1.7: Properties of Exponents
- Page ID
- 45002
Multiplication Properties
Recall from chapter 1 that \(5^{3}=5 \cdot 5 \cdot 5 .\) In the same way we have
\[x^{3}=x \cdot x \cdot x\nonumber\]
Example 5.1
Perform the given operations:
- \(x^{3} \cdot x^{5}=\underbrace{x \cdot x \cdot x}_{=x^{3}} \cdot \underbrace{x \cdot x \cdot x \cdot x \cdot x}_{=x^{5}}=x^{8}\)
- \(\left(x^{3}\right)^{4}=x^{3} \cdot x^{3} \cdot x^{3} \cdot x^{3}=\underbrace{x \cdot x \cdot x} \cdot x \cdot x \cdot x \cdot \cdot \underbrace{x \cdot x \cdot x} \cdot x \cdot x \cdot x=x^{12}\)
- \((x \cdot y)^{4}=\underbrace{x \cdot y} \cdot \underbrace{x \cdot y} \cdot \underbrace{x \cdot y} \cdot \underbrace{x \cdot y}=x^{4} y^{4}\)
We can summarize these examples into the following useful rules:
Multiplication and Exponentiation Rules
For any integers \(n\) and \(m\)
- \(x^{n} x^{m}=x^{n+m}\)
- \((x^{n})^{m}=x^{nm}\)
- \((x y)^{n}=x^{n} y^{n}\)
Example 5.2
Perform the given operation using the multiplication properties of exponents and write your answer in simplest form:
- \(b^{2} \cdot b^{3}=b^{2+3}=b^{5}\) (recall the meaning of exponents)
- \(x^{8} x^{7}=x^{8+7}=x^{15}\) (note that juxtaposition indicates multiplication)
- \(a^{8} a^{9} a^{14}=a^{8+9+14}=a^{31}\)
- \(4 x^{4} \cdot 7 x^{6}=(4 \cdot 7)\left(x^{4} \cdot x^{6}\right)=28 x^{10}\)
- \(5 x y^{3} \cdot 6 y=(5 \cdot 6)(x)\left(y^{3} \cdot y\right)=30 x y^{4}\)
- \(\left(5 x^{4} y^{2}\right)\left(2 x^{7} y^{3}\right)=10 x^{4+7} y^{2+3}=10 x^{11} y^{5}\)
- \((2 x)^{3}=2^{3} x^{3}=8 x^{3}\)
- \(\left(-4 a^{2} b^{5}\right)^{3}=(-4)^{3} a^{2 \cdot 3} b^{5 \cdot 3}=-64 a^{6} b^{15}\)
- \(\begin{align*}\left(-5 r^{3} s\right)^{2} \cdot\left(2 r^{4} s^{3}\right)^{3} \cdot\left(-r^{2} s^{2}\right) &=(-5)^{2} r^{3 \cdot 2} s^{2} \cdot 2^{3} r^{4 \cdot 3} s^{3 \cdot 3} \cdot(-1) r^{2} s^{2} \\&=25 r^{6} s^{2} \cdot 8 r^{12} s^{9} \cdot(-1) r^{2} s^{2} \\&=-200 r^{20} s^{13}\end{align*}\)
Division Properties
Example 5.3
- \(\dfrac{x^{5}}{x^{3}}=\dfrac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x \cdot x}=\dfrac{x \cdot x}{1}=\dfrac{x^{2}}{1}=x^{2}\)
- \(\dfrac{x^{3}}{x^{5}}=\dfrac{x \cdot x \cdot x}{x \cdot x \cdot x \cdot x \cdot x}=\dfrac{1}{x \cdot x}=\dfrac{1}{x^{2}}\)
- \(\left(\dfrac{x}{y}\right)^{5}=\dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y} \cdot \dfrac{x}{y}=\dfrac{x \cdot x \cdot x \cdot x \cdot x}{y \cdot y \cdot y \cdot y \cdot y}=\dfrac{x^{5}}{y^{5}}\)
We can summarize these examples into the following useful rules:
Division Rules
For any integers \(n\) and \(m\)
- \(\dfrac{x^{n}}{x^{m}}=x^{n-m}\)
- \(\left(\dfrac{x}{y}\right)^{n}=\dfrac{x^{n}}{y^{n}}\)
Example 5.4
Perform the given operation using the division properties of exponents and state your answer in simplest form:
- \(\dfrac{b^{8}}{b^{7}}=b^{8-7}=b^{8+(-7)}=b^{1}=b\)
- \(\dfrac{x^{12} y^{2}}{x^{8} y}=x^{12-8} y^{2-1}=x^{12+(-8)} y^{2+(-1)}=x^{4} y^{1}=x^{4} y\)
- \(\dfrac{8 x^{6} y^{2}}{6 x^{5} y^{7}}=\dfrac{4 x^{6-5}}{3 y^{7-2}}=\dfrac{4 x^{1}}{3 y^{5}}=\dfrac{4 x}{3 y^{5}}\)
- \(\dfrac{\left(m^{2}\right)^{3}\left(n^{4}\right)^{5}}{\left(m^{3}\right)^{3}}=\dfrac{m^{6} n^{20}}{m^{9}}=\dfrac{n^{20}}{m^{3}}\)
- \(\left(\dfrac{3 a^{2} b^{4}}{9 c^{3}}\right)^{2}=\left(\dfrac{a^{2} b^{4}}{3 c^{3}}\right)^{2}=\dfrac{a^{2 \cdot 2} b^{4 \cdot 2}}{3^{2} c^{3 \cdot 2}}=\dfrac{a^{4} b^{8}}{9 c^{6}}\)Another way to simplify this correctly is this: \(\left(\dfrac{3 a^{2} b^{4}}{9 c^{3}}\right)^{2}=\dfrac{3^{2} a^{2 \cdot 2} b^{4 \cdot 2}}{9^{2} c^{3 \cdot 2}}=\dfrac{9 a^{4} b^{8}}{81 c^{6}}=\dfrac{a^{4} b^{8}}{9 c^{6}}\)
Zero Exponent
Recall from chapter 1 that \((-7)^{0}=1\) and \(8^{0}=1 .\) In the same way we have \(x^{0}=1\)
Zero Exponent
\(a^{0}=1 \nonumber\)
Example 5.5
Evaluate
- \(15^{0}=1\)
- \((-15)^{0}=1\)
- \(-15^{0}=-1\)
- \((15 x)^{0}=1\)
- \(15 x^{0}=15 \cdot 1=15\)
Negative Exponents
Negative Exponents
For any integer \(n\)
- \(a^{-n}=\dfrac{1}{a^{n}}\)
- \(\dfrac{1}{a^{-n}}=a^{n}\)
- \(\left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^{n}\)
Note 5.6. The negative exponent rules can be used to switch terms from numerator to denominator or vice versa, and is useful to write expressions using positive exponents only.
Example 5.7
- \(\dfrac{x^{-3} y^{2}}{z^{4}}=\dfrac{y^{2}}{x^{3} z^{4}}\)
- \(\dfrac{x^{4}}{y^{-5} z^{2}}=\dfrac{x^{4} y^{5}}{z^{2}}\)
Example 5.8
Perform the given operation and write your answer using positive exponents only
- \(\dfrac{x^{4} y^{2}}{x^{3} y^{-3}}=x^{4-3} y^{2-(-3)}=x^{4+(-3)} y^{2+3}=x y^{5}\)
- \(\left(x^{2} y\right) \cdot\left(x y^{-4}\right)=x^{2+1} y^{1+(-4)}=x^{3} y^{-3}=\dfrac{x^{3}}{y^{3}}\)
Exit Probelm
Simplify: \(\left(\dfrac{8 y}{3 x^{3}}\right)^{3}\)