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Study Guides > College Algebra

Zero and Negative Exponents

Learning Objectives

  • 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\frac{t^{8}}{t^{8}}=\frac{\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\frac{{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

Example: Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.
  1. c3c3\frac{{c}^{3}}{{c}^{3}}
  2. −3x5x5\frac{-3{x}^{5}}{{x}^{5}}
  3. (j2k)4(j2k)⋅(j2k)3\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}
  4. 5(rs2)2(rs2)2\frac{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{array}\text{ }\frac{c^{3}}{c^{3}} \hfill& =c^{3-3} \\ \hfill& =c^{0} \\ \hfill& =1\end{array}
  2. −3x5x5=−3⋅x5x5=−3⋅x5−5=−3⋅x0=−3⋅1=−3\begin{array}{ccc}\hfill \frac{-3{x}^{5}}{{x}^{5}}& =& -3\cdot \frac{{x}^{5}}{{x}^{5}}\hfill \\ & =& -3\cdot {x}^{5 - 5}\hfill \\ & =& -3\cdot {x}^{0}\hfill \\ & =& -3\cdot 1\hfill \\ & =& -3\hfill \end{array}
  3. (j2k)4(j2k)⋅(j2k)3=(j2k)4(j2k)1+3Use the product rule in the denominator.=(j2k)4(j2k)4Simplify.=(j2k)4−4Use the quotient rule.=(j2k)0Simplify.=1\begin{array}{cccc}\hfill \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}}\hfill & \text{Use the product rule in the denominator}.\hfill \\ & =& \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}}\hfill & \text{Simplify}.\hfill \\ & =& {\left({j}^{2}k\right)}^{4 - 4}\hfill & \text{Use the quotient rule}.\hfill \\ & =& {\left({j}^{2}k\right)}^{0}\hfill & \text{Simplify}.\hfill \\ & =& 1& \end{array}
  4. 5(rs2)2(rs2)2=5(rs2)2−2Use the quotient rule.=5(rs2)0Simplify.=5⋅1Use the zero exponent rule.=5Simplify.\begin{array}{cccc}\hfill \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& =& 5{\left(r{s}^{2}\right)}^{2 - 2}\hfill & \text{Use the quotient rule}.\hfill \\ & =& 5{\left(r{s}^{2}\right)}^{0}\hfill & \text{Simplify}.\hfill \\ & =& 5\cdot 1\hfill & \text{Use the zero exponent rule}.\hfill \\ & =& 5\hfill & \text{Simplify}.\hfill \end{array}

Try It

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

Answer:

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

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\frac{{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\frac{{h}^{3}}{{h}^{5}}.
h3h5=h⋅h⋅hh⋅h⋅h⋅h⋅h=h⋅h⋅hh⋅h⋅h⋅h⋅h=1h⋅h=1h2\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h}\hfill \\ & =& \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h}\hfill \\ & =& \frac{1}{h\cdot h}\hfill \\ & =& \frac{1}{{h}^{2}}\hfill \end{array}
If we were to simplify the original expression using the quotient rule, we would have
h3h5=h3−5= h−2\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\hfill \\ & =& \text{ }{h}^{-2}\hfill \end{array}
Putting the answers together, we have h−2=1h2{h}^{-2}=\frac{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=1anandan=1a−n\begin{array}{ccc}{a}^{-n}=\frac{1}{{a}^{n}}& \text{and}& {a}^{n}=\frac{1}{{a}^{-n}}\end{array}
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{a}^{-n}=\frac{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\frac{{\theta }^{3}}{{\theta }^{10}}
  2. z2⋅zz4\frac{{z}^{2}\cdot z}{{z}^{4}}
  3. (−5t3)4(−5t3)8\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}

Answer:

  1. θ3θ10=θ3−10=θ−7=1θ7\frac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\frac{1}{{\theta }^{7}}
  2. z2⋅zz4=z2+1z4=z3z4=z3−4=z−1=1z\frac{{z}^{2}\cdot z}{{z}^{4}}=\frac{{z}^{2+1}}{{z}^{4}}=\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\frac{1}{z}
  3. (−5t3)4(−5t3)8=(−5t3)4−8=(−5t3)−4=1(−5t3)4\frac{{\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}=\frac{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\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}
  2. f47f49⋅f\frac{{f}^{47}}{{f}^{49}\cdot f}
  3. 2k45k7\frac{2{k}^{4}}{5{k}^{7}}

Answer:

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

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=p⋅q⋅p⋅q⋅p⋅q=p⋅p⋅p3 factors⋅q⋅q⋅q3 factors=p3⋅q3\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, (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{\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\frac{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\frac{1}{{\left(-7z\right)}^{4}}=\frac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{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}=\frac{{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\frac{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\frac{1}{{a}^{18}{b}^{21}}
  5. r12s8\frac{{r}^{12}}{{s}^{8}}

In the following video we show more examples of how to find hte 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}=\frac{{f}^{14}}{{e}^{14}}
Let’s rewrite the original problem differently and look at the result.
(e−2f2)7=(f2e2)7=f14e14\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.
(e−2f2)7=(f2e2)7=(f2)7(e2)7=f2⋅7e2⋅7=f14e14\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 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: 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. (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(\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=p1⋅6q3⋅6=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=−1t2⋅27=−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. (j3k−2)4=(j3k2)4=(j3)4(k2)4=j3⋅4k2⋅4=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. (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(\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

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(\frac{{b}^{5}}{c}\right)}^{3}
  2. (5u8)4{\left(\frac{5}{{u}^{8}}\right)}^{4}
  3. (−1w3)35{\left(\frac{-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\frac{{b}^{15}}{{c}^{3}}
  2. 625u32\frac{625}{{u}^{32}}
  3. −1w105\frac{-1}{{w}^{105}}
  4. q24p32\frac{{q}^{24}}{{p}^{32}}
  5. 1c20d12\frac{1}{{c}^{20}{d}^{12}}

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

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