We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > College Algebra CoRequisite Course

Zero and Negative Exponents

Learning Outcomes

  • Simplify expressions with exponents equal to zero.
  • Simplify expressions with negative exponents.
  • Simplify exponential expressions.
Return to the quotient rule. We made the condition that m>nm>n so that the difference m−nm-n would never be zero or negative. What would happen if m=nm=n? In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.

t8t8=t8t8=1\dfrac{t^{8}}{t^{8}}=\dfrac{\cancel{t^{8}}}{\cancel{t^{8}}}=1

If we were to simplify the original expression using the quotient rule, we would have
t8t8=t8−8=t0\dfrac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}
If we equate the two answers, the result is t0=1{t}^{0}=1. This is true for any nonzero real number, or any variable representing a real number.
a0=1{a}^{0}=1
The sole exception is the expression 00{0}^{0}. This appears later in more advanced courses, but for now, we will consider the value to be undefined.

A General Note: The Zero Exponent Rule of Exponents

For any nonzero real number aa, the zero exponent rule of exponents states that
a0=1{a}^{0}=1
 

using order of operations with fractions

When simplifying expressions with exponents, it is sometimes helpful to rely on the rule for multiplying fractions to separate the factors before doing work on them. For example, to simplify the expression 5amz2amz\dfrac{5 a^m z^2}{a^mz} using exponent rules, you may find it helpful to break the fraction up into a product of fractions, then simplify.

5amz2amz=5⋅amam⋅z2z=5⋅am−m⋅z2−1=5⋅a0⋅z1=5z\dfrac{5 a^m z^2}{a^mz}\quad=\quad 5\cdot\dfrac{a^m}{a^m}\cdot\dfrac{z^2}{z} \quad=\quad 5 \cdot a^{m-m}\cdot z^{2-1}\quad=\quad 5\cdot a^0 \cdot z^1 \quad=\quad 5z

 

Example: Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.
  1. c3c3\dfrac{{c}^{3}}{{c}^{3}}
  2. −3x5x5\dfrac{-3{x}^{5}}{{x}^{5}}
  3. (j2k)4(j2k)⋅(j2k)3\dfrac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}
  4. 5(rs2)2(rs2)2\dfrac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}

Answer: Use the zero exponent and other rules to simplify each expression.

  1. \begin{align}\frac{c^{3}}{c^{3}} & =c^{3-3} \\ & =c^{0} \\ & =1\end{align}
  2. \begin{align} \frac{-3{x}^{5}}{{x}^{5}}& = -3\cdot \frac{{x}^{5}}{{x}^{5}} \\ & = -3\cdot {x}^{5 - 5} \\ & = -3\cdot {x}^{0} \\ & = -3\cdot 1 \\ & = -3 \end{align}
  3. \begin{align} \frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}& = \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{1+3}} && \text{Use the product rule in the denominator}. \\ & = \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}} && \text{Simplify}. \\ & = {\left({j}^{2}k\right)}^{4 - 4} && \text{Use the quotient rule}. \\ & = {\left({j}^{2}k\right)}^{0} && \text{Simplify}. \\ & = 1 \end{align}
  4. \begin{align} \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& = 5{\left(r{s}^{2}\right)}^{2 - 2} && \text{Use the quotient rule}. \\ & = 5{\left(r{s}^{2}\right)}^{0} && \text{Simplify}. \\ & = 5\cdot 1 && \text{Use the zero exponent rule}. \\ & = 5 && \text{Simplify}. \end{align}

Try It

Simplify each expression using the zero exponent rule of exponents.
  1. t7t7\dfrac{{t}^{7}}{{t}^{7}}
  2. (de2)112(de2)11\dfrac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}
  3. w4⋅w2w6\dfrac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}
  4. t3⋅t4t2⋅t5\dfrac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}

Answer:

  1. 11
  2. 12\dfrac{1}{2}
  3. 11
  4. 11

[embed] [embed]
In this video we show more examples of how to simplify expressions with zero exponents. https://youtu.be/rpoUg32utlc

Using the Negative Rule of Exponents

Another useful result occurs if we relax the condition that m>nm>n in the quotient rule even further. For example, can we simplify h3h5\dfrac{{h}^{3}}{{h}^{5}}? When m<nm<n—that is, where the difference m−nm-n is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, h3h5\dfrac{{h}^{3}}{{h}^{5}}.
\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}
If we were to simplify the original expression using the quotient rule, we would have
\begin{align} \frac{{h}^{3}}{{h}^{5}}& = {h}^{3 - 5} \\ & = {h}^{-2} \end{align}
Putting the answers together, we have h−2=1h2{h}^{-2}=\dfrac{1}{{h}^{2}}. 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.
a−n=1an and an=1a−n{a}^{-n}=\dfrac{1}{{a}^{n}} \text{ and } {a}^{n}=\dfrac{1}{{a}^{-n}}
We have shown that the exponential expression an{a}^{n} is defined when nn is a natural number, 0, or the negative of a natural number. That means that an{a}^{n} is defined for any integer nn. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer nn.

A General Note: The Negative Rule of Exponents

For any nonzero real number aa and natural number nn, the negative rule of exponents states that
a−n=1an and an=1a−n{a}^{-n}=\dfrac{1}{{a}^{n}} \text{ and } {a}^{n}=\dfrac{1}{{a}^{-n}}

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.
  1. θ3θ10\dfrac{{\theta }^{3}}{{\theta }^{10}}
  2. z2⋅zz4\dfrac{{z}^{2}\cdot z}{{z}^{4}}
  3. (−5t3)4(−5t3)8\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}

Answer:

  1. θ3θ10=θ3−10=θ−7=1θ7\dfrac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\dfrac{1}{{\theta }^{7}}
  2. z2⋅zz4=z2+1z4=z3z4=z3−4=z−1=1z\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}
  3. (−5t3)4(−5t3)8=(−5t3)4−8=(−5t3)−4=1(−5t3)4\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}}

Try It

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
  1. (−3t)2(−3t)8\dfrac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}
  2. f47f49⋅f\dfrac{{f}^{47}}{{f}^{49}\cdot f}
  3. 2k45k7\dfrac{2{k}^{4}}{5{k}^{7}}

Answer:

  1. 1(−3t)6\dfrac{1}{{\left(-3t\right)}^{6}}
  2. 1f3\dfrac{1}{{f}^{3}}
  3. 25k3\dfrac{2}{5{k}^{3}}

[embed] [embed] [embed]
Watch this video to see more examples of simplifying expressions with negative exponents. https://youtu.be/Gssi4dBtAEI

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.
  1. b2⋅b−8{b}^{2}\cdot {b}^{-8}
  2. (−x)5⋅(−x)−5{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}
  3. −7z(−7z)5\dfrac{-7z}{{\left(-7z\right)}^{5}}

Answer:

  1. b2⋅b−8=b2−8=b−6=1b6{b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}
  2. (−x)5⋅(−x)−5=(−x)5−5=(−x)0=1{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1
  3. −7z(−7z)5=(−7z)1(−7z)5=(−7z)1−5=(−7z)−4=1(−7z)4\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}}

Try It

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
  1. t−11⋅t6{t}^{-11}\cdot {t}^{6}
  2. 25122513\dfrac{{25}^{12}}{{25}^{13}}

Answer:

  1. t−5=1t5{t}^{-5}=\dfrac{1}{{t}^{5}}
  2. 125\dfrac{1}{25}

[embed]

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.
\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}
In other words, (pq)3=p3⋅q3{\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\large{\left(ab\right)}^{n}={a}^{n}{b}^{n}

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.
  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\dfrac{1}{{\left(-7z\right)}^{4}}
  5. (e−2f2)7{\left({e}^{-2}{f}^{2}\right)}^{7}

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

  1. (ab2)3=(a)3⋅(b2)3=a1⋅3⋅b2⋅3=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=−8⋅w3⋅3=−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\dfrac{1}{{\left(-7z\right)}^{4}}=\dfrac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\dfrac{1}{2,401{z}^{4}}
  5. (e−2f2)7=(e−2)7⋅(f2)7=e−2⋅7⋅f2⋅7=e−14f14=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}=\dfrac{{f}^{14}}{{e}^{14}}

Try It

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
  1. (g2h3)5{\left({g}^{2}{h}^{3}\right)}^{5}
  2. (5t)3{\left(5t\right)}^{3}
  3. (−3y5)3{\left(-3{y}^{5}\right)}^{3}
  4. 1(a6b7)3\dfrac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}
  5. (r3s−2)4{\left({r}^{3}{s}^{-2}\right)}^{4}

Answer:

  1. g10h15{g}^{10}{h}^{15}
  2. 125t3125{t}^{3}
  3. −27y15-27{y}^{15}
  4. 1a18b21\dfrac{1}{{a}^{18}{b}^{21}}
  5. r12s8\dfrac{{r}^{12}}{{s}^{8}}

[embed] [embed] [embed]
In the following video we show more examples of how to find the power of a product. https://youtu.be/p-2UkpJQWpo

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−2f2)7=f14e14{\left({e}^{-2}{f}^{2}\right)}^{7}=\dfrac{{f}^{14}}{{e}^{14}}
Let’s rewrite the original problem differently and look at the result.
\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}
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{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}

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\large{\left(\dfrac{a}{b}\right)}^{n}=\dfrac{{a}^{n}}{{b}^{n}}

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.
  1. (4z11)3{\left(\dfrac{4}{{z}^{11}}\right)}^{3}
  2. (pq3)6{\left(\dfrac{p}{{q}^{3}}\right)}^{6}
  3. (−1t2)27{\left(\dfrac{-1}{{t}^{2}}\right)}^{27}
  4. (j3k−2)4{\left({j}^{3}{k}^{-2}\right)}^{4}
  5. (m−2n−2)3{\left({m}^{-2}{n}^{-2}\right)}^{3}

Answer:

  1. (4z11)3=(4)3(z11)3=64z11⋅3=64z33{\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}}
  2. (pq3)6=(p)6(q3)6=p1⋅6q3⋅6=p6q18{\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}}
  3. (−1t2)27=(−1)27(t2)27=−1t2⋅27=−1t54=−1t54{\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}}
  4. (j3k−2)4=(j3k2)4=(j3)4(k2)4=j3⋅4k2⋅4=j12k8{\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}}
  5. (m−2n−2)3=(1m2n2)3=(1)3(m2n2)3=1(m2)3(n2)3=1m2⋅3⋅n2⋅3=1m6n6{\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}}

Try It

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
  1. (b5c)3{\left(\dfrac{{b}^{5}}{c}\right)}^{3}
  2. (5u8)4{\left(\dfrac{5}{{u}^{8}}\right)}^{4}
  3. (−1w3)35{\left(\dfrac{-1}{{w}^{3}}\right)}^{35}
  4. (p−4q3)8{\left({p}^{-4}{q}^{3}\right)}^{8}
  5. (c−5d−3)4{\left({c}^{-5}{d}^{-3}\right)}^{4}

Answer:

  1. b15c3\dfrac{{b}^{15}}{{c}^{3}}
  2. 625u32\dfrac{625}{{u}^{32}}
  3. −1w105\dfrac{-1}{{w}^{105}}
  4. q24p32\dfrac{{q}^{24}}{{p}^{32}}
  5. 1c20d12\dfrac{1}{{c}^{20}{d}^{12}}

[embed] [embed] [embed]

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.
  1. (6m2n−1)3{\left(6{m}^{2}{n}^{-1}\right)}^{3}
  2. 175⋅17−4⋅17−3{17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}
  3. (u−1vv−1)2{\left(\dfrac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}
  4. (−2a3b−1)(5a−2b2)\left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)
  5. (x22)4(x22)−4{\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}
  6. (3w2)5(6w−2)2\dfrac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}

Answer:

  1. \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}
  2. \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}
  3. \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}
  4. \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}
  5. \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}
  6. \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}

Try It

Simplify each expression and write the answer with positive exponents only.
  1. (2uv−2)−3{\left(2u{v}^{-2}\right)}^{-3}
  2. x8⋅x−12⋅x{x}^{8}\cdot {x}^{-12}\cdot x
  3. (e2f−3f−1)2{\left(\frac{{e}^{2}{f}^{-3}}{{f}^{-1}}\right)}^{2}
  4. (9r−5s3)(3r6s−4)\left(9{r}^{-5}{s}^{3}\right)\left(3{r}^{6}{s}^{-4}\right)
  5. (49tw−2)−3(49tw−2)3{\left(\frac{4}{9}t{w}^{-2}\right)}^{-3}{\left(\frac{4}{9}t{w}^{-2}\right)}^{3}
  6. (2h2k)4(7h−1k2)2\dfrac{{\left(2{h}^{2}k\right)}^{4}}{{\left(7{h}^{-1}{k}^{2}\right)}^{2}}

Answer:

  1. v68u3\dfrac{{v}^{6}}{8{u}^{3}}
  2. 1x3\dfrac{1}{{x}^{3}}
  3. e4f4\dfrac{{e}^{4}}{{f}^{4}}
  4. 27rs\dfrac{27r}{s}
  5. 11
  6. 16h1049\dfrac{16{h}^{10}}{49}

[embed] [embed] [embed] [embed]
In the following video we show more examples of how to find the power of a quotient. https://youtu.be/BoBe31pRxFM

Licenses & Attributions