5.8: Power of a quotient rule for exponents
The power of a quotient rule for exponents will focus on what happens to a quotient when it is raised to some power.
For any real number \(a\) and \(b\) and any integer \(n\), the power of a quotient rule for exponents is the following:
\(\left( \dfrac{a }{b} \right)^n = \dfrac{a^n }{b^n }\),
where \(b \neq 0\).
Simplify the following using power of a quotient rule for exponents.
Simplify the following using power of a quotient rule for exponents.
\(\left( \dfrac{a }{b} \right)^4\)
Solution
\(\begin{aligned} &\left( \dfrac{a}{ b} \right)^4 && \text{Given} \\ &= \dfrac{a }{b} \cdot \dfrac{a }{b} \cdot \dfrac{a }{b} \cdot \dfrac{a }{b} &&\text{Expand using the exponent definition} \\ &= \dfrac{a^4 }{b^4} && \text{Multiply as needed to simplify} \end{aligned}\)
\(\left(\dfrac{x^2 }{3y^5} \right)^3\)
Solution
\(\begin{aligned} &\left( \dfrac{x^2 }{3y^5 }\right)^3 && \text{Given} \\ &= \dfrac{x^{2\cdot 3 }}{3^3 \cdot y^{5\cdot 3 }} && \text{power of quotient rule for exponents applied} \\ &= \dfrac{x^6 }{3^3 \cdot y^{15 }} &&\text{Simplify exponent product} \\ &= \dfrac{x^6 }{27y^{15 }} && \text{Multiply as needed to simplify.} \end{aligned}\)
\(\left( \dfrac{2x }{y }\right)^{−3}\)
Solution
\(\begin{aligned} &\left( \dfrac{2x }{y }\right)^{−3 } &&\text{Given} \\ &= \left( \dfrac{y }{2x} \right)^3 && \text{Negative exponent rule applied} \\ &= \dfrac{y^3 }{2^3 \cdot x^3} && \text{Power of a quotient rule for exponents applied.} \\ &= \dfrac{y^3 }{8x^3 } && \text{Multiply as needed to simplify.} \end{aligned}\)
The order in which rules of exponents are applied does not matter. In example three, steps 2 and 3 can be done in any order. The results will be the same.
Simplify the expression using the power of a quotient rule for exponents.
- \(\left( \dfrac{p^4 }{p^7 }\right) ^3\)
- \(−\left(\dfrac{ x^2 \cdot x^3 }{x \cdot y^3} \right) ^2\)
- \(\left( \dfrac{5x^3 }{2y^{13 }}\right) ^{−2}\)
- \(\left( \dfrac{2c^3}{ c^4} \right) ^3\)
- \(\left( \dfrac{a ^{−7}b }{a^2b^{−4 }}\right)^3\)
- \(\left( \dfrac{f^{−7 }}{f^5 }\right)^9\)
- \(\left(\dfrac{ xy^2z^3}{ x^3y^2z} \right) ^5\)