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$\begin{array}{ll}\frac{{x}^{5}}{{x}^{2}}=\frac{xxxxx}{xx}=\frac{(\overline{)xx})xxx}{(\overline{)xx})}=xxx={x}^{3}.\hfill & \text{Notice}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}5-2=3.\hfill \end{array}$
$\begin{array}{ll}\frac{{a}^{8}}{{a}^{3}}=\frac{aaaaaaaa}{aaa}=\frac{(\overline{)aaa})aaaaa}{(\overline{)aaa})}=aaaaa={a}^{5}.\hfill & \text{Notice}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}8-3=5.\hfill \end{array}$
Find the following quotients. All exponents are natural numbers.
$\begin{array}{cc}\frac{{x}^{5}}{{x}^{2}}=\begin{array}{||}\hline {\text{x}}^{\text{5-2}}\\ \hline\end{array}={x}^{3}& \text{The}\text{\hspace{0.17em}}\text{part}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{box}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{usally}\text{\hspace{0.17em}}\text{done}\text{\hspace{0.17em}}\text{mentally}\text{.}\end{array}$
$$\frac{27{a}^{3}{b}^{6}{c}^{2}}{3{a}^{2}bc}=\begin{array}{||}\hline 9{a}^{3-2}{b}^{6-1}{c}^{2-1}\\ \hline\end{array}=9a{b}^{5}c$$
$$\frac{15{x}^{\square}}{3{x}^{\u25b3}}=5{x}^{\square -\u25b3}$$
The bases are the same, so we subtract the exponents. Although we don’t know exactly what $\square -\u25b3$ is, the notation $\square -\u25b3$ indicates the subtraction.
Find each quotient
$\frac{26{x}^{4}{y}^{6}{z}^{2}}{13{x}^{2}{y}^{2}z}$
$2{x}^{2}{y}^{4}z$
When we make the subtraction, $n-m$ , in the division $\frac{{x}^{n}}{{x}^{m}}$ , there are three possibilities for the values of the exponents:
In Sample Set C, the exponents of the numerators were greater than the exponents of the denominators. Let’s study the case when the exponents are the same.
When the exponents are the same, say
$n$ , the subtraction
$n-n$ produces 0.
Thus, by the second rule of exponents,
$\frac{{x}^{n}}{{x}^{n}}={x}^{n-n}={x}^{0}$ .
But what real number, if any, does
${x}^{0}$ represent? Let’s think for a moment about our experience with division in arithmetic. We know that any nonzero number divided by itself is one.
$\begin{array}{ccc}\frac{8}{8}=1,& \frac{43}{43}=1,& \frac{258}{258}\end{array}=1$
Since the letter
$x$ represents some nonzero real number, so does
${x}^{n}$ . Thus,
$\frac{{x}^{n}}{{x}^{n}}$
represents some nonzero real number divided by itself. Then
$\frac{{x}^{n}}{{x}^{n}}=1$ .
But we have also established that if
$x\ne 0,\text{\hspace{0.17em}}\frac{{x}^{n}}{{x}^{n}}={x}^{0}$ . We now have that
$\text{\hspace{0.17em}}\frac{{x}^{n}}{{x}^{n}}={x}^{0}$
and
$\text{\hspace{0.17em}}\frac{{x}^{n}}{{x}^{n}}=1$ . This implies that
${x}^{0}=1,x\ne 0$ .
Exponents can now be natural numbers and zero. We have enlarged our collection of numbers that can be used as exponents from the collection of natural numbers to the collection of whole numbers.
Any number, other than 0, raised to the power of 0, is 1. ${0}^{0}$ has no meaning (it does not represent a number).
Find each value. Assume the base is not zero.
$${6}^{0}=1$$
${247}^{0}=1$
${(2a+5)}^{0}=1$
$$4{y}^{0}=4\cdot 1=4$$
$$\frac{{y}^{6}}{{y}^{6}}={y}^{0}=1$$
$$\frac{2{x}^{2}}{{x}^{2}}=2{x}^{0}=2\cdot 1=2$$
$$\begin{array}{lll}\frac{5{(x+4)}^{8}{(x-1)}^{5}}{5{(x+4)}^{3}{(x-1)}^{5}}\hfill & =\hfill & {(x+4)}^{8-3}{(x-1)}^{5-5}\hfill \\ \hfill & =\hfill & {(x+4)}^{5}{(x-1)}^{0}\hfill \\ \hfill & =\hfill & {(x+4)}^{5}\hfill \end{array}$$
Find each value. Assume the base is not zero.
$\frac{36{a}^{4}{b}^{3}{c}^{8}}{8a{b}^{3}{c}^{6}}$
$\frac{9}{2}{a}^{3}{c}^{2}$
$\frac{52{a}^{7}{b}^{3}{(a+b)}^{8}}{26{a}^{2}b{(a+b)}^{8}}$
$2{a}^{5}{b}^{2}$
$\frac{14{x}^{r}{y}^{p}{z}^{q}}{2{x}^{r}{y}^{h}{z}^{5}}$
$7{y}^{p-h}{z}^{q-5}$
We will study the case where the exponent of the denominator is greater than the exponent of the numerator in Section [link] .
Use the product rule and quotient rule of exponents to simplify the following problems. Assume that all bases are nonzero and that all exponents are whole numbers.
${5}^{2}\cdot {5}^{4}$
${7}^{3}\cdot {7}^{0}$
${x}^{5}{x}^{4}$
${a}^{9}{a}^{7}$
${m}^{10}{m}^{2}$
${y}^{3}{y}^{4}{y}^{6}$
${a}^{2}{a}^{3}{a}^{8}$
$2{a}^{3}{b}^{2}\cdot 3ab$
$12x{y}^{3}{z}^{2}\cdot 4{x}^{2}{y}^{2}z\cdot 3x$
$144{x}^{4}{y}^{5}{z}^{3}$
$(3ab)(2{a}^{2}b)$
$(2xy)(3y)(4{x}^{2}{y}^{5})$
$\left(\frac{1}{4}{a}^{2}{b}^{4}\right)\text{\hspace{0.17em}}\left(\frac{1}{2}{b}^{4}\right)$
$\frac{1}{8}{a}^{2}{b}^{8}$
$\left(\frac{3}{8}\right)\text{\hspace{0.17em}}\left(\frac{16}{21}{x}^{2}{y}^{3}\right)\text{\hspace{0.17em}}({x}^{3}{y}^{2})$
$\frac{{6}^{4}}{{6}^{3}}$
$\frac{{4}^{16}}{{4}^{13}}$
$\frac{{y}^{4}}{{y}^{3}}$
$\frac{{k}^{16}}{{k}^{13}}$
$\frac{{y}^{5}}{{y}^{2}}$
$\frac{{a}^{9}{b}^{6}}{{a}^{5}{b}^{2}}$
$\frac{{m}^{17}{n}^{12}}{{m}^{16}{n}^{10}}$
$\frac{15{x}^{20}{y}^{24}{z}^{4}}{5{x}^{19}yz}$
$\frac{6{r}^{4}}{6{r}^{4}}$
$\frac{{a}^{0}{b}^{0}}{{c}^{0}}$
$\frac{24{x}^{4}{y}^{4}{z}^{0}{w}^{8}}{9xy{w}^{7}}$
${x}^{3}\left(\frac{{x}^{6}}{{x}^{2}}\right)$
${a}^{4}{b}^{6}\left(\frac{{a}^{10}{b}^{16}}{{a}^{5}{b}^{7}}\right)$
${a}^{9}{b}^{15}$
$3{a}^{2}{b}^{3}\left(\frac{14{a}^{2}{b}^{5}}{2b}\right)$
$\frac{{(x+3y)}^{11}{(2x-1)}^{4}}{{(x+3y)}^{3}(2x-1)}$
${\left(x+3y\right)}^{8}{\left(2x-1\right)}^{3}$
$\frac{40{x}^{5}{z}^{10}{(z-{x}^{4})}^{12}{(x+z)}^{2}}{10{z}^{7}{(z-{x}^{4})}^{5}}$
${a}^{x}{b}^{y}{c}^{5z}$
$\frac{{x}^{n+3}}{{x}^{n}}$
${y}^{\Delta}{y}^{\nabla}$
${a}^{\Delta}{a}^{\nabla}{b}^{\square}{b}^{\diamond}$
${a}^{\Delta +\nabla}{b}^{\square +\diamond}$
( [link] ) What natural numbers can replace $x$ so that the statement $-5<x\le 3$ is true?
( [link] ) Use the distributive property to expand $4x(2a+3b)$ .
$8ax+12bx$
( [link] ) Express $xxxyyyy(a+b)(a+b)$ using exponents.
( [link] ) Find the value of ${4}^{2}+{3}^{2}\cdot {2}^{3}-10\cdot 8$ .
8
( [link] ) Find the value of $\frac{{4}^{2}+{(3+2)}^{2}-1}{{2}^{3}\cdot 5}+\frac{{2}^{4}({3}^{2}-{2}^{3})}{{4}^{2}}$ .
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