Learning Outcomes
Simplify expressions with negative exponents.
Simplify exponential expressions.
Using the Negative Rule of Exponents
Another useful result occurs if we relax the condition that [latex]m>n[/latex] in the quotient rule even further. For example, can we simplify [latex]\dfrac{{h}^{3}}{{h}^{5}}[/latex]? When [latex]mnegative rule of exponents to simplify the expression to its reciprocal.
Divide one exponential expression by another with a larger exponent. Use our example, [latex]\dfrac{{h}^{3}}{{h}^{5}}[/latex].
[latex]\begin{align} \frac{{h}^{3}}{{h}^{5}}& = \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h} \\ & = \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h} \\ & = \frac{1}{h\cdot h} \\ & = \frac{1}{{h}^{2}} \end{align}[/latex]
If we were to simplify the original expression using the quotient rule, we would have
[latex]\begin{align} \frac{{h}^{3}}{{h}^{5}}& = {h}^{3 - 5} \\ & = {h}^{-2} \end{align}[/latex]
Putting the answers together, we have [latex]{h}^{-2}=\dfrac{1}{{h}^{2}}[/latex]. This is true for any nonzero real number, or any variable representing a nonzero real number.
A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.
[latex]{a}^{-n}=\dfrac{1}{{a}^{n}} \text{ and } {a}^{n}=\dfrac{1}{{a}^{-n}}[/latex]
We have shown that the exponential expression [latex]{a}^{n}[/latex] is defined when [latex]n[/latex] is a natural number, 0, or the negative of a natural number. That means that [latex]{a}^{n}[/latex] is defined for any integer [latex]n[/latex]. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer [latex]n[/latex].
A General Note: The Negative Rule of Exponents
For any nonzero real number [latex]a[/latex] and natural number [latex]n[/latex], the negative rule of exponents states that
[latex]{a}^{-n}=\dfrac{1}{{a}^{n}} \text{ and } {a}^{n}=\dfrac{1}{{a}^{-n}}[/latex]
Example: Using the Negative Exponent Rule
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
[latex]\dfrac{{\theta }^{3}}{{\theta }^{10}}[/latex]
[latex]\dfrac{{z}^{2}\cdot z}{{z}^{4}}[/latex]
[latex]\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}[/latex]
Show Solution
[latex]\dfrac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\dfrac{1}{{\theta }^{7}}[/latex]
[latex]\dfrac{{z}^{2}\cdot z}{{z}^{4}}=\dfrac{{z}^{2+1}}{{z}^{4}}=\dfrac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\dfrac{1}{z}[/latex]
[latex]\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4 - 8}={\left(-5{t}^{3}\right)}^{-4}=\dfrac{1}{{\left(-5{t}^{3}\right)}^{4}}[/latex]
Try It
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
[latex]\dfrac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}[/latex]
[latex]\dfrac{{f}^{47}}{{f}^{49}\cdot f}[/latex]
[latex]\dfrac{2{k}^{4}}{5{k}^{7}}[/latex]
Show Solution
[latex]\dfrac{1}{{\left(-3t\right)}^{6}}[/latex]
[latex]\dfrac{1}{{f}^{3}}[/latex]
[latex]\dfrac{2}{5{k}^{3}}[/latex]
Watch this video to see more examples of simplifying expressions with negative exponents.VIDEO
Example: Using the Product and Quotient Rules
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
[latex]{b}^{2}\cdot {b}^{-8}[/latex]
[latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}[/latex]
[latex]\dfrac{-7z}{{\left(-7z\right)}^{5}}[/latex]
Show Solution
[latex]{b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}[/latex]
[latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1[/latex]
[latex]\dfrac{-7z}{{\left(-7z\right)}^{5}}=\dfrac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}={\left(-7z\right)}^{1 - 5}={\left(-7z\right)}^{-4}=\dfrac{1}{{\left(-7z\right)}^{4}}[/latex]
Try It
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
[latex]{t}^{-11}\cdot {t}^{6}[/latex]
[latex]\dfrac{{25}^{12}}{{25}^{13}}[/latex]
Show Solution
[latex]{t}^{-5}=\dfrac{1}{{t}^{5}}[/latex]
[latex]\dfrac{1}{25}[/latex]
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 [latex]{\left(pq\right)}^{3}[/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors.
[latex]\begin{align} {\left(pq\right)}^{3}& = \stackrel{3\text{ factors}}{{\left(pq\right)\cdot \left(pq\right)\cdot \left(pq\right)}} \\ & = p\cdot q\cdot p\cdot q\cdot p\cdot q \\ & = \stackrel{3\text{ factors}}{{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{{q\cdot q\cdot q}} \\ & = {p}^{3}\cdot {q}^{3} \end{align}[/latex]
In other words, [latex]{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}[/latex].
A General Note: The Power of a Product Rule of Exponents
For any real numbers [latex]a[/latex] and [latex]b[/latex] and any integer [latex]n[/latex], the power of a product rule of exponents states that
[latex]\large{\left(ab\right)}^{n}={a}^{n}{b}^{n}[/latex]
Example: 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.
[latex]{\left(a{b}^{2}\right)}^{3}[/latex]
[latex]{\left(2t\right)}^{15}[/latex]
[latex]{\left(-2{w}^{3}\right)}^{3}[/latex]
[latex]\dfrac{1}{{\left(-7z\right)}^{4}}[/latex]
[latex]{\left({e}^{-2}{f}^{2}\right)}^{7}[/latex]
Show Solution
Use the product and quotient rules and the new definitions to simplify each expression.
[latex]{\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}[/latex]
[latex]2{t}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}[/latex]
[latex]{\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}[/latex]
[latex]\dfrac{1}{{\left(-7z\right)}^{4}}=\dfrac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\dfrac{1}{2,401{z}^{4}}[/latex]
[latex]{\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}=\dfrac{{f}^{14}}{{e}^{14}}[/latex]
Try It
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
[latex]{\left({g}^{2}{h}^{3}\right)}^{5}[/latex]
[latex]{\left(5t\right)}^{3}[/latex]
[latex]{\left(-3{y}^{5}\right)}^{3}[/latex]
[latex]\dfrac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}[/latex]
[latex]{\left({r}^{3}{s}^{-2}\right)}^{4}[/latex]
Show Solution
[latex]{g}^{10}{h}^{15}[/latex]
[latex]125{t}^{3}[/latex]
[latex]-27{y}^{15}[/latex]
[latex]\dfrac{1}{{a}^{18}{b}^{21}}[/latex]
[latex]\dfrac{{r}^{12}}{{s}^{8}}[/latex]
In the following video we show more examples of how to find hte power of a product.
VIDEO
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.
[latex]{\left({e}^{-2}{f}^{2}\right)}^{7}=\dfrac{{f}^{14}}{{e}^{14}}[/latex]
Let’s rewrite the original problem differently and look at the result.
[latex]\begin{align} {\left({e}^{-2}{f}^{2}\right)}^{7}& = {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7} \\[1mm] & = \frac{{f}^{14}}{{e}^{14}} \\ \text{ } \end{align}[/latex]
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.
[latex]\begin{align} {\left({e}^{-2}{f}^{2}\right)}^{7}& = {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7} \\[1mm] & = \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}} \\[1mm] & = \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}} \\[1mm] & = \frac{{f}^{14}}{{e}^{14}} \\ \text{ } \end{align}[/latex]
A General Note: The Power of a Quotient Rule of Exponents
For any real numbers [latex]a[/latex] and [latex]b[/latex] and any integer [latex]n[/latex], the power of a quotient rule of exponents states that
[latex]\large{\left(\dfrac{a}{b}\right)}^{n}=\dfrac{{a}^{n}}{{b}^{n}}[/latex]
Example: 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.
[latex]{\left(\dfrac{4}{{z}^{11}}\right)}^{3}[/latex]
[latex]{\left(\dfrac{p}{{q}^{3}}\right)}^{6}[/latex]
[latex]{\left(\dfrac{-1}{{t}^{2}}\right)}^{27}[/latex]
[latex]{\left({j}^{3}{k}^{-2}\right)}^{4}[/latex]
[latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}[/latex]
Show Solution
[latex]{\left(\dfrac{4}{{z}^{11}}\right)}^{3}=\dfrac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\dfrac{64}{{z}^{11\cdot 3}}=\dfrac{64}{{z}^{33}}[/latex]
[latex]{\left(\dfrac{p}{{q}^{3}}\right)}^{6}=\dfrac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\dfrac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\dfrac{{p}^{6}}{{q}^{18}}[/latex]
[latex]{\left(\dfrac{-1}{{t}^{2}}\right)}^{27}=\dfrac{{\left(-1\right)}^{27}}{{\left({t}^{2}\right)}^{27}}=\dfrac{-1}{{t}^{2\cdot 27}}=\dfrac{-1}{{t}^{54}}=-\dfrac{1}{{t}^{54}}[/latex]
[latex]{\left({j}^{3}{k}^{-2}\right)}^{4}={\left(\dfrac{{j}^{3}}{{k}^{2}}\right)}^{4}=\dfrac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\dfrac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\dfrac{{j}^{12}}{{k}^{8}}[/latex]
[latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}={\left(\dfrac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\dfrac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\dfrac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\dfrac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\dfrac{1}{{m}^{6}{n}^{6}}[/latex]
Try It
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
[latex]{\left(\dfrac{{b}^{5}}{c}\right)}^{3}[/latex]
[latex]{\left(\dfrac{5}{{u}^{8}}\right)}^{4}[/latex]
[latex]{\left(\dfrac{-1}{{w}^{3}}\right)}^{35}[/latex]
[latex]{\left({p}^{-4}{q}^{3}\right)}^{8}[/latex]
[latex]{\left({c}^{-5}{d}^{-3}\right)}^{4}[/latex]
Show Solution
[latex]\dfrac{{b}^{15}}{{c}^{3}}[/latex]
[latex]\dfrac{625}{{u}^{32}}[/latex]
[latex]\dfrac{-1}{{w}^{105}}[/latex]
[latex]\dfrac{{q}^{24}}{{p}^{32}}[/latex]
[latex]\dfrac{1}{{c}^{20}{d}^{12}}[/latex]
Simplifying Exponential Expressions
Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.
Example: Simplifying Exponential Expressions
Simplify each expression and write the answer with positive exponents only.
[latex]{\left(6{m}^{2}{n}^{-1}\right)}^{3}[/latex]
[latex]{17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}[/latex]
[latex]{\left(\dfrac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}[/latex]
[latex]\left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)[/latex]
[latex]{\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}[/latex]
[latex]\dfrac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}[/latex]
Show Solution
[latex]\begin{align} {\left(6{m}^{2}{n}^{-1}\right)}^{3}& = {\left(6\right)}^{3}{\left({m}^{2}\right)}^{3}{\left({n}^{-1}\right)}^{3}&& \text{The power of a product rule} \\ & = {6}^{3}{m}^{2\cdot 3}{n}^{-1\cdot 3}&& \text{The power rule} \\ & = 216{m}^{6}{n}^{-3}&& \text{Simplify}. \\ & = \frac{216{m}^{6}}{{n}^{3}}&& \text{The negative exponent rule} \end{align}[/latex]
[latex]\begin{align} {17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}& =& {17}^{5 - 4-3}&& \text{The product rule} \\ & = {17}^{-2}&& \text{Simplify}. \\ & = \frac{1}{{17}^{2}}\text{ or }\frac{1}{289}&& \text{The negative exponent rule} \end{align}[/latex]
[latex]\begin{align} {\left(\frac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}& = \frac{{\left({u}^{-1}v\right)}^{2}}{{\left({v}^{-1}\right)}^{2}}&& \text{The power of a quotient rule} \\ & = \frac{{u}^{-2}{v}^{2}}{{v}^{-2}}&& \text{The power of a product rule} \\ & = {u}^{-2}{v}^{2-\left(-2\right)}&& \text{The quotient rule} \\ & = {u}^{-2}{v}^{4}&& \text{Simplify}. \\ & = \frac{{v}^{4}}{{u}^{2}}&& \text{The negative exponent rule} \end{align}[/latex]
[latex]\begin{align} \left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)& =& -2\cdot 5\cdot {a}^{3}\cdot {a}^{-2}\cdot {b}^{-1}\cdot {b}^{2}&& \text{Commutative and associative laws of multiplication} \\ & = -10\cdot {a}^{3 - 2}\cdot {b}^{-1+2}&& \text{The product rule} \\ & = -10ab&& \text{Simplify}. \end{align}[/latex]
[latex]\begin{align} {\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}& = {\left({x}^{2}\sqrt{2}\right)}^{4 - 4} && \text{The product rule} \\ & = {\left({x}^{2}\sqrt{2}\right)}^{0}&& \text{Simplify}. \\ & = 1&& \text{The zero exponent rule} \end{align}[/latex]
[latex]\begin{align} \frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}& = \frac{{\left(3\right)}^{5}\cdot {\left({w}^{2}\right)}^{5}}{{\left(6\right)}^{2}\cdot {\left({w}^{-2}\right)}^{2}}&& \text{The power of a product rule} \\ & = \frac{{3}^{5}{w}^{2\cdot 5}}{{6}^{2}{w}^{-2\cdot 2}}&& \text{The power rule} \\ & = \frac{243{w}^{10}}{36{w}^{-4}} && \text{Simplify}. \\ & = \frac{27{w}^{10-\left(-4\right)}}{4}&& \text{The quotient rule and reduce fraction} \\ & = \frac{27{w}^{14}}{4}&& \text{Simplify}. \end{align}[/latex]
Try It
Simplify each expression and write the answer with positive exponents only.
[latex]{\left(2u{v}^{-2}\right)}^{-3}[/latex]
[latex]{x}^{8}\cdot {x}^{-12}\cdot x[/latex]
[latex]{\left(\frac{{e}^{2}{f}^{-3}}{{f}^{-1}}\right)}^{2}[/latex]
[latex]\left(9{r}^{-5}{s}^{3}\right)\left(3{r}^{6}{s}^{-4}\right)[/latex]
[latex]{\left(\frac{4}{9}t{w}^{-2}\right)}^{-3}{\left(\frac{4}{9}t{w}^{-2}\right)}^{3}[/latex]
[latex]\dfrac{{\left(2{h}^{2}k\right)}^{4}}{{\left(7{h}^{-1}{k}^{2}\right)}^{2}}[/latex]
Show Solution
[latex]\dfrac{{v}^{6}}{8{u}^{3}}[/latex]
[latex]\dfrac{1}{{x}^{3}}[/latex]
[latex]\dfrac{{e}^{4}}{{f}^{4}}[/latex]
[latex]\dfrac{27r}{s}[/latex]
[latex]1[/latex]
[latex]\dfrac{16{h}^{10}}{49}[/latex]
In the following video we show more examples of how to find the power of a quotient.VIDEO