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}


e.
(r3s2)4{\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.

(e2f2)7=f14e14{\left({e}^{-2}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}
Let’s rewrite the original problem differently and look at the result.

(e2f2)7=(f2e2)7=f14e14\begin{array}{ccc}\qquad {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\qquad \\ & =& \frac{{f}^{14}}{{e}^{14}}\qquad \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.

(e2f2)7=(f2e2)7=(f2)7(e2)7=f27e27=f14e14\begin{array}{ccc}\qquad {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\qquad \\ & =& \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}}\qquad \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\qquad \\ & =& \frac{{f}^{14}}{{e}^{14}}\qquad \end{array}

A General Note: The Power of a Quotient Rule of Exponents

For any real numbers
aa
and
bb
and any integer
nn
, the power of a quotient rule of exponents states that

(ab)n=anbn{\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.

  1. (4z11)3{\left(\frac{4}{{z}^{11}}\right)}^{3}
  2. (pq3)6{\left(\frac{p}{{q}^{3}}\right)}^{6}
  3. (1t2)27{\left(\frac{-1}{{t}^{2}}\right)}^{27}
  4. (j3k2)4{\left({j}^{3}{k}^{-2}\right)}^{4}
  5. (m2n2)3{\left({m}^{-2}{n}^{-2}\right)}^{3}


Solution

  1. (4z11)3=(4)3(z11)3=64z113=64z33{\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}}
  2. (pq3)6=(p)6(q3)6=p16q36=p6q18{\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}}
  3. left(1t2right)27=left(1right)27left(t2right)27=1t227=1t54=1t54{\\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}}
  4. (j3k2)4=(j3k2)4=(j3)4(k2)4=j34k24=j12k8{\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}}
  5. (m2n2)3=(1m2n2)3=(1)3(m2n2)3=1(m2)3(n2)3=1m23n23=1m6n6{\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.

(b5c)3{\left(\frac{{b}^{5}}{c}\right)}^{3}


b.
(5u8)4{\left(\frac{5}{{u}^{8}}\right)}^{4}


c.
(1w3)35{\left(\frac{-1}{{w}^{3}}\right)}^{35}


d.
(p4q3)8{\left({p}^{-4}{q}^{3}\right)}^{8}


e.
(c5d3)4{\left({c}^{-5}{d}^{-3}\right)}^{4}

Solution

Licenses and Attributions