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The following examples suggest a rule for raising a power to a power:
${({a}^{2})}^{3}={a}^{2}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}\text{}{a}^{2}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}{a}^{2}$
Using the product rule we get
$\begin{array}{l}{({a}^{2})}^{3}={a}^{2+2+2}\hfill \\ {({a}^{2})}^{3}={a}^{3\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}2}\hfill \\ {({a}^{2})}^{3}={a}^{6}\hfill \end{array}$
$\begin{array}{lll}{({x}^{9})}^{4}\hfill & =\hfill & {x}^{9}\cdot {x}^{9}\cdot {x}^{9}\cdot {x}^{9}\hfill \\ {({x}^{9})}^{4}\hfill & =\hfill & {x}^{9+9+9+9}\hfill \\ {({x}^{9})}^{4}\hfill & =\hfill & {x}^{4\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}9}\hfill \\ {({x}^{9})}^{4}\hfill & =\hfill & {x}^{36}\hfill \end{array}$
To raise a power to a power, multiply the exponents.
Simplify each expression using the power rule for powers. All exponents are natural numbers.
$\begin{array}{cc}{({x}^{3})}^{4}=\begin{array}{||}\hline {x}^{3\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}4}\\ \hline\end{array}\text{\hspace{0.17em}}{x}^{12}& \text{The}\text{\hspace{0.17em}}\text{box}\text{\hspace{0.17em}}\text{represents}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{step}\text{\hspace{0.17em}}\text{done}\text{\hspace{0.17em}}\text{mentally.}\end{array}$
$${({y}^{5})}^{3}=\begin{array}{||}\hline {y}^{5\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}3}\\ \hline\end{array}={y}^{15}$$
$${({d}^{20})}^{6}=\begin{array}{||}\hline {d}^{20\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}6}\\ \hline\end{array}={d}^{120}$$
${({x}^{\square})}^{\u25b3}=\text{\hspace{0.17em}}{x}^{\square \u25b3}$
Although we don’t know exactly what number $\square \u25b3$ is, the notation $\square \u25b3$ indicates the multiplication.
Simplify each expression using the power rule for powers.
The following examples suggest a rule for raising a product to a power:
$\begin{array}{llll}{(ab)}^{3}\hfill & =\hfill & ab\cdot ab\cdot ab\hfill & \text{Use}\text{\hspace{0.17em}}\text{}\text{the}\text{\hspace{0.17em}}\text{commutative}\text{\hspace{0.17em}}\text{property}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{multiplication}.\hfill \\ \hfill & =\hfill & aaabbb\hfill & \hfill \\ \hfill & =\hfill & {a}^{3}{b}^{3}\hfill & \hfill \end{array}$
$\begin{array}{lll}{(xy)}^{5}\hfill & =\hfill & xy\cdot xy\cdot xy\cdot xy\cdot xy\hfill \\ \hfill & =\hfill & xxxxx\cdot yyyyy\hfill \\ \hfill & =\hfill & {x}^{5}{y}^{5}\hfill \end{array}$
$\begin{array}{lll}{(4xy\mathrm{z})}^{2}\hfill & =\hfill & 4xyz\cdot 4xyz\hfill \\ \hfill & =\hfill & 4\cdot 4\cdot xx\cdot yy\cdot zz\hfill \\ \hfill & =\hfill & 16{x}^{2}{y}^{2}{z}^{2}\hfill \end{array}$
To raise a product to a power, apply the exponent to each and every factor.
Make use of either or both the power rule for products and power rule for powers to simplify each expression.
${(ab)}^{7}={a}^{7}{b}^{7}$
${(axy)}^{4}={a}^{4}{x}^{4}{y}^{4}$
$\begin{array}{cc}{(3ab)}^{2}={3}^{2}{a}^{2}{b}^{2}=9{a}^{2}{b}^{2}& \text{Don't}\text{\hspace{0.17em}}\text{forget}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{apply}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{exponent}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{3!}\end{array}$
$\begin{array}{c}{(2st)}^{5}={2}^{5}{s}^{5}{t}^{5}=32{s}^{5}{t}^{5}\end{array}$
$\begin{array}{ll}{(a{b}^{3})}^{2}={a}^{2}{({b}^{3})}^{2}={a}^{2}{b}^{6}\hfill & \text{We}\text{\hspace{0.17em}}\text{used}\text{\hspace{0.17em}}\text{two}\text{\hspace{0.17em}}\text{rules}\text{\hspace{0.17em}}\text{here}\text{.}\text{\hspace{0.17em}}\text{First,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{power}\text{\hspace{0.17em}}\text{rule}\text{\hspace{0.17em}}\text{for}\hfill \\ \hfill & \text{products}\text{.}\text{\hspace{0.17em}}\text{Second,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{power}\text{\hspace{0.17em}}\text{rule}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}\text{powers.}\hfill \end{array}$
$\begin{array}{ll}{(7{a}^{4}{b}^{2}{c}^{8})}^{2}\hfill & ={7}^{2}{({a}^{4})}^{2}{({b}^{2})}^{2}{({c}^{8})}^{2}\hfill \\ \hfill & =49{a}^{8}{b}^{4}{c}^{16}\hfill \end{array}$
$\begin{array}{cc}\text{If}\text{\hspace{0.17em}}6{a}^{3}{c}^{7}\ne 0,\text{\hspace{0.17em}}\text{then}\text{\hspace{0.17em}}{(6{a}^{3}{c}^{7})}^{0}=1& \text{Recall}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}{x}^{0}=1\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}x\ne 0.\end{array}$
$\begin{array}{ll}{\left[2{\left(x+1\right)}^{4}\right]}^{6}\hfill & ={2}^{6}{(x+1)}^{24}\hfill \\ \hfill & =64{(x+1)}^{24}\hfill \end{array}$
Make use of either or both the power rule for products and the power rule for powers to simplify each expression.
${\left[4t\left(s-5\right)\right]}^{3}$
$64{t}^{3}{\left(s-5\right)}^{3}$
${(1{a}^{5}{b}^{8}{c}^{3}d)}^{6}$
${a}^{30}{b}^{48}{c}^{18}{d}^{6}$
${\left[\left(a+8\right)\left(a+5\right)\right]}^{4}$
${\left(a+8\right)}^{4}{\left(a+5\right)}^{4}$
${\left[(12{c}^{4}{u}^{3}{(w-3)}^{2}\right]}^{5}$
${12}^{5}{c}^{20}{u}^{15}{(w-3)}^{10}$
${\left[10{t}^{4}{y}^{7}{j}^{3}{d}^{2}{v}^{6}{n}^{4}{g}^{8}{\left(2-k\right)}^{17}\right]}^{4}$
${10}^{4}{t}^{16}{y}^{28}{j}^{12}{d}^{8}{v}^{24}{n}^{16}{g}^{32}{(2-k)}^{68}$
${({x}^{3}{x}^{5}{y}^{2}{y}^{6})}^{9}$
${\left({x}^{8}{y}^{8}\right)}^{9}={x}^{72}{y}^{72}$
${({10}^{6}\cdot {10}^{12}\cdot {10}^{5})}^{10}$
${10}^{230}$
The following example suggests a rule for raising a quotient to a power.
${\left(\frac{a}{b}\right)}^{3}=\frac{a}{b}\cdot \frac{a}{b}\cdot \frac{a}{b}=\frac{a\cdot a\cdot a}{b\cdot b\cdot b}=\frac{{a}^{3}}{{b}^{3}}$
To raise a quotient to a power, distribute the exponent to both the numerator and denominator.
Make use of the power rule for quotients, the power rule for products, the power rule for powers, or a combination of these rules to simplify each expression. All exponents are natural numbers.
${\left(\frac{x}{y}\right)}^{6}=\frac{{x}^{6}}{{y}^{6}}$
${\left(\frac{a}{c}\right)}^{2}=\frac{{a}^{2}}{{c}^{2}}$
${\left(\frac{2x}{b}\right)}^{4}=\frac{{\left(2x\right)}^{4}}{{b}^{4}}=\frac{{2}^{4}{x}^{4}}{{b}^{4}}=\frac{16{x}^{4}}{{b}^{4}}$
${\left(\frac{{a}^{3}}{{b}^{5}}\right)}^{7}=\frac{{({a}^{3})}^{7}}{{({b}^{5})}^{7}}=\frac{{a}^{21}}{{b}^{35}}$
$\begin{array}{ccc}{\left(\frac{3{c}^{4}{r}^{2}}{{2}^{3}{g}^{5}}\right)}^{3}=\frac{{3}^{3}{c}^{12}{r}^{6}}{{2}^{9}{g}^{15}}=\frac{27{c}^{12}{r}^{6}}{{2}^{9}{g}^{15}}& \text{or}& \frac{27{c}^{12}{r}^{6}}{512{g}^{15}}\end{array}$
${\left[\frac{\left(a-2\right)}{\left(a+7\right)}\right]}^{4}=\frac{{\left(a-2\right)}^{4}}{{\left(a+7\right)}^{4}}$
${\left[\frac{6x{\left(4-x\right)}^{4}}{2a{\left(y-4\right)}^{6}}\right]}^{2}=\frac{{6}^{2}{x}^{2}{\left(4-x\right)}^{8}}{{2}^{2}{a}^{2}{\left(y-4\right)}^{12}}=\frac{36{x}^{2}{\left(4-x\right)}^{8}}{4{a}^{2}{\left(y-4\right)}^{12}}=\frac{9{x}^{2}{\left(4-x\right)}^{8}}{{a}^{2}{\left(y-4\right)}^{12}}$
$\begin{array}{llll}{\left(\frac{{a}^{3}{b}^{5}}{{a}^{2}b}\right)}^{3}\hfill & =\hfill & {({a}^{3-2}{b}^{5-1})}^{3}\hfill & \text{We}\text{\hspace{0.17em}}\text{can}\text{\hspace{0.17em}}\text{simplify}\text{\hspace{0.17em}}\text{within}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{parentheses}\text{.}\text{\hspace{0.17em}}\text{We}\hfill \\ \hfill & \hfill & \hfill & \text{have}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{rule}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}\text{tells}\text{\hspace{0.17em}}\text{us}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{proceed}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{way}\text{.}\hfill \\ \hfill & =\hfill & {(a{b}^{4})}^{3}\hfill & \hfill \\ \hfill & =\hfill & {a}^{3}{b}^{12}\hfill & \hfill \\ {\left(\frac{{a}^{3}{b}^{5}}{{a}^{2}b}\right)}^{3}\hfill & =\hfill & \frac{{a}^{9}{b}^{15}}{{a}^{6}{b}^{3}}={a}^{9-6}{b}^{15-3}={a}^{3}{b}^{12}\hfill & \text{We}\text{\hspace{0.17em}}\text{could}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{actually}\text{\hspace{0.17em}}\text{used}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{power}\text{\hspace{0.17em}}\text{rule}\text{\hspace{0.17em}}\text{for}\hfill \\ \hfill & \hfill & \hfill & \text{quotients}\text{\hspace{0.17em}}\text{first}\text{.}\text{\hspace{0.17em}}\text{Distribute}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{exponent,}\text{\hspace{0.17em}}\text{then}\hfill \\ \hfill & \hfill & \hfill & \text{simplify}\text{\hspace{0.17em}}\text{using}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{other}\text{\hspace{0.17em}}\text{rules}\text{.}\hfill \\ \hfill & \hfill & \hfill & \text{It}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{probably}\text{\hspace{0.17em}}\text{better,}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{sake}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{consistency,}\hfill \\ \hfill & \hfill & \hfill & \text{to}\text{\hspace{0.17em}}\text{work}\text{\hspace{0.17em}}\text{inside}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{parentheses}\text{\hspace{0.17em}}\text{first.}\hfill \end{array}$
${\left(\frac{{a}^{r}{b}^{s}}{{c}^{t}}\right)}^{w}=\frac{{a}^{rw}{b}^{sw}}{{c}^{tw}}\text{\hspace{0.17em}}$
Make use of the power rule for quotients, the power rule for products, the power rule for powers, or a combination of these rules to simplify each expression.
${\left(\frac{2x}{3y}\right)}^{3}$
$\frac{8{x}^{3}}{27{y}^{3}}$
${\left(\frac{{x}^{2}{y}^{4}{z}^{7}}{{a}^{5}b}\right)}^{9}$
$\frac{{x}^{18}{y}^{36}{z}^{63}}{{a}^{45}{b}^{9}}$
${\left[\frac{2{a}^{4}\left(b-1\right)}{3{b}^{3}\left(c+6\right)}\right]}^{4}$
$\frac{16{a}^{16}{\left(b-1\right)}^{4}}{81{b}^{12}{\left(c+6\right)}^{4}}$
${\left(\frac{8{a}^{3}{b}^{2}{c}^{6}}{4{a}^{2}b}\right)}^{3}$
$8{a}^{3}{b}^{3}{c}^{18}$
${\left[\frac{{\left(9+w\right)}^{2}}{{\left(3+w\right)}^{5}}\right]}^{10}$
$\frac{{\left(9+w\right)}^{20}}{{\left(3+w\right)}^{50}}$
${\left[\frac{5{x}^{4}\left(y+1\right)}{5{x}^{4}\left(y+1\right)}\right]}^{6}$
$1,\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{x}^{4}(y+1)\ne 0$
${\left(\frac{16{x}^{3}{v}^{4}{c}^{7}}{12{x}^{2}v{c}^{6}}\right)}^{0}$
$1,\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{x}^{2}v{c}^{6}\ne 0$
Use the power rules for exponents to simplify the following problems. Assume that all bases are nonzero and that all variable exponents are natural numbers.
${\left(nm\right)}^{7}$
${\left(2a\right)}^{5}$
${\left(2xy\right)}^{5}$
${\left(6mn\right)}^{2}$
${(3{m}^{3})}^{4}$
${(5{x}^{2})}^{3}$
${(8{x}^{2}{y}^{3})}^{2}$
${(2{a}^{5}{b}^{11})}^{0}$
${({m}^{6}{n}^{2}{p}^{5})}^{5}$
${({a}^{4}{b}^{7}{c}^{6}{d}^{8})}^{8}$
${a}^{32}{b}^{56}{c}^{48}{d}^{64}$
${({x}^{2}{y}^{3}{z}^{9}{w}^{7})}^{3}$
${\left(\frac{1}{2}{f}^{2}{r}^{6}{s}^{5}\right)}^{4}$
${\left(\frac{1}{8}{c}^{10}{d}^{8}{e}^{4}{f}^{9}\right)}^{2}$
$\frac{1}{64}{c}^{20}{d}^{16}{e}^{8}{f}^{18}$
${\left(\frac{3}{5}{a}^{3}{b}^{5}{c}^{10}\right)}^{3}$
${(2{a}^{2})}^{4}{(3{a}^{5})}^{2}$
${({a}^{2}{b}^{3})}^{3}{({a}^{3}{b}^{3})}^{4}$
${a}^{18}{b}^{21}$
${({h}^{3}{k}^{5})}^{2}{({h}^{2}{k}^{4})}^{3}$
${({x}^{4}{y}^{3}z)}^{4}{({x}^{5}y{z}^{2})}^{2}$
${x}^{26}{y}^{14}{z}^{8}$
${(a{b}^{3}{c}^{2})}^{5}{({a}^{2}{b}^{2}c)}^{2}$
$\frac{{(6{a}^{2}{b}^{8})}^{2}}{{(3a{b}^{5})}^{2}}$
$4{a}^{2}{b}^{6}$
$\frac{{({a}^{3}{b}^{4})}^{5}}{{({a}^{4}{b}^{4})}^{3}}$
$\frac{{\left({x}^{6}{y}^{5}\right)}^{3}}{{\left({x}^{2}{y}^{3}\right)}^{5}}$
${x}^{8}$
$\frac{{({a}^{8}{b}^{10})}^{3}}{{({a}^{7}{b}^{5})}^{3}}$
$\frac{{\left({m}^{5}{n}^{6}{p}^{4}\right)}^{4}}{{\left({m}^{4}{n}^{5}p\right)}^{4}}$
${m}^{4}{n}^{4}{p}^{12}$
$\frac{{\left({x}^{8}{y}^{3}{z}^{2}\right)}^{5}}{{\left({x}^{6}yz\right)}^{6}}$
$\frac{{\left(10{x}^{4}{y}^{5}{z}^{11}\right)}^{3}}{{\left(x{y}^{2}\right)}^{4}}$
$1000{x}^{8}{y}^{7}{z}^{33}$
$\frac{(9{a}^{4}{b}^{5})(2{b}^{2}c)}{(3{a}^{3}b)(6bc)}$
$\frac{{\left(2{x}^{3}{y}^{3}\right)}^{4}{\left(5{x}^{6}{y}^{8}\right)}^{2}}{{\left(4{x}^{5}{y}^{3}\right)}^{2}}$
$25{x}^{14}{y}^{22}$
${\left(\frac{3x}{5y}\right)}^{2}$
${\left(\frac{3ab}{4xy}\right)}^{3}$
$\frac{27{a}^{3}{b}^{3}}{64{x}^{3}{y}^{3}}$
${\left(\frac{{x}^{2}{y}^{2}}{2{z}^{3}}\right)}^{5}$
${\left(\frac{3{a}^{2}{b}^{3}}{{c}^{4}}\right)}^{3}$
$\frac{27{a}^{6}{b}^{9}}{{c}^{12}}$
${\left(\frac{{4}^{2}{a}^{3}{b}^{7}}{{b}^{5}{c}^{4}}\right)}^{2}$
${\left[\frac{{x}^{2}{\left(y-1\right)}^{3}}{\left(x+6\right)}\right]}^{4}$
$\frac{{x}^{8}{\left(y-1\right)}^{12}}{{\left(x+6\right)}^{4}}$
${\left({x}^{n}{t}^{2m}\right)}^{4}$
${\left(xy\right)}^{\u25b3}$
$\frac{{4}^{3}{a}^{\Delta}{a}^{\square}}{4{a}^{\nabla}}$
${\left(\frac{4{x}^{\Delta}}{2{y}^{\nabla}}\right)}^{\square}$
( [link] ) Is there a smallest integer? If so, what is it?
no
( [link] ) Use the distributive property to expand $5a\left(2x+8\right)$ .
( [link] ) Find the value of $\frac{{\left(5-3\right)}^{2}+{\left(5+4\right)}^{3}+2}{{4}^{2}-2\cdot 5-1}$ .
147
( [link] ) Assuming the bases are not zero, find the value of $(4{a}^{2}{b}^{3})(5a{b}^{4})$ .
( [link] ) Assuming the bases are not zero, find the value of $\frac{36{x}^{10}{y}^{8}{z}^{3}{w}^{0}}{9{x}^{5}{y}^{2}z}$ .
$4{x}^{5}{y}^{6}{z}^{2}$
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