# 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.

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

### 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

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

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

### Solution

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

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

### 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.

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

$\begin{array}{ccc}\qquad {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\qquad \\ & =& \frac{{f}^{14}}{{e}^{14}}\qquad \end{array}$
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.

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

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

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

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

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

### Solution

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

### 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
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