Using the Zero Exponent Rule of Exponents

Return to the quotient rule. We made the condition that $m>n$ so that the difference $m-n$ would never be zero or negative. What would happen if $m=n$ ? 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.

$\frac{{t}^{8}}{{t}^{8}}=\frac{\cancel{t}^{8}}{\cancel{t}^{8}}=1$

If we were to simplify the original expression using the quotient rule, we would have

$\frac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}$

If we equate the two answers, the result is ${t}^{0}=1$ . This is true for any nonzero real number, or any variable representing a real number.

${a}^{0}=1$

The sole exception is the expression ${0}^{0}$ . 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 $a$ , the zero exponent rule of exponents states that

${a}^{0}=1$

Example 4: Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.

$\frac{{c}^{3}}{{c}^{3}}$
$\frac{-3{x}^{5}}{{x}^{5}}$
$\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}$
$\frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}$

Solution

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

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

Try It 4

Simplify each expression using the zero exponent rule of exponents.

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

Solution

Exponents and Scientific Notation