2.5 Rules of exponents

$$\begin{array}{l}\begin{array}{ccccccccccc}{x}^2& \cdot & x^4& =& \underset{}{\underbrace{xx}}& \cdot & \underset{}{\underbrace{xxxx}}& =& \underset{}{\underbrace{xxxxxx}}& =& x^6\\ & & & & 2& +& 4& =& 6& & \end{array}\\ \text{factors}\begin{array}{ccccc}& & & \text{factors}& \end{array}\end{array}$$

$$\begin{array}{l}\begin{array}{ccccccccccc}a& \cdot & a^2& =& \underset{}{\underbrace{a}}& \cdot & \underset{}{\underbrace{aa}}& =& \underset{}{\underbrace{aaa}}& =& a^3\\ & & & & 1& +& 2& =& 3& & \end{array}\\ \begin{array}{cccc}\text{factors}& & \text{factors}& \end{array}\end{array}$$

$x^3\cdot x^5=x^{3+5}=x^8$

$a^6\cdot a^{14}=a^{6+14}=a^{20}$

$y^5\cdot y=y^5\cdot y^1=y^{5+1}=y^6$

${(x-2y)}^8{(x-2y)}^5={(x-2y)}^{8+5}={(x-2y)}^{13}$

$\begin{array}{ll}{x}^3{y}^4\ne {(xy)}^{3+4}\hfill & \text{Since}\text{the}\text{bases}\text{are}\text{not}\text{the}\text{same,}\text{the}\hfill \\ \hfill & \text{product}\text{rule}\text{does}\text{not}\text{apply}\text{.}\hfill \end{array}$

$2x^3\cdot 7x^5=\begin{array}{||}\hline 2\cdot 7\cdot x^{3+5}\\ \hline\end{array}=14x^8$

We used the commutative and associative properties of multiplication. In practice, we use these properties “mentally” (as signified by the drawing of the box). We don’t actually write the second step.

$4y^3\cdot 6y^2=\begin{array}{||}\hline 4\cdot 6\cdot y^{3+2}\\ \hline\end{array}=24y^5$

$9a^2{b}^6(8a{b}^{4}2b^3)=\begin{array}{||}\hline 9\cdot 8\cdot 2a^{2+1}{b}^{6+4+3}\\ \hline\end{array}=144a^3{b}^{13}$

$5{(a+6)}^2\cdot 3{(a+6)}^8=\begin{array}{||}\hline 5\cdot 3{(a+6)}^{2+8}\\ \hline\end{array}=15{(a+6)}^{10}$

$4x^3\cdot 12\cdot y^2=48x^3{y}^2$

The bases are the same, so we add the exponents. Although we don’t know exactly what number is, the notation indicates the addition.