5.3: The Quotient Rule of Exponents
For any real number \(a\) and positive numbers \(m\) and \(n\), where \(m > n\).
The Quotient Rule For Exponents is the following.
\(\dfrac{a^m }{a^n} = a^{ m−n}\)
Note: Bases MUST be the same. Result will have the same base.
Idea:
From the last section,
\(x^3 = \textcolor{blue}{x \cdot x \cdot x} \qquad x^5 = \textcolor{red}{x \cdot x \cdot x \cdot x \cdot x}\)
Their quotient
\(\dfrac{x^ 5 }{x^3} = \dfrac{\textcolor{red}{x \cdot x \cdot x \cdot x \cdot x }}{\textcolor{blue}{x \cdot x \cdot x }}= \dfrac{\textcolor{red}{\cancel{x \cdot x\cdot x \cdot x }\cdot x }}{\textcolor{blue}{\cancel{x \cdot x\cdot x }}}= \dfrac{\textcolor{red}{x \cdot x }}{1} = \textcolor{red}{x \cdot x}\).
So, \(\dfrac{x^5 }{x^3 }= x^{5−3 }= x^2\)
Using the quotient rule of exponents to simplify expressions.
- \(\dfrac{k^3 }{k^2}\)
- \(\dfrac{r^{32} }{r^{21}}\)
- \(\dfrac{\sqrt{2}^ 7 }{\sqrt{2 }^4}\)
- \(\dfrac{(−7)^9 }{(−7)^6}\)
- \(\dfrac{(x \sqrt{5})^8 }{x\sqrt{ 5}}\)
- \(\dfrac{(xy)^{18} }{(xy)^{17}}\)
Solution
| Expression | Quotient Rule | Base |
| \(\dfrac{k^3 }{k^2}\) | \(k^{3−2 }= k\) | \(k\) |
| \(\dfrac{r^{32} }{r^{21}}\) | \(r^{32−21 }= r^{11}\) | \(r\) |
| \(\dfrac{\sqrt{2}^ 7 }{\sqrt{2 }^4}\) | \(\sqrt{2 }^{7−4 }= \sqrt{2 }^3\) | \(\sqrt{2}\) |
| \(\dfrac{(−7)^9 }{(−7)^6}\) | \((−7)^{9−6 }= (−7)^3\) | \(-7\) |
| \(\dfrac{(x \sqrt{5})^8 }{x\sqrt{ 5}}\) | \((x \sqrt{5})^{8−1 }= (x \sqrt{5})^7\) | \(x\sqrt{5}\) |
| \(\dfrac{(xy)^{18} }{(xy)^{17}}\) | \((xy)^{18−17 }= xy\) | \(xy\) |
Note: In this section the exponent of the numerator was greater than the exponent of the denominator. That won’t always be the case. The case where the exponent in the denominator is greater than the exponent in the numerator will be discussed in a later section.
Use the quotient rule of exponents to simplify the given expression.
- \(\dfrac{−y ^{13} }{−y^7}\)
- \(\dfrac{(2x)^{25}}{ 2x}\)
- \(\dfrac{\sqrt{7 }^{17 }}{\sqrt{7 }^{12}}\)
- \(\dfrac{(−7)^9 }{(−7)^6}\)
- \(\dfrac{(x + y) ^{78}}{ (x + y)^{43}}\)
- \(\dfrac{\sqrt{xy }^{15 }}{\sqrt{xy }^{11}}\)