# 2.6 The power rules for exponents

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The symbols, notations, and properties of numbers that form the basis of algebra, as well as exponents and the rules of exponents, are introduced in this chapter. Each property of real numbers and the rules of exponents are expressed both symbolically and literally. Literal explanations are included because symbolic explanations alone may be difficult for a student to interpret.Objectives of this module: understand the power rules for powers, products, and quotients.

## Overview

• The Power Rule for Powers
• The Power Rule for Products
• The Power Rule for quotients

## The power rule for powers

The following examples suggest a rule for raising a power to a power:

${\left({a}^{2}\right)}^{3}={a}^{2}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}\text{​}{a}^{2}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}{a}^{2}$

Using the product rule we get

$\begin{array}{l}{\left({a}^{2}\right)}^{3}={a}^{2+2+2}\hfill \\ {\left({a}^{2}\right)}^{3}={a}^{3\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}2}\hfill \\ {\left({a}^{2}\right)}^{3}={a}^{6}\hfill \end{array}$

$\begin{array}{lll}{\left({x}^{9}\right)}^{4}\hfill & =\hfill & {x}^{9}\cdot {x}^{9}\cdot {x}^{9}\cdot {x}^{9}\hfill \\ {\left({x}^{9}\right)}^{4}\hfill & =\hfill & {x}^{9+9+9+9}\hfill \\ {\left({x}^{9}\right)}^{4}\hfill & =\hfill & {x}^{4\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}9}\hfill \\ {\left({x}^{9}\right)}^{4}\hfill & =\hfill & {x}^{36}\hfill \end{array}$

## Power rule for powers

If $x$ is a real number and $n$ and $m$ are natural numbers,
${\left({x}^{n}\right)}^{m}={x}^{n\cdot m}$

To raise a power to a power, multiply the exponents.

## Sample set a

Simplify each expression using the power rule for powers. All exponents are natural numbers.

$\begin{array}{cc}{\left({x}^{3}\right)}^{4}=\begin{array}{||}\hline {x}^{3\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}4}\\ \hline\end{array}\text{\hspace{0.17em}}{x}^{12}& \text{The}\text{\hspace{0.17em}}\text{box}\text{\hspace{0.17em}}\text{represents}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{step}\text{\hspace{0.17em}}\text{done}\text{\hspace{0.17em}}\text{mentally.}\end{array}$

${\left({y}^{5}\right)}^{3}=\begin{array}{||}\hline {y}^{5\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}3}\\ \hline\end{array}={y}^{15}$

${\left({d}^{20}\right)}^{6}=\begin{array}{||}\hline {d}^{20\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}6}\\ \hline\end{array}={d}^{120}$

${\left({x}^{\square }\right)}^{△}=\text{\hspace{0.17em}}{x}^{\square △}$

Although we don’t know exactly what number $\square △$ is, the notation $\square △$ indicates the multiplication.

## Practice set a

Simplify each expression using the power rule for powers.

${\left({x}^{5}\right)}^{4}$

${x}^{20}$

${\left({y}^{7}\right)}^{7}$

${y}^{49}$

## The power rule for products

The following examples suggest a rule for raising a product to a power:

$\begin{array}{llll}{\left(ab\right)}^{3}\hfill & =\hfill & ab\cdot ab\cdot ab\hfill & \text{Use}\text{\hspace{0.17em}}\text{​}\text{the}\text{\hspace{0.17em}}\text{commutative}\text{\hspace{0.17em}}\text{property}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{multiplication}.\hfill \\ \hfill & =\hfill & aaabbb\hfill & \hfill \\ \hfill & =\hfill & {a}^{3}{b}^{3}\hfill & \hfill \end{array}$

$\begin{array}{lll}{\left(xy\right)}^{5}\hfill & =\hfill & xy\cdot xy\cdot xy\cdot xy\cdot xy\hfill \\ \hfill & =\hfill & xxxxx\cdot yyyyy\hfill \\ \hfill & =\hfill & {x}^{5}{y}^{5}\hfill \end{array}$

$\begin{array}{lll}{\left(4xyz\right)}^{2}\hfill & =\hfill & 4xyz\cdot 4xyz\hfill \\ \hfill & =\hfill & 4\cdot 4\cdot xx\cdot yy\cdot zz\hfill \\ \hfill & =\hfill & 16{x}^{2}{y}^{2}{z}^{2}\hfill \end{array}$

## Power rule for products

If $x$ and $y$ are real numbers and $n$ is a natural number,
${\left(xy\right)}^{n}={x}^{n}{y}^{n}$

To raise a product to a power, apply the exponent to each and every factor.

## Sample set b

Make use of either or both the power rule for products and power rule for powers to simplify each expression.

${\left(ab\right)}^{7}={a}^{7}{b}^{7}$

${\left(axy\right)}^{4}={a}^{4}{x}^{4}{y}^{4}$

$\begin{array}{cc}{\left(3ab\right)}^{2}={3}^{2}{a}^{2}{b}^{2}=9{a}^{2}{b}^{2}& \text{Don't}\text{\hspace{0.17em}}\text{forget}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{apply}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{exponent}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{3!}\end{array}$

$\begin{array}{c}{\left(2st\right)}^{5}={2}^{5}{s}^{5}{t}^{5}=32{s}^{5}{t}^{5}\end{array}$

$\begin{array}{ll}{\left(a{b}^{3}\right)}^{2}={a}^{2}{\left({b}^{3}\right)}^{2}={a}^{2}{b}^{6}\hfill & \text{We}\text{\hspace{0.17em}}\text{used}\text{\hspace{0.17em}}\text{two}\text{\hspace{0.17em}}\text{rules}\text{\hspace{0.17em}}\text{here}\text{.}\text{\hspace{0.17em}}\text{First,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{power}\text{\hspace{0.17em}}\text{rule}\text{\hspace{0.17em}}\text{for}\hfill \\ \hfill & \text{products}\text{.}\text{\hspace{0.17em}}\text{Second,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{power}\text{\hspace{0.17em}}\text{rule}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}\text{powers.}\hfill \end{array}$

$\begin{array}{ll}{\left(7{a}^{4}{b}^{2}{c}^{8}\right)}^{2}\hfill & ={7}^{2}{\left({a}^{4}\right)}^{2}{\left({b}^{2}\right)}^{2}{\left({c}^{8}\right)}^{2}\hfill \\ \hfill & =49{a}^{8}{b}^{4}{c}^{16}\hfill \end{array}$

$\begin{array}{cc}\text{If}\text{\hspace{0.17em}}6{a}^{3}{c}^{7}\ne 0,\text{\hspace{0.17em}}\text{then}\text{\hspace{0.17em}}{\left(6{a}^{3}{c}^{7}\right)}^{0}=1& \text{Recall}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}{x}^{0}=1\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}x\ne 0.\end{array}$

$\begin{array}{ll}{\left[2{\left(x+1\right)}^{4}\right]}^{6}\hfill & ={2}^{6}{\left(x+1\right)}^{24}\hfill \\ \hfill & =64{\left(x+1\right)}^{24}\hfill \end{array}$

## Practice set b

Make use of either or both the power rule for products and the power rule for powers to simplify each expression.

${\left(ax\right)}^{4}$

${a}^{4}{x}^{4}$

${\left(3bxy\right)}^{2}$

$9{b}^{2}{x}^{2}{y}^{2}$

${\left[4t\left(s-5\right)\right]}^{3}$

$64{t}^{3}{\left(s-5\right)}^{3}$

${\left(9{x}^{3}{y}^{5}\right)}^{2}$

$81{x}^{6}{y}^{10}$

${\left(1{a}^{5}{b}^{8}{c}^{3}d\right)}^{6}$

${a}^{30}{b}^{48}{c}^{18}{d}^{6}$

${\left[\left(a+8\right)\left(a+5\right)\right]}^{4}$

${\left(a+8\right)}^{4}{\left(a+5\right)}^{4}$

${\left[\left(12{c}^{4}{u}^{3}{\left(w-3\right)}^{2}\right]}^{5}$

${12}^{5}{c}^{20}{u}^{15}{\left(w-3\right)}^{10}$

${\left[10{t}^{4}{y}^{7}{j}^{3}{d}^{2}{v}^{6}{n}^{4}{g}^{8}{\left(2-k\right)}^{17}\right]}^{4}$

${10}^{4}{t}^{16}{y}^{28}{j}^{12}{d}^{8}{v}^{24}{n}^{16}{g}^{32}{\left(2-k\right)}^{68}$

${\left({x}^{3}{x}^{5}{y}^{2}{y}^{6}\right)}^{9}$

${\left({x}^{8}{y}^{8}\right)}^{9}={x}^{72}{y}^{72}$

${\left({10}^{6}\cdot {10}^{12}\cdot {10}^{5}\right)}^{10}$

${10}^{230}$

## The power rule for quotients

The following example suggests a rule for raising a quotient to a power.

${\left(\frac{a}{b}\right)}^{3}=\frac{a}{b}\cdot \frac{a}{b}\cdot \frac{a}{b}=\frac{a\cdot a\cdot a}{b\cdot b\cdot b}=\frac{{a}^{3}}{{b}^{3}}$

## Power rule for quotients

If $x$ and $y$ are real numbers and $n$ is a natural number,
${\left(\frac{x}{y}\right)}^{n}=\frac{{x}^{n}}{{y}^{n}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\ne 0$

To raise a quotient to a power, distribute the exponent to both the numerator and denominator.

## Sample set c

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.

${\left(\frac{x}{y}\right)}^{6}=\frac{{x}^{6}}{{y}^{6}}$

${\left(\frac{a}{c}\right)}^{2}=\frac{{a}^{2}}{{c}^{2}}$

${\left(\frac{2x}{b}\right)}^{4}=\frac{{\left(2x\right)}^{4}}{{b}^{4}}=\frac{{2}^{4}{x}^{4}}{{b}^{4}}=\frac{16{x}^{4}}{{b}^{4}}$

${\left(\frac{{a}^{3}}{{b}^{5}}\right)}^{7}=\frac{{\left({a}^{3}\right)}^{7}}{{\left({b}^{5}\right)}^{7}}=\frac{{a}^{21}}{{b}^{35}}$

$\begin{array}{ccc}{\left(\frac{3{c}^{4}{r}^{2}}{{2}^{3}{g}^{5}}\right)}^{3}=\frac{{3}^{3}{c}^{12}{r}^{6}}{{2}^{9}{g}^{15}}=\frac{27{c}^{12}{r}^{6}}{{2}^{9}{g}^{15}}& \text{or}& \frac{27{c}^{12}{r}^{6}}{512{g}^{15}}\end{array}$

${\left[\frac{\left(a-2\right)}{\left(a+7\right)}\right]}^{4}=\frac{{\left(a-2\right)}^{4}}{{\left(a+7\right)}^{4}}$

${\left[\frac{6x{\left(4-x\right)}^{4}}{2a{\left(y-4\right)}^{6}}\right]}^{2}=\frac{{6}^{2}{x}^{2}{\left(4-x\right)}^{8}}{{2}^{2}{a}^{2}{\left(y-4\right)}^{12}}=\frac{36{x}^{2}{\left(4-x\right)}^{8}}{4{a}^{2}{\left(y-4\right)}^{12}}=\frac{9{x}^{2}{\left(4-x\right)}^{8}}{{a}^{2}{\left(y-4\right)}^{12}}$

$\begin{array}{llll}{\left(\frac{{a}^{3}{b}^{5}}{{a}^{2}b}\right)}^{3}\hfill & =\hfill & {\left({a}^{3-2}{b}^{5-1}\right)}^{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 & {\left(a{b}^{4}\right)}^{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}$

${\left(\frac{{a}^{r}{b}^{s}}{{c}^{t}}\right)}^{w}=\frac{{a}^{rw}{b}^{sw}}{{c}^{tw}}\text{\hspace{0.17em}}$

## Practice set c

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.

${\left(\frac{a}{c}\right)}^{5}$

$\frac{{a}^{5}}{{c}^{5}}$

${\left(\frac{2x}{3y}\right)}^{3}$

$\frac{8{x}^{3}}{27{y}^{3}}$

${\left(\frac{{x}^{2}{y}^{4}{z}^{7}}{{a}^{5}b}\right)}^{9}$

$\frac{{x}^{18}{y}^{36}{z}^{63}}{{a}^{45}{b}^{9}}$

${\left[\frac{2{a}^{4}\left(b-1\right)}{3{b}^{3}\left(c+6\right)}\right]}^{4}$

$\frac{16{a}^{16}{\left(b-1\right)}^{4}}{81{b}^{12}{\left(c+6\right)}^{4}}$

${\left(\frac{8{a}^{3}{b}^{2}{c}^{6}}{4{a}^{2}b}\right)}^{3}$

$8{a}^{3}{b}^{3}{c}^{18}$

${\left[\frac{{\left(9+w\right)}^{2}}{{\left(3+w\right)}^{5}}\right]}^{10}$

$\frac{{\left(9+w\right)}^{20}}{{\left(3+w\right)}^{50}}$

${\left[\frac{5{x}^{4}\left(y+1\right)}{5{x}^{4}\left(y+1\right)}\right]}^{6}$

$1,\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{x}^{4}\left(y+1\right)\ne 0$

${\left(\frac{16{x}^{3}{v}^{4}{c}^{7}}{12{x}^{2}v{c}^{6}}\right)}^{0}$

$1,\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{x}^{2}v{c}^{6}\ne 0$

## Exercises

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.

${\left(ac\right)}^{5}$

${a}^{5}{c}^{5}$

${\left(nm\right)}^{7}$

${\left(2a\right)}^{3}$

$8{a}^{3}$

${\left(2a\right)}^{5}$

${\left(3xy\right)}^{4}$

$81{x}^{4}{y}^{4}$

${\left(2xy\right)}^{5}$

${\left(3ab\right)}^{4}$

$81{a}^{4}{b}^{4}$

${\left(6mn\right)}^{2}$

${\left(7{y}^{3}\right)}^{2}$

$49{y}^{6}$

${\left(3{m}^{3}\right)}^{4}$

${\left(5{x}^{6}\right)}^{3}$

$125{x}^{18}$

${\left(5{x}^{2}\right)}^{3}$

${\left(10{a}^{2}b\right)}^{2}$

$100{a}^{4}{b}^{2}$

${\left(8{x}^{2}{y}^{3}\right)}^{2}$

${\left({x}^{2}{y}^{3}{z}^{5}\right)}^{4}$

${x}^{8}{y}^{12}{z}^{20}$

${\left(2{a}^{5}{b}^{11}\right)}^{0}$

${\left({x}^{3}{y}^{2}{z}^{4}\right)}^{5}$

${x}^{15}{y}^{10}{z}^{20}$

${\left({m}^{6}{n}^{2}{p}^{5}\right)}^{5}$

${\left({a}^{4}{b}^{7}{c}^{6}{d}^{8}\right)}^{8}$

${a}^{32}{b}^{56}{c}^{48}{d}^{64}$

${\left({x}^{2}{y}^{3}{z}^{9}{w}^{7}\right)}^{3}$

${\left(9x{y}^{3}\right)}^{0}$

1

${\left(\frac{1}{2}{f}^{2}{r}^{6}{s}^{5}\right)}^{4}$

${\left(\frac{1}{8}{c}^{10}{d}^{8}{e}^{4}{f}^{9}\right)}^{2}$

$\frac{1}{64}{c}^{20}{d}^{16}{e}^{8}{f}^{18}$

${\left(\frac{3}{5}{a}^{3}{b}^{5}{c}^{10}\right)}^{3}$

${\left(xy\right)}^{4}\left({x}^{2}{y}^{4}\right)$

${x}^{6}{y}^{8}$

${\left(2{a}^{2}\right)}^{4}{\left(3{a}^{5}\right)}^{2}$

${\left({a}^{2}{b}^{3}\right)}^{3}{\left({a}^{3}{b}^{3}\right)}^{4}$

${a}^{18}{b}^{21}$

${\left({h}^{3}{k}^{5}\right)}^{2}{\left({h}^{2}{k}^{4}\right)}^{3}$

${\left({x}^{4}{y}^{3}z\right)}^{4}{\left({x}^{5}y{z}^{2}\right)}^{2}$

${x}^{26}{y}^{14}{z}^{8}$

${\left(a{b}^{3}{c}^{2}\right)}^{5}{\left({a}^{2}{b}^{2}c\right)}^{2}$

$\frac{{\left(6{a}^{2}{b}^{8}\right)}^{2}}{{\left(3a{b}^{5}\right)}^{2}}$

$4{a}^{2}{b}^{6}$

$\frac{{\left({a}^{3}{b}^{4}\right)}^{5}}{{\left({a}^{4}{b}^{4}\right)}^{3}}$

$\frac{{\left({x}^{6}{y}^{5}\right)}^{3}}{{\left({x}^{2}{y}^{3}\right)}^{5}}$

${x}^{8}$

$\frac{{\left({a}^{8}{b}^{10}\right)}^{3}}{{\left({a}^{7}{b}^{5}\right)}^{3}}$

$\frac{{\left({m}^{5}{n}^{6}{p}^{4}\right)}^{4}}{{\left({m}^{4}{n}^{5}p\right)}^{4}}$

${m}^{4}{n}^{4}{p}^{12}$

$\frac{{\left({x}^{8}{y}^{3}{z}^{2}\right)}^{5}}{{\left({x}^{6}yz\right)}^{6}}$

$\frac{{\left(10{x}^{4}{y}^{5}{z}^{11}\right)}^{3}}{{\left(x{y}^{2}\right)}^{4}}$

$1000{x}^{8}{y}^{7}{z}^{33}$

$\frac{\left(9{a}^{4}{b}^{5}\right)\left(2{b}^{2}c\right)}{\left(3{a}^{3}b\right)\left(6bc\right)}$

$\frac{{\left(2{x}^{3}{y}^{3}\right)}^{4}{\left(5{x}^{6}{y}^{8}\right)}^{2}}{{\left(4{x}^{5}{y}^{3}\right)}^{2}}$

$25{x}^{14}{y}^{22}$

${\left(\frac{3x}{5y}\right)}^{2}$

${\left(\frac{3ab}{4xy}\right)}^{3}$

$\frac{27{a}^{3}{b}^{3}}{64{x}^{3}{y}^{3}}$

${\left(\frac{{x}^{2}{y}^{2}}{2{z}^{3}}\right)}^{5}$

${\left(\frac{3{a}^{2}{b}^{3}}{{c}^{4}}\right)}^{3}$

$\frac{27{a}^{6}{b}^{9}}{{c}^{12}}$

${\left(\frac{{4}^{2}{a}^{3}{b}^{7}}{{b}^{5}{c}^{4}}\right)}^{2}$

${\left[\frac{{x}^{2}{\left(y-1\right)}^{3}}{\left(x+6\right)}\right]}^{4}$

$\frac{{x}^{8}{\left(y-1\right)}^{12}}{{\left(x+6\right)}^{4}}$

${\left({x}^{n}{t}^{2m}\right)}^{4}$

$\frac{{\left({x}^{n+2}\right)}^{3}}{{x}^{2n}}$

${x}^{n+6}$

${\left(xy\right)}^{△}$

$\frac{{4}^{3}{a}^{\Delta }{a}^{\square }}{4{a}^{\nabla }}$

${\left(\frac{4{x}^{\Delta }}{2{y}^{\nabla }}\right)}^{\square }$

## Exercises for review

( [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 $\left(4{a}^{2}{b}^{3}\right)\left(5a{b}^{4}\right)$ .

( [link] ) Assuming the bases are not zero, find the value of $\frac{36{x}^{10}{y}^{8}{z}^{3}{w}^{0}}{9{x}^{5}{y}^{2}z}$ .

$4{x}^{5}{y}^{6}{z}^{2}$

#### Questions & Answers

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Maira Reply
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Maira Reply
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Damian
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Professor
I think
Professor
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Brian Reply
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Rafiq
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Damian
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LITNING Reply
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LITNING
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Bob
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brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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?
Kyle
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biomolecules are e building blocks of every organics and inorganic materials.
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Smarajit Reply
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