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Study Guides > College Algebra

Using the Zero Exponent Rule of Exponents

Return to the quotient rule. We made the condition that m>nm>n so that the difference mnm-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=t88=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 4: 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}}

Solution

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=3x5x5=3x55=3x0=31=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)44Use 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)22Use the quotient rule.=5(rs2)0Simplify.=51Use 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 4

Simplify each expression using the zero exponent rule of exponents.

a. t7t7\frac{{t}^{7}}{{t}^{7}} b. (de2)112(de2)11\frac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}} c. w4w2w6\frac{{w}^{4}\cdot {w}^{2}}{{w}^{6}} d. t3t4t2t5\frac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}

Solution

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