5.5: The Negative Exponent Rule
In section 5.3, the exponent of the number in the numerator was greater than the exponent of the number in the denominator. In section 5.4, the exponent of the number in the numerator was equal to the exponent of the number in the denominator. In section 5.5, the exponent of the number in the denominator may be greater than the exponent of the number in the numerator.
For any non zero real number a and any integer n, the negative exponent rule is the following
\(a^{−n}= \dfrac{1 }{a^n} or \dfrac{1 }{a^{−n}} = a^n\)
It is poor form in mathematics to leave negative exponents in the answer. All answers will always be simplified to show positive exponents.
How does this work?
Recall:
\[\begin{align*} 2^3 &= 2 \cdot 2 \cdot 2 = 8 \\[4pt] 2^2 &= 2 \cdot 2 = 4 \\[4pt] 2^1 &= 2 = 2 \\[4pt] 2^0 &= 1 \end{align*} \nonumber \]
What happens with negative exponents?
\[\begin{align*} 2^{−1 } &= \dfrac{1 }{2^1 }= \dfrac{1 }{2} \\[4pt] 2^{−2 } &= \dfrac{1}{2^2} = \dfrac{1}{ 4} \end{align*} \nonumber \]
Recall: From the last section,
\[\begin{align*} x^3 = \textcolor{blue}{x \cdot x \cdot x} \\[4pt] x^5 &= \textcolor{red}{x \cdot x \cdot x \cdot x \cdot x} \end{align*} \nonumber \]
Their quotient:
\(\dfrac{x^3 }{x^5} = \dfrac{x \cdot x \cdot x }{x \cdot x \cdot x \cdot x \cdot x }= \dfrac{\textcolor{blue}{\cancel{x \cdot x\cdot x }}}{\textcolor{red}{\cancel{x \cdot x\cdot x \cdot x \cdot x }}}=\dfrac{ 1 }{\textcolor{red}{x \cdot x }}= \dfrac{1 }{x^2}\)
Apply the quotient rule to obtain an equivalent result.
\(\dfrac{x^3 }{x^5} = x^{3−5 }= x^{−2}\)
Using the negative exponent rule.
\(x^{−2 }= \dfrac{1 }{x^2}\).
Review the following examples to help understand the process of simplifying using the quotient rule of exponents and the negative exponent rule.
Hint: Be patient, take your time and be careful when simplifying!
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{t^5}{ t^11}\)
Solution
\(t^{5−11 }= t^{−6 }= \dfrac{1 }{t^6}\)
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{x^{3} \cdot x^{11} }{x \cdot x^{15}}\)
Solution
\(\dfrac{x^{ 3+11 }}{x^{1+15 }}= \dfrac{x^14 }{x^16 }= x^{14−16 }= x^{−2 }= \dfrac{1}{x^2}\)
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{2y^3 }{7y^7}\)
Solution
\(\dfrac{2 }{7} \cdot \dfrac{y^3 }{y^7 }= \dfrac{2 }{7} \cdot y^{3−7 }= \dfrac{2 }{7 }\cdot y^{−4} = \dfrac{2 }{7 }\cdot \dfrac{1 }{y^4 }= \dfrac{2 }{7y^4}\)
Simplify the following expression to a single base with only positive exponents.
\(-\dfrac{\sqrt{3}z^6}{ z^7}\)
Solution
\(− \sqrt{3} \cdot \dfrac{z^6 }{z^7} = − \sqrt{3} \cdot z^{6−7 }= − \sqrt{3} \cdot z^{−1} = − \sqrt{3} \cdot \dfrac{1 }{z} = − \dfrac{\sqrt{3}}{z}\)
In Examples 3 and 4, factor out the constant to see the common bases clearly.
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{1}{ a^{−9}}\)
Solution
\(a^9\)
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{x^3 }{x^{−5}}\)
Solution
\(x^{3−(−5) }= x^{3+5 }= x^{8}\)
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{c^{−7 }}{c^{−3}}\)
Solution
\(c^{(−7)−(−3) }= c^{−7+3 }= c ^{−4} = \dfrac{1 }{c^4}\)
In examples 6 and 7, the quotient rule of exponents was used before changing exponents to positive exponents. The same results are obtained by expanding and changing exponents to positive exponents first and then applying the quotient rule of exponents.
Simplify the following expressions to a single base with only positive exponents.
- \(\dfrac{p^4}{ p^{13}}\)
- \(-\dfrac{k^2 \cdot k^3 }{k^7 \cdot k^8}\)
- \(\dfrac{5(x + y)^3 }{2(x + y)^{13}}\)
- −\(\dfrac{\sqrt{8}y^3}{ y^{−3}}\)
- \(\dfrac{a^{−7}}{ a^2 \cdot a^{−5}}\)
- \(\dfrac{x^{−7}}{ x^5}\)