\hfill \thepage} %} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.4in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\large \color{blue}Lecture Notes} \lfoot{} \cfoot{} \chead{\Large Integer Exponents} \rhead{\small page \thepage} \textwidth 7.7in \textheight 9.7in \setlength{\headheight}{27pt} \lfoot{\footnotesize \copyright \; Hidegkuti, Powell, 2008} \rfoot{\footnotesize Last revised: October 1, 2018} \newcommand{\defstart}{ \begin{boxedminipage}{7in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\rightskip}{20pt} \setlength{\leftskip}{20pt} \color{white} . \color{black} } \newcommand{\defcont}{ \setlength{\hangindent}{57pt} } \newcommand{\definside}{ \setlength{\hangindent}{57pt} \setlength{\parindent}{57pt} } \newcommand{\defend}{ \end{boxedminipage} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.06in} \setstretch{1.08} } \newcommand{\exstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{60pt} \setlength{\parskip}{0.04in} \setstretch{1.08} } \newcommand{\excont}{ \setlength{\parindent}{60pt} \setlength{\hangindent}{75pt} } \newcommand{\solstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{63pt} } \newcommand{\solcont}{ \setlength{\parindent}{63pt} \setlength{\hangindent}{63pt} } \newcommand{\solrestart}{ \setlength{\parindent}{48pt} \setlength{\hangindent}{63pt} } \newcommand{\sol}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{50pt} } \newcommand{\solinside}{ \setlength{\parindent}{50pt} \setlength{\hangindent}{50pt} \setstretch{1.08} } \newcommand{\vissza}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.1in} \setstretch{1.08} \normalsize \setcounter{enumi}{0}\RESUME } \newcommand{\matek}{ \setstretch{1.04} \normalsize \setlength{\parskip}{0.04in} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setcounter{enumi}{0}\RESUME } \setstretch{1.06} \begin{document} \begin{center} {\Large Part 1 - The History Thus Far and the Problem} \end{center} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Recall what we know about exponentiation thus far. \ Exponential notation expresses repeated multiplication.\vspace{0.1in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{\TeXButton{\defcont}{\defcont}}% %BeginExpansion \defcont% %EndExpansion \textbf{Definition}: \ $\ $We define $2^{7}$ to denote the factor $2$ multiplied by itself repeatedly, such as% \begin{equation*} \underset{\text{7 factors}}{\underbrace{~2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2~}}=2^{7} \end{equation*}% %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.1in} When mathematicians agreed to this definition, that was a free choice. They could have gone with other definitions. \ Once this definition exists, however, certain properties are automatically true, and we have no other option but to recognize them as true. \ They just fell into our laps.\vspace{% 0.1in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem 1.} \ $\ $If $a$ is any number and $m$, $n$ are any positive integers, then $a^{n}\cdot a^{m}=a^{n+m}$\vspace{0.1in} \textbf{Theorem 2.} \ $\ $If $a$ is any non-zero number and $m$, $n$ are any positive integers, then $\dfrac{a^{n}}{a^{m}}=a^{n-m}$\vspace{0.1in} \textbf{Theorem 3.} \ $\ $If $a\ $is any number and $m$, $n$ are any positive integers, then $\left( a^{n}\right) ^{m}=a^{nm}$\vspace{0.1in} \textbf{Theorem 4.} \ $\ $If $a,$ $b$ are any numbers and $n$ is any positive integer, then $\left( ab\right) ^{n}=a^{n}b^{n}$\vspace{0.1in} \textbf{Theorem 5.} \ $\ $If $a,$ $b$ are any numbers, $b\not=0$, and $n$ is any positive integer, then $\left( \dfrac{a}{b}\right) ^{n}=\dfrac{a^{n}}{% b^{n}}$\vspace{0.1in} %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.1in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Again, the definition, immediately followed by the theorems. \ And then there was a quiet. \ \ Another opening for a free choice. \ \vspace{0.1in} Consider the expression $2^{x}$. \ The problem is that the definition of exponentiation only allows for a positive integer value of $x$. \ The expression $2^{x}$ is meaningful for $x=2$ or $9$ or $100$, but it is not meaningful for values of $x$ such as $-3$ or $\dfrac{3}{5}$ or $3.2$. \ In short, the world of exponents was just the set of all natural numbers. \ Mathematicians usually don't like that. \ The best case scenario, the ultimate hope is that the definition of exponents could be extended to any number for $x$. \ That way, $2^{x}$ would be meaningful, no matter what the value of $x$ is.\vspace{0.1in} So, one of the issues was the desire to grow our world of exponents beyond the set of all natural numbers. \ This will be achieved in several steps. \ Today, we are only focusing on enlarging the world of exponents from $% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $ to $% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $ (i.e. from the set of all natural numbers to the set of all integers). The other issue was that \ as we enlarge our world, we pay especial attention that the new definitions will not conflict with the mathematics we already have. \ This principle comes up often in our choices, and it is sometimes called the \textbf{expansion principle}.\vspace{0.1in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{\TeXButton{\defcont}{\defcont}}% %BeginExpansion \defcont% %EndExpansion \textbf{Definition}: \ $\ $In many situations, mathematicians attempt to increase, to enlarge our world. \ The \textbf{expansion principle}\ is that when we enlarge our mathematics by adding new definitions, we do so in such a way that the new definitions never create conflicts with the mathematics we already have. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.1in} \pagebreak \begin{center} {\Large Part 2 - Integer Exponents} \end{center} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Suppose we want to define $2^{0}$. \ The repeated multiplication definition can not be applied to zero, so we have complete freedom to define $2^{0}$. \ As it turns out, if we insist on a definition that does not conflict with Rule 2, $\dfrac{a^{n}}{a^{m}}=a^{n-m}$, then we do not have all that many choices for $2^{0}$. \ Let us think of zero as the result of the subtraction $3-3$, and that we would like to define $2^{0}$ so that Rule 2 is still true. \ \begin{equation*} 2^{0}=2^{3-3}~~\overset{\text{rule 2}}{=}~~\dfrac{2^{3}}{2^{3}}=\dfrac{8}{8}% =1 \end{equation*}% This is an expansion principle proof. \ It did not prove that the value of $% 2^{0}$ is or must be zero. \ It showed much less; that if we wanted to define $2^{0}$ without harming Rule 2 in the example given, then the only possible value for $2^{0}$ is $1$. \ The reader should imagine a team of mathematicians making first sure that no part of our good old math is hurt if we define $2^{0}=1$. \ And as it turned out, this is exactly the case.% \vspace{0.1in} This computation can be repeated with many different bases. \ For example,% \begin{equation*} 5^{0}=5^{2-2}~~\overset{\text{rule 2}}{=}~~\dfrac{5^{2}}{5^{2}}=\dfrac{25}{25% }=1\text{ \ \ \ or \ \ \ \ }\left( -3\right) ^{0}=\left( -3\right) ^{2-2}~~% \overset{\text{rule 2}}{=}~~\dfrac{\left( -3\right) ^{2}}{\left( -3\right) ^{2}}=\dfrac{9}{9}=1 \end{equation*}% The only base that is problematic is $0$. \ Indeed, division by zero is not allowed and Rule 2, $\dfrac{a^{n}}{a^{m}}=a^{n-m}$ does not work with $a=0$. \ If we try to perform the same computation with zero, we ultimately end up in $\dfrac{0}{0}$ which is undefined. \vspace{0.1in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem 6.} \ If $a$ is any non-zero number, then $a^{0}=1$. \ \vspace{0.08in} $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0^{0}$ is undefined. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.1in} Please note that as we extend our world of exponents, old issues might re-surface. \ For example, $\left( -3\right) ^{0}=1$ but $-3^{0}=-1$ is an important distinction, but not a new one. \ Now that we have defined zero exponent, we will similarly try to define negative integer exponents such as $2^{-3}$. Again, the original definition can not be applied. \ We cannot write down the factor two negative three times. \ So we have a freedom here to define $% 2^{-3}$ in any way we wish. \ In this decision, we will again use the expansion principle: that we would like to keep our old rules after having $% 2^{-3}$ defined. \ We will again use Rule 2, $\dfrac{a^{n}}{a^{m}}=a^{n-m}$ \ and write $-3$ as a subtraction between two positive integers.% \begin{equation*} 2^{-3}=2^{1-4}~\overset{\text{Rule 2}}{=}~\dfrac{2^{1}}{2^{4}}=\dfrac{2}{16}=% \dfrac{1}{8}=\dfrac{1}{2^{3}}\text{ \ or, more elegantly, }2^{-3}=2^{1-4}~% \overset{\text{Rule 2}}{=}~\dfrac{2^{1}}{2^{4}}=\dfrac{\NEG{2}}{\NEG{2}\cdot 2\cdot 2\cdot 2}=\dfrac{1}{2^{3}} \end{equation*}% When we discovered this rule, we saw that it was true because of cancellation. \ In case of a negative exponent, we have the same cancellation, it's just that we run out of factors in the numerator first. \ The computation can be repated with any base except for zero.\vspace{0.1in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem 7.} \ If $a$ is any non-zero number, and $n$ is any positive integer, then $a^{-n}=\dfrac{1}{a^{n}}$.\vspace{0.08in} $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0^{-n}$ is undefined. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.1in} \pagebreak %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 1.} \ Simplify each of the following expressions. \ Use only positive exponents in your answer. %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $5^{-2}$ \ \ \ \ \ \ \ \ b) \ $a^{-5}$ \ \ \ \ \ \ \ \ \ c) \ $\dfrac{1% }{3^{-2}}$ \ \ \ \ \ \ \ \ \ d) \ $\left( \dfrac{2}{3}\right) ^{-3}$ \ \ \ \ \ \ \ \ \ \ e) \ $\dfrac{1}{x^{-3}}$ \ \ \ \ \ \ \ \ \ f) \ $2x^{-3}$ %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ Recall our new rule, $a^{-n}=\dfrac{1}{a^{n}}$. \ \ We apply this rule: \ \ $5^{-2}=\dfrac{1}{5^{2}}=\,$\fbox{$\dfrac{1}{25}$}.% \vspace{0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ We can use the same rule again:\ $a^{-5}=\,$\fbox{$\dfrac{1}{a^{5}}$}.% \vspace{0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ In this case, the expression with the negative exponent is in the denominator. \ \ %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion The short story is that $\dfrac{1}{3^{-2}}=3^{2}=9$.\vspace{0.03in} \ The long story is that we apply our new rule $a^{-n}=\dfrac{1}{a^{n}}$ and then we divide by mutiplying by the reciprocal. \ \vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{1}{3^{-2}}=\dfrac{~~1~~}{\dfrac{1}{3^{2}% }}=\dfrac{~~\dfrac{1}{1}~~}{\dfrac{1}{3^{2}}}=\dfrac{1}{1}\cdot \dfrac{3^{2}% }{1}=\dfrac{9}{1}=\,$\fbox{$9$}\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion So, $\dfrac{1}{a^{-n}}$ can be re-written as $a^{n}$.\vspace{0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion d) \ In this case, the expression with the negative exponent is already a fraction.\vspace{0.03in} \ \ %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion The short story is that $\ \left( \dfrac{2}{3}\right) ^{-3}=\left( \dfrac{3}{% 2}\right) ^{3}=\dfrac{27}{8}$.\vspace{0.03in} \ The long story is that we apply our new rule $a^{-n}=\dfrac{1}{a^{n}}$ and then we divide by mutiplying by the reciprocal. \vspace{0.03in}\ %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \dfrac{2}{3}\right) ^{-3}=\dfrac{~~1~~}{% \left( \dfrac{2}{3}\right) ^{3}}=\dfrac{~~\dfrac{1}{1}~~}{\dfrac{8}{27}}=% \dfrac{1}{1}\cdot \dfrac{27}{8}=\,$\fbox{$\dfrac{27}{8}$}\vspace{0.03in}% \vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion This computation shows that $\left( \dfrac{a}{b}\right) ^{-n}=\left( \dfrac{b% }{a}\right) ^{n}$.\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion e) \ The short story is that $\dfrac{1}{a^{-n}}$ can be re-written as $a^{n}$% . \ The computation below justifies this step.\ \vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{1}{x^{-3}}=\dfrac{~~1~~}{\dfrac{1}{x^{3}% }}=\dfrac{~~\dfrac{1}{1}~~}{\dfrac{1}{x^{3}}}=\dfrac{1}{1}\cdot \dfrac{x^{3}% }{1}=\dfrac{x^{3}}{1}=\,$\fbox{$x^{3}$}\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion So, $\dfrac{1}{a^{-n}}$ can be re-written as $a^{n}$.\vspace{0.03in}\vspace{% 0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion f) \ It is a common mistake to interpret $2x^{-3}$ as $\left( 2x\right) ^{-3} $. \ Without the parentheses, we perform the exponentiation before the multiplication. \ Therefore, the correct computation is\vspace{0.03in}% \vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x^{-3}=2\cdot x^{-3}=\dfrac{2}{1}\cdot \dfrac{1}{x^{3}}=\,$\fbox{$\dfrac{2}{x^{3}}$}.\vspace{0.03in}\vspace{0.03in} \pagebreak %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{\TeXButton{\defcont}{\defcont}}% %BeginExpansion \defcont% %EndExpansion \textbf{Theorem:} \ The following statements are practical applications of the rule $a^{-n}=\dfrac{1}{a^{n}}$ and frequently occur in computations. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{1}{a^{-n}}=a^{n}$ \ \ \ \ \ \ and \ \ $\left( \dfrac{a}{b}\right) ^{-n}=\left( \dfrac{b}{a}% \right) ^{n}$ %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \textit{Proof}: \ As the computation shows, we apply the rule $a^{-n}=\dfrac{% 1}{a^{n}}$ and then perform the division by multiplying by the reciprocal. $\ \ \ \ \ \ \ \ \dfrac{1}{a^{-n}}=\dfrac{~~1~~}{\dfrac{1}{a^{n}}}=\dfrac{~~% \dfrac{1}{1}~~}{\dfrac{1}{a^{n}}}=\dfrac{1}{1}\cdot \dfrac{a^{n}}{1}=\dfrac{% a^{n}}{1}=a^{n}$ \ \ and \ $\left( \dfrac{a}{b}\right) ^{-n}=\dfrac{1}{% \left( \dfrac{a}{b}\right) ^{n}}=\dfrac{\dfrac{1}{1}}{\dfrac{a^{n}}{b^{n}}}=% \dfrac{1}{1}\cdot \dfrac{b^{n}}{a^{n}}=\dfrac{b^{n}}{a^{n}}=\left( \dfrac{b}{% a}\right) ^{n}$ \ $\blacksquare $\ (\textit{end of proof}) %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \bigskip %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 2.} \ Re-write the expression $\dfrac{a^{3}b^{-5}}{% c^{-2}d^{4}}$ using only positive exponents. \vspace{0.03in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ We re-write the expressions with negative exponents using the rule $a^{-n}=\dfrac{1}{a^{n}}$.\vspace{0.03in} %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{% a^{3}b^{-5}}{c^{-2}d^{4}}=\dfrac{a^{3}\cdot \dfrac{1}{b^{5}}}{\dfrac{1}{c^{2}% }\cdot d^{4}}=\dfrac{\dfrac{a^{3}}{1}\cdot \dfrac{1}{b^{5}}}{\dfrac{1}{c^{2}}% \cdot \dfrac{d^{4}}{1}}=\dfrac{~~\dfrac{a^{3}}{b^{5}}~~}{\dfrac{d^{4}}{c^{2}}% }=\dfrac{a^{3}}{b^{5}}\cdot \dfrac{c^{2}}{d^{4}}=\,\fbox{$\dfrac{a^{3}c^{2}}{% b^{5}d^{4}}$}$.\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Notice the pattern here. If\ a factor with a negative exponent is in the numerator, we can re-write it with a positive exponent in the denominator. \ Also, if\ a factor with a negative exponent is in the denominator, we can re-write it with a positive exponent in the numerator. %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{\TeXButton{\defcont}{\defcont}}% %BeginExpansion \defcont% %EndExpansion \textbf{Theorem:} \ $\dfrac{a^{-n}b^{m}}{c^{p}d^{-q}}=\dfrac{b^{m}d^{q}}{% a^{n}c^{p}}$ \ \ where $a,c,d$ are any non-zero numbers and $n$, $m$, $p$, $% q $ are positive integers. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion The definitions of $a^{0}$ \ and $a^{-n}$ were developed with the intention that the previous rules (1 through 5) will remain true. \ Keep that in mind in case of computations with more complex exponential expressions. %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 3.} \ Simplify each of the given expressions. \ Present your answer using only positive exponents.\vspace{0.03in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $\left( a^{-2}\right) ^{-5}$ \ \ \ \ \ \ \ \ b) \ $\dfrac{\left( -x^{-2}\right) ^{-3}}{x^{-6}\left( -x\right) ^{-4}}$ \ \ \ \ \ \ \ \ \ \ c) \ $\dfrac{a^{-3}}{a^{-8}}$ \ \ \ \ \ \ \ \ \ \ \ \ d) \ $\dfrac{a^{-2}b^{-3}% }{a^{-5}b^{3}}$ \ \ \ \ \ \ \ e) \ $\dfrac{\left( 2a^{-4}b^{3}\right) ^{-5}}{% \left( 3a^{3}b^{-2}\right) ^{0}}$\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ It is much preferred to first simplify the exponent. \ Repeated exponentiation means multiplication in the exponent.% \vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $\left( a^{-2}\right) ^{-5}=a^{-2\left( -5\right) }=\,$\fbox{$a^{10}$}% \vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ Let us re-write the solo negative signs as multiplications by $-1$. \ Then we will use the rules of exponents to simplify the exponents. \ \ Only after that will we address negative exponents.\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{\left( -x^{-2}\right) ^{-3}}{x^{-6}\left( -x\right) ^{-4}}=\dfrac{\left( -1\cdot x^{-2}\right) ^{-3}}{x^{-6}\left( -1\cdot x\right) ^{-4}}=\dfrac{\left( -1\right) ^{-3}\left( x^{-2}\right) ^{-3}}{x^{-6}\left( -1\right) ^{-4}x^{-4}% }=\dfrac{\left( -1\right) ^{-3}x^{6}}{x^{-6}\left( -1\right) ^{-4}x^{-4}}$% \vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion Now we get rid of all negative exponents by moving the factors. \ A factor with exponent $-5$ in the numerator can be re-written as a factor with exponent $5$ in the denominator, and vica versa.\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{\left( -1\right) ^{-3}x^{6}}{x^{-6}\left( -1\right) ^{-4}x^{-4}}=\dfrac{\left( -1\right) ^{4}x^{6}x^{6}x^{4}}{\left( -1\right) ^{3}}=\dfrac{1\cdot x^{16}}{% -1}=\,$\fbox{$-x^{16}$}\vspace{0.03in}\vspace{0.03in} \pagebreak %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ Solution 1: \ apply the rule $\dfrac{a^{n}}{a^{m}}=a^{n-m}$. \ \ \ $% \dfrac{a^{-3}}{a^{-8}}=a^{-3-\left( -8\right) }=a^{-3+8}=\,$\fbox{$a^{5}$}\ \vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion Solution 2: \ First we get rid of negative exponents and then apply the rule $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{a^{-3}}{a^{-8}% }=\dfrac{a^{8}}{a^{3}}=a^{8-3}=\,$\fbox{$a^{5}$}\ \vspace{0.03in}\vspace{% 0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion d) \ First we get rid of negative exponents.\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{a^{-2}b^{-3}}{% a^{-5}b^{3}}=\dfrac{a^{5}}{a^{2}b^{3}b^{3}}=\dfrac{a^{5}}{a^{2}b^{6}}=\,$% \fbox{$\dfrac{a^{3}}{b^{6}}$}\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion e) \ We can save a lot of work\ by noticing that the denominator is just $1,$ because any non-zero quantity raised to the power zero is $1$, and so $% \left( 3a^{3}b^{-2}\right) ^{0}=1$.\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{\left( 2a^{-4}b^{3}\right) ^{-5}}{\left( 3a^{3}b^{-2}\right) ^{0}}=% \dfrac{2^{-5}\left( a^{-4}\right) ^{-5}\left( b^{3}\right) ^{-5}}{1}=\dfrac{% 2^{-5}a^{20}b^{-15}}{1}=\dfrac{a^{20}}{2^{5}b^{15}}=\,$\fbox{$\dfrac{a^{20}}{% 32b^{15}}$}\vspace{0.03in}\vspace{0.03in}\vspace{0.03in}\vspace{0.03in}% \vspace{0.1in} \begin{center} {\Large Part 3 - Scientific Notation Revisited} \end{center} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion When we first saw scientific notation, we learned to use it to handle uncomfortbly large numbers. Recall the definition of scientific notation: %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{\TeXButton{\defcont}{\defcont}}% %BeginExpansion \defcont% %EndExpansion \textbf{Definition:} \ We can write numbers in scientific notation. \ This means to write a number as a product of two numbers. \ The first number is between $1$ and $10$ (can be $1$ but must be less than $10$), and the second number is a $10-$power. \ For example, the scientific notation for \newline $428\,\allowbreak 600\,000\,000$ is $4.286\times 10^{11}$. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion With negative exponents, we can also use scientific notation to handle extremely small numbers. \ For example, the mass of an electron is $% 0.00000000000000000000000000091094$ grams. \ Instead of hurds of trailing zeroes, now we are faced with many zeroes after the decimal point. \ This number can be re-written as $9.\,\allowbreak 109\,4\cdot 10^{-28}$.\vspace{% 0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 4. \ }Re-write the number $0.0000000317$ using scientific notation.\vspace{0.03in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ The first number in scientific notation needs to be between $1$ and $10$. \ In this case, this number is $3.17$. \ We just need to figure out the $10-$power in the second part. \ We count how many decimal places we move the decimal from $0.0000000317$ to $3.17$. \ We count $8$ decimal places. \ So the correct answer is $\,$\fbox{$3.17\cdot 10^{-8}$}.% \vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 5. }\ Suppose that $A=3.8\cdot 10^{15}$ \ and \ $B=6.5\cdot 10^{-8}$. \ Perform each of the following operations. \ Present your answer using scientific notation.\vspace{0.03in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $B^{2}\ \ \ \ \ \ \ \ \ $ \ \ \ \ \ \ \ b) \ $AB^{2}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c) \ $\dfrac{B}{A}$ \ \vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ We will apply rules of exponents.\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $B^{2}=\left( 6.5\cdot 10^{-8}\right) ^{2}=6.5^{2}\cdot \left( 10^{-8}\right) ^{2}=42.25\cdot 10^{-16}$\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion This number is not in scientific notation because $42.25$ is too large for the first part of scientific notation. \ Recall that the first factor must be between $1$ and $10$. \ So we re-write $42.25$ as $4.225\cdot 10$.\vspace{% 0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $B^{2}=42.25\cdot 10^{-16}=4.225\cdot 10\cdot 10^{-16}=4.225\cdot 10^{1+\left( -16\right) }=\,$\fbox{$4.225\cdot 10^{-15}$}\vspace{0.03in}% \vspace{0.03in} \pagebreak %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ We will apply rules of exponents.\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $AB^{2}=\left( 3.8\cdot 10^{15}\right) \left( 6.5\cdot 10^{-8}\right) ^{2}=3.8\cdot 10^{15}\cdot 6.5^{2}\cdot \left( 10^{-8}\right) ^{2}=3.8\cdot 10^{15}\cdot 42.25\cdot 10^{-16}$\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion $\ \ \ \ \ \ =\left( 3.8\cdot 42.25\right) \cdot \left( 10^{15}\cdot 10^{-16}\right) =160.\,\allowbreak 55\cdot 10^{15+\left( -16\right) }=160.\,\allowbreak 55\cdot 10^{-1}=16.055$\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion This number is not in scientific notation because $16.055$ is too large for the first part of scientific notation.\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $AB^{2}=16.055=\,\fbox{$1.6055\cdot 10$}$.\vspace{0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ $\dfrac{B}{A}=\dfrac{6.5\cdot 10^{-8}}{3.8\cdot 10^{15}}=\left( \dfrac{% 6.5}{3.8}\right) \cdot 10^{-8-15}=\,\fbox{$1.\,\allowbreak 710\,5\cdot 10^{-23}$}$.\vspace{0.03in}\vspace{0.07in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \FRAME{itbpF}{0.8553in}{0.6962in}{0.2811in}{}{}{sample.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.8553in;height 0.6962in;depth 0.2811in;original-width 8.4267in;original-height 6.8441in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'sample.jpg';file-properties "XNPEU";}}\ \ {\LARGE Sample Problems} Simplify each of the following. \ Assume that all variables represent positive numbers. \ \ Present your answer without negative exponents.\medskip\ \begin{enumerate} \item %TCIMACRO{\TeXButton{4col begin}{\begin{multicols}{4}}}% %BeginExpansion \begin{multicols}{4}% %EndExpansion $3^{-2}$\vspace{0.03in} \item $\dfrac{1}{2^{-3}}$\vspace{0.03in} \item $m^{-4}$\vspace{0.03in} \item $\dfrac{1}{x^{-5}}$\vspace{0.03in} \item $a^{8}\cdot a^{-1}$\vspace{0.03in} \item $p^{3}\left( p^{-7}\right) p^{8}$\vspace{0.03in} \item $\dfrac{x^{-4}}{x^{-9}}$\vspace{0.03in} \item $\dfrac{50a^{12}}{10a^{-3}}$\vspace{0.03in} \item $\dfrac{t^{-3}}{t^{4}}$\vspace{0.03in} \item $x^{0}$\vspace{0.03in} \item $-x^{0}$\vspace{0.03in} \item $\left( -x\right) ^{0}$\vspace{0.03in} \item $\left( b^{-5}\right) \left( b^{2}\right) \left( b^{-1}\right) $% \vspace{0.03in} \item $\dfrac{1}{\left( b^{-5}\right) \left( b^{2}\right) \left( b^{-1}\right) }$\vspace{0.03in} \item $\dfrac{m^{-2}}{m^{-5}}$\vspace{0.03in} \item $\dfrac{x^{3}y^{-5}}{z^{-4}}$\vspace{0.03in} \item $\dfrac{18q^{3}}{6q^{-3}}$\vspace{0.03in} \item $\left( \dfrac{2}{3}\right) ^{-3}$\vspace{0.03in} \item $2y^{-3}$\vspace{0.03in} \item $\left( 2y\right) ^{-3}$\vspace{0.03in} \item $\left( -\dfrac{3}{5}\right) ^{-2}$\vspace{0.03in} \item $\dfrac{a^{3}b^{-5}}{a^{-2}b^{3}}$\vspace{0.03in} \item $\left( 3m^{3}\right) ^{-2}$\vspace{0.03in} \item $\left( -2ab^{-3}\right) ^{-3}$\vspace{0.03in} \item $\dfrac{\left( k^{3}\right) ^{-3}}{\left( k^{-5}\right) ^{2}}$\vspace{% 0.03in}% %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \item $\left( \dfrac{2a^{-3}b^{5}}{-3a^{3}b^{-2}}\right) ^{-2}\left( a^{3}b^{-5}\right) ^{-3}$\vspace{0.03in} \item $\left( -2a^{-3}\right) \left( -2a^{-2}b\right) ^{-4}$\vspace{0.03in} \item $\dfrac{\left( -3p^{3}q^{5}\right) ^{2}}{\left( 2q^{0}p^{3}\right) ^{-1}}$\vspace{0.03in} \item $\left( \dfrac{2a^{-2}b^{3}}{-2^{2}\left( a^{-1}b\right) ^{-3}}\right) ^{-2}$\vspace{0.03in} \item $\left( -\dfrac{x^{3}y^{0}x^{-5}}{y^{-3}}\right) ^{-2}$\vspace{0.03in} \item $\left( -\dfrac{x^{3}y^{7}x^{-5}}{y^{-3}}\right) ^{0}$\vspace{0.03in} \item $\dfrac{x^{-1}+y^{-1}}{x^{-2}-y^{-2}}$\vspace{0.03in} \item $\dfrac{\left( -2a^{-2}\right) ^{-2}b^{3}a^{0}\left( -aba^{-2}b^{-2}\right) ^{-3}}{2a^{2}\left( -2a^{-2}b\right) ^{-2}ab^{0}}$% \vspace{0.03in} \item $\left( \dfrac{-a^{2}\left( b^{-1}a\right) ^{-5}}{b^{7}\left( -ab^{2}\right) ^{-3}}\right) ^{-2}$\vspace{0.03in} \item $\dfrac{\left( x^{-2}\right) ^{-2}y^{3}x^{0}\left( -2yx^{0}y^{-2}x^{-2}\right) ^{0}}{yx^{5}\left( y^{-2}x\right) ^{-3}\left( 2x^{-1}yx^{3}\right) ^{-1}}$% %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \ \ \item Suppose that $x=8.5\cdot 10^{-12}$ \ and \ $y=7.5\cdot 10^{7}$. \ Perform each of the following operations. \ Present your answer using scientific notation.\vspace{0.03in} a) \ $xy$ \ \ \ \ \ \ \ \ b) \ $x^{3}\ \ \ \ \ \ \ \ \ \ \ \ $c) \ $xy^{2}\ \ \ \ \ \ \ \ \ \ \ $d) \ $\dfrac{x}{y}\ \ \ \ \ \ \ \ \ \ \ \ $e) \ $\dfrac{% y}{x^{5}}$ \end{enumerate} \ \pagebreak \bigskip {\LARGE \FRAME{itbpF}{0.9055in}{0.9055in}{0.1609in}{}{}{answers.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.9055in;height 0.9055in;depth 0.1609in;original-width 8.6455in;original-height 8.6455in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'answers.jpg';file-properties "XNPEU";}} \ \ \ Answers}\vspace{0.2in} {\Large Sample Problems}{\large \ }% %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \begin{enumerate} \item[1.] $\dfrac{1}{9}$ \ \ \ \ \ \ \ 2. $8$ \ \ \ \ \ \ \ 3. \ $\dfrac{1}{% m^{4}}$ \ \ \ \ \ \ \ \ 4. \ $x^{5}$ \ \ \ \ \ \ \ 5. \ $a^{7}$ \ \ \ \ \ \ \ \ 6. \ $p^{4}$ \ \ \ \ \ \ \ 7. \ $x^{5}$ \ \ \ \ \ \ \ 8. \ $5a^{15}$ \ \ \ \ \ \ 9. \ $\dfrac{1}{t^{7}}$ \ \ \ \ \ \ 10. \ $1$ \item[11.] $-1$ \ \ \ \ \ 12. \ $1$ \ \ \ \ \ \ \ 13. \ $\dfrac{1}{b^{4}}$ \ \ \ \ \ 14. \ $b^{4}$ \ \ \ \ \ 15. \ $m^{3}$ \ \ \ \ \ \ 16. \ $\dfrac{% x^{3}z^{4}}{y^{5}}$ \ \ \ \ \ \ \ 17. \ $3q^{6}$ \ \ \ \ \ 18. \ $\dfrac{27}{% 8}$ \ \ \ \ \ 19. \ $\dfrac{2}{y^{3}}$ \item[20.] $\dfrac{1}{8y^{3}}$ \ \ \ \ \ \ \ 21. \ $\dfrac{25}{9}$ \ \ \ \ \ \ \ 22. \ $\dfrac{a^{5}}{b^{8}}$ \ \ \ \ \ 23. \ $\dfrac{1}{9m^{6}}$ \ \ \ \ \ \ 24. \ $-\dfrac{b^{9}}{8a^{3}}$ \ \ \ \ \ \ \ 25. \ $k$ \ \ \ \ \ \ \ 26. \ $\dfrac{9}{4}a^{3}b$ \ \ \ \ \ \ \ 27. \ $-\dfrac{a^{5}}{8b^{4}}$ \item[28.] $18p^{9}q^{10}$ \ \ \ \ \ \ \ 29. \ $\dfrac{4a^{10}}{b^{12}}$ \ \ \ \ \ \ 30. \ $\dfrac{x^{4}}{y^{6}}$ \ \ \ \ \ \ 31. \ $1$ \ \ \ \ \ \ 32. \ $\dfrac{xy}{y-x}$ \ \ \ \ \ \ \ \ 33.$-\dfrac{b^{8}}{2}$ \ \ \ \ \ \ \ 34. \ $\dfrac{1}{b^{8}}$ \ \ \ \ \ 35. \ $\dfrac{2x^{4}}{y^{3}}$ \item[36.] a) \ $6.\,\allowbreak 375\cdot 10^{-4}$ \ \ \ \ \ \ \ b) \ $% 6.\,\allowbreak 141\,3\cdot 10^{-34}$ \ \ \ \ \ \ c) \ $4.7813\cdot 10^{4}$ \ \ \ \ \ \ \ d) \ $1.\,\allowbreak 133\,3\cdot 10^{-19}$ \ \ \ \ \ \ \ e) \ $1.\,\allowbreak 690\,3\cdot 10^{63}$ \end{enumerate} \bigskip \pagebreak \FRAME{itbpF}{1.1096in}{0.6512in}{0.1807in}{}{}{pencil.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.1096in;height 0.6512in;depth 0.1807in;original-width 1.9735in;original-height 1.1467in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pencil.bmp';file-properties "XNPEU";}}{\LARGE Sample Problems - Solutions}% \bigskip \bigskip Simplify each of the following. \ Assume that all variables represent positive numbers. \ \ Present your answer without negative exponents.\medskip\ \medskip %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \begin{enumerate} \item $3^{-2}$\medskip Solution: \ We just apply the rule $a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} 3^{-2}=\dfrac{1}{3^{2}}=\dfrac{1}{9} \end{equation*} \item $\dfrac{1}{2^{-3}}$\medskip Solution: \ We apply the rule $a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \dfrac{1}{2^{-3}}=\dfrac{1}{~~\dfrac{1}{2^{3}}~~}=\dfrac{1}{~~\dfrac{1}{8}~~} \end{equation*}% To divide is to multiply by the reciprocal:% \begin{equation*} \dfrac{1}{~~\dfrac{1}{8}~~}=1\cdot \dfrac{8}{1}=8 \end{equation*}% This is true in general: \ \ $\dfrac{1}{a^{-n}}=a^{n}$ \begin{equation*} \dfrac{1}{a^{-n}}=\dfrac{1}{~~\dfrac{1}{a^{n}}~~}=1\cdot \dfrac{a^{n}}{1}% =a^{n} \end{equation*} \item $m^{-4}$\medskip Solution: \ We apply the rule $a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} m^{-4}=\dfrac{1}{m^{4}} \end{equation*} \item $\dfrac{1}{x^{-5}}$\medskip Solution: \ We have already proven that $\dfrac{1}{a^{-n}}=a^{n}$% \begin{equation*} \dfrac{1}{x^{-5}}=x^{5} \end{equation*} \item $a^{8}\cdot a^{-1}$\medskip Solution 1: \ We can apply the rule $a^{n}\cdot a^{m}=a^{n+m}$% \begin{equation*} a^{8}\cdot a^{-1}=a^{8+\left( -1\right) }=a^{7} \end{equation*}% Solution 2: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$ and then the rule $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} a^{8}\cdot a^{-1}=a^{8}\cdot \dfrac{1}{a^{1}}=\dfrac{a^{8}}{1}\cdot \dfrac{1% }{a}=\dfrac{a^{8}}{a}=\dfrac{a^{8}}{a^{1}}=a^{8-1}=a^{7} \end{equation*} \item $p^{3}\left( p^{-7}\right) p^{8}$\medskip Solution 1: \ We can apply the rule $a^{n}\cdot a^{m}=a^{n+m}$% \begin{equation*} p^{3}\left( p^{-7}\right) p^{8}=p^{3+\left( -7\right) +8}=p^{4} \end{equation*}% Solution 2: \ We can apply the rules $a^{-n}=\dfrac{1}{a^{n}}$ and $% a^{n}\cdot a^{m}=a^{n+m}$ and $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} p^{3}\left( p^{-7}\right) p^{8}=p^{3}\cdot \dfrac{1}{p^{7}}\cdot p^{8}=% \dfrac{p^{3}}{1}\cdot \dfrac{1}{p^{7}}\cdot \dfrac{p^{8}}{1}=\dfrac{% p^{3}\cdot p^{8}}{p^{7}}=\dfrac{p^{3+8}}{p^{7}}=\dfrac{p^{11}}{p^{7}}% =p^{11-7}=p^{4} \end{equation*} \item $\dfrac{x^{-4}}{x^{-9}}$\medskip Solution 1: \ We can apply the rule $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{x^{-4}}{x^{-9}}=x^{-4-\left( -9\right) }=x^{-4+9}=x^{5} \end{equation*}% Solution 2: \ We can apply the rules $a^{-n}=\dfrac{1}{a^{n}}$ and $\dfrac{% a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{x^{-4}}{x^{-9}}=\dfrac{x^{9}}{x^{4}}=x^{9-4}=x^{5} \end{equation*} \item $\dfrac{50a^{12}}{10a^{-3}}$\medskip Solution 1: \ We can apply the rule $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{50a^{12}}{10a^{-3}}=5a^{12-\left( -3\right) }=5a^{12+3}=5a^{15} \end{equation*}% Solution 2: \ We can apply the rules $a^{-n}=\dfrac{1}{a^{n}}$ and $\dfrac{% a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{50a^{12}}{10a^{-3}}=\dfrac{50a^{12}a^{3}}{10}=5a^{12+3}=5a^{15} \end{equation*} \item $\dfrac{t^{-3}}{t^{4}}$\medskip Solution 1: \ We can apply the rules $\dfrac{a^{n}}{a^{m}}=a^{n-m}$ and then $a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \dfrac{t^{-3}}{t^{4}}=t^{-3-4}=t^{-7}=\dfrac{1}{t^{7}} \end{equation*}% Solution 2: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$ and then $% a^{n}\cdot a^{m}=a^{n+m}$.% \begin{equation*} \dfrac{t^{-3}}{t^{4}}=\dfrac{1}{t^{4}\cdot t^{3}}=\dfrac{1}{t^{7}} \end{equation*} \item $x^{0}$\medskip Solution: \ There is a separate rule stating that as long as $x$ is not zero, then $x^{0}=1$. \ So the answer is $1$. \item $-x^{0}$\medskip Solution: \ This is the opposite of $x^{0}$ and so the answer is $-1$.% \begin{equation*} -x^{0}=-1\cdot x^{0}=-1\cdot 1=-1 \end{equation*} \item $\left( -x\right) ^{0}$\medskip Solution: \ This is again $1$ because \ any non-zero riased to the power zero is $1$. \item $\left( b^{-5}\right) \left( b^{2}\right) \left( b^{-1}\right) $% \medskip Solution 1: \ We can apply the rules $a^{n}\cdot a^{m}=a^{n+m}$ and then $% a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \left( b^{-5}\right) \left( b^{2}\right) \left( b^{-1}\right) =b^{-5+2+\left( -1\right) }=b^{-4}=\dfrac{1}{b^{4}} \end{equation*}% Solution 2: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$ and then just cancel.% \begin{equation*} \left( b^{-5}\right) \left( b^{2}\right) \left( b^{-1}\right) =\dfrac{1}{% b^{5}}\cdot b^{2}\cdot \dfrac{1}{b^{1}}=\dfrac{1}{b^{5}}\cdot \dfrac{b^{2}}{1% }\cdot \dfrac{1}{b^{1}}=\dfrac{b^{2}}{b^{6}}=\dfrac{\NEG{b}\cdot \NEG{b}}{% \NEG{b}\cdot \NEG{b}\cdot b\cdot b\cdot b\cdot b}=\dfrac{1}{b^{4}} \end{equation*} \item $\dfrac{1}{\left( b^{-5}\right) \left( b^{2}\right) \left( b^{-1}\right) }$\medskip Solution 1: \ We can apply the rules $a^{n}\cdot a^{m}=a^{n+m}$ and then $% a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \dfrac{1}{\left( b^{-5}\right) \left( b^{2}\right) \left( b^{-1}\right) }=% \dfrac{1}{b^{-5+2+\left( -1\right) }}=\dfrac{1}{b^{-4}}=\dfrac{1}{~~\dfrac{1% }{b^{4}}~~}=1\cdot \dfrac{b^{4}}{1}=b^{4} \end{equation*}% Solution 2: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$ and then $% \dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{1}{\left( b^{-5}\right) \left( b^{2}\right) \left( b^{-1}\right) }=% \dfrac{b^{5}\cdot b^{1}}{b^{2}}=\dfrac{b^{6}}{b^{2}}=b^{6-2}=b^{4} \end{equation*} \item $\dfrac{m^{-2}}{m^{-5}}$\medskip Solution 1: \ We can apply the rules $\dfrac{a^{n}}{a^{m}}=a^{n-m}$ and then $a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \dfrac{m^{-2}}{m^{-5}}=m^{-2-\left( -5\right) }=m^{-2+5}=m^{3} \end{equation*}% Solution 2: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$ and then $% \dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{m^{-2}}{m^{-5}}=\dfrac{m^{5}}{m^{2}}=m^{5-2}=m^{3} \end{equation*} \item $\dfrac{x^{3}y^{-5}}{z^{-4}}$\medskip Solution: \ Each variable occurs only once and so this problem is just about bringing it to the form required. \ We can apply the rule $a^{-n}=\dfrac{1}{% a^{n}}$. \ We hve alread shown that $\dfrac{1}{a^{-n}}=a^{n}$.% \begin{equation*} \dfrac{x^{3}y^{-5}}{z^{-4}}=\dfrac{x^{3}z^{4}}{y^{5}} \end{equation*} \item $\dfrac{18q^{3}}{6q^{-3}}$\medskip Solution 1: \ We can apply the rule $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{18q^{3}}{6q^{-3}}=\dfrac{\NEG{6}\cdot 3q^{3-\left( -3\right) }}{\NEG% {6}\cdot 1}=\dfrac{3q^{3+3}}{1}=3q^{6} \end{equation*}% Solution 2: \ We can apply the rules $a^{-n}=\dfrac{1}{a^{n}}$ and then $% a^{n}\cdot a^{m}=a^{n+m}$.% \begin{equation*} \dfrac{18q^{3}}{6q^{-3}}=\dfrac{\NEG{6}\cdot 3q^{3}q^{3}}{\NEG{6}\cdot 1}% =3q^{6} \end{equation*} \item $\left( \dfrac{2}{3}\right) ^{-3}$\medskip Solution: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \left( \dfrac{2}{3}\right) ^{-3}=\dfrac{1}{\left( \dfrac{2}{3}\right) ^{3}}=% \dfrac{1}{\dfrac{2}{3}\cdot \dfrac{2}{3}\cdot \dfrac{2}{3}}=\dfrac{1}{~~% \dfrac{8}{27}~~}=1\cdot \dfrac{27}{8}=\dfrac{27}{8} \end{equation*}% Note that we basically proved here that $\left( \dfrac{a}{b}\right) ^{-n}=\left( \dfrac{b}{a}\right) ^{n}$. \item $2y^{-3}$\medskip Solution: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$. \ It is important to note that the base of exponentiation is $y$ and not $2y$.% \begin{equation*} 2y^{-3}=2\cdot \dfrac{1}{y^{3}}=\dfrac{2}{1}\cdot \dfrac{1}{y^{3}}=\dfrac{2}{% y^{3}} \end{equation*} \pagebreak \item $\left( 2y\right) ^{-3}$\medskip Solution: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$. \ This time the base of exponentiation is $2y$. \ So we will apply the rule $\left( ab\right) ^{n}=a^{n}b^{n}$.% \begin{equation*} \left( 2y\right) ^{-3}=\dfrac{1}{\left( 2y\right) ^{3}}=\dfrac{1}{2^{3}y^{3}}% =\dfrac{1}{8y^{3}} \end{equation*} \item $\left( -\dfrac{3}{5}\right) ^{-2}$\medskip Solution 1: \ We can apply the rule $a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \left( -\dfrac{3}{5}\right) ^{-2}=\dfrac{1}{\left( -\dfrac{3}{5}\right) ^{2}}% =\dfrac{1}{\left( -\dfrac{3}{5}\right) \left( -\dfrac{3}{5}\right) }=\dfrac{1% }{\dfrac{-3}{5}\cdot \dfrac{-3}{5}}=\dfrac{1}{~~~\dfrac{9}{25}~~~}=1\cdot \dfrac{25}{9}=\dfrac{25}{9} \end{equation*}% Solution 2: \ We proved previously that $\left( \dfrac{a}{b}\right) ^{-n}=\left( \dfrac{b}{a}\right) ^{n}$. \ Using that,% \begin{equation*} \left( -\dfrac{3}{5}\right) ^{-2}=\left( -\dfrac{5}{3}\right) ^{2}=\left( -% \dfrac{5}{3}\right) \left( -\dfrac{5}{3}\right) =\dfrac{25}{9} \end{equation*} \item $\dfrac{a^{3}b^{-5}}{a^{-2}b^{3}}$\medskip Solution 1: \ We can apply the rule $\dfrac{a^{n}}{a^{m}}=a^{n-m}$ and then $% a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \dfrac{a^{3}b^{-5}}{a^{-2}b^{3}}=a^{3-\left( -2\right) }b^{-5-3}=a^{3+2}b^{-5-3}=a^{5}b^{-8}=a^{5}\cdot \dfrac{1}{b^{8}}=\dfrac{% a^{5}}{1}\cdot \dfrac{1}{b^{8}}=\dfrac{a^{5}}{b^{8}} \end{equation*}% Solution 2: \ We can apply the rules $a^{-n}=\dfrac{1}{a^{n}}$ and $% a^{n}\cdot a^{m}=a^{n+m}$.% \begin{equation*} \dfrac{a^{3}b^{-5}}{a^{-2}b^{3}}=\dfrac{a^{3}a^{2}}{b^{3}b^{5}}=\dfrac{a^{5}% }{b^{8}} \end{equation*} \item $\left( 3m^{3}\right) ^{-2}$\medskip Solution: \ We can apply the rule $\ a^{-n}=\dfrac{1}{a^{n}}$ and then $% \left( ab\right) ^{n}=a^{n}b^{n}$ and also $\left( a^{n}\right) ^{m}=a^{nm}$.% \begin{equation*} \left( 3m^{3}\right) ^{-2}=\dfrac{1}{\left( 3m^{3}\right) ^{2}}=\dfrac{1}{% 3^{2}\left( m^{3}\right) ^{2}}=\dfrac{1}{9m^{3\cdot 2}}=\dfrac{1}{9m^{6}} \end{equation*} \item $\left( -2ab^{-3}\right) ^{-3}$\medskip Solution: \ We can apply the rule $\ \left( ab\right) ^{n}=a^{n}b^{n}$ and then $\left( a^{n}\right) ^{m}=a^{nm}$.% \begin{equation*} \left( -2ab^{-3}\right) ^{-3}=\left( -2\right) ^{-3}a^{-3}\left( b^{-3}\right) ^{-3}=\left( -2\right) ^{-3}a^{-3}b^{-3\left( -3\right) }=\left( -2\right) ^{-3}a^{-3}b^{9} \end{equation*}% We now apply $a^{-n}=\dfrac{1}{a^{n}}$.% \begin{equation*} \left( -2\right) ^{-3}a^{-3}b^{9}=\dfrac{1}{\left( -2\right) ^{3}}\cdot \dfrac{1}{a^{3}}\cdot b^{9}=\dfrac{1}{-8}\cdot \dfrac{1}{a^{3}}\cdot \dfrac{% b^{9}}{1}=\dfrac{b^{9}}{-8a^{3}}=-\dfrac{b^{9}}{8a^{3}} \end{equation*} \item $\dfrac{\left( k^{3}\right) ^{-3}}{\left( k^{-5}\right) ^{2}}$\medskip Solution: \ We can apply the rule $\ \left( a^{n}\right) ^{m}=a^{nm}$ and then $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{\left( k^{3}\right) ^{-3}}{\left( k^{-5}\right) ^{2}}=\dfrac{% k^{3\left( -3\right) }}{k^{-5\cdot 2}}=\dfrac{k^{-9}}{k^{-10}}=k^{-9-\left( -10\right) }=k^{-9+10}=k^{1}=k \end{equation*} \item $\left( \dfrac{2a^{-3}b^{5}}{-3a^{3}b^{-2}}\right) ^{-2}\left( a^{3}b^{-5}\right) ^{-3}$\medskip Solution: \ \begin{eqnarray*} E &=&\left( \dfrac{2a^{-3}b^{5}}{-3a^{3}b^{-2}}\right) ^{-2}\left( a^{3}b^{-5}\right) ^{-3}=\left( \dfrac{2a^{-3-3}b^{5-\left( -2\right) }}{-3}% \right) ^{-2}\left( a^{3}b^{-5}\right) ^{-3}\text{ \ \ \ \ \ \ \ \ \ \ apply }\dfrac{a^{n}}{a^{m}}=a^{n-m} \\ &=&\left( \dfrac{2a^{-6}b^{5+2}}{-3}\right) ^{-2}\left( a^{3}b^{-5}\right) ^{-3} \\ &=&\left( \dfrac{2a^{-6}b^{7}}{-3}\right) ^{-2}\left( a^{3}b^{-5}\right) ^{-3}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply \ }\left( \dfrac{a}{b}\right) ^{-n}=\left( \dfrac{b}{a}\right) ^{n} \\ &=&\left( \dfrac{-3}{2a^{-6}b^{7}}\right) ^{2}\left( a^{3}b^{-5}\right) ^{-3}% \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( \dfrac{a}{% b}\right) ^{n}=\dfrac{a^{n}}{b^{n}} \\ &=&\dfrac{\left( -3\right) ^{2}}{\left( 2a^{-6}b^{7}\right) ^{2}}\left( a^{3}b^{-5}\right) ^{-3}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply \ }\left( ab\right) ^{n}=a^{n}b^{n}\text{ and }a^{-n}=\dfrac{1}{% a^{n}} \\ &=&\dfrac{9}{2^{2}\left( a^{-6}\right) ^{2}\left( b^{7}\right) ^{2}}\cdot \dfrac{1}{\left( a^{3}b^{-5}\right) ^{3}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( a^{n}\right) ^{m}=a^{nm}\text{ and }\left( ab\right) ^{n}=a^{n}b^{n} \\ &=&\dfrac{9}{4a^{-12}b^{14}}\cdot \dfrac{1}{\left( a^{3}\right) ^{3}\left( b^{-5}\right) ^{3}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }% a^{-n}=\dfrac{1}{a^{n}}\text{ and }\left( ab\right) ^{n}=a^{n}b^{n} \\ &=&\dfrac{9a^{12}}{4b^{14}}\cdot \dfrac{1}{a^{3\cdot 3}b^{\left( -5\right) 3}% } \\ &=&\dfrac{9a^{12}}{4b^{14}}\cdot \dfrac{1}{a^{9}b^{-15}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }a^{-n}=% \dfrac{1}{a^{n}} \\ &=&\dfrac{9a^{12}}{4b^{14}}\cdot \dfrac{b^{15}}{a^{9}}=\dfrac{9a^{12}b^{15}}{% 4b^{14}a^{9}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }% \dfrac{a^{n}}{a^{m}}=a^{n-m} \\ &=&\dfrac{9a^{12-9}b^{15-14}}{4}=\dfrac{9a^{3}b^{1}}{4}=\dfrac{9}{4}a^{3}b \end{eqnarray*} \pagebreak \item $\left( -2a^{-3}\right) \left( -2a^{-2}b\right) ^{-4}$\medskip Solution: \ \begin{eqnarray*} E &=&\left( -2a^{-3}\right) \left( -2a^{-2}b\right) ^{-4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }a^{-n}=\dfrac{1}{a^{n}} \\ &=&\left( -2\cdot \dfrac{1}{a^{3}}\right) \dfrac{1}{\left( -2a^{-2}b\right) ^{4}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( ab\right) ^{n}=a^{n}b^{n} \\ &=&\left( \dfrac{-2}{1}\cdot \dfrac{1}{a^{3}}\right) \dfrac{1}{\left( -2\right) ^{4}\left( a^{-2}\right) ^{4}b^{4}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( a^{n}\right) ^{m}=a^{nm} \\ &=&\dfrac{-2}{a^{3}}\cdot \dfrac{1}{16a^{-8}b^{4}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }a^{-n}=\dfrac{1}{% a^{n}} \\ &=&\dfrac{-2}{a^{3}}\cdot \dfrac{a^{8}}{16b^{4}} \\ &=&\dfrac{-2a^{8}}{a^{3}\cdot 16b^{4}}=\dfrac{-2a^{8}}{16a^{3}b^{4}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\dfrac{a^{n}}{a^{m}}% =a^{n-m} \\ &=&\dfrac{-1\cdot \NEG{2}a^{8-3}}{8\cdot \NEG{2}b^{4}}=\dfrac{-a^{5}}{8b^{4}} \end{eqnarray*} \item $\dfrac{\left( -3p^{3}q^{5}\right) ^{2}}{\left( 2q^{0}p^{3}\right) ^{-1}}$\medskip Solution: \ \begin{eqnarray*} E &=&\dfrac{\left( -3p^{3}q^{5}\right) ^{2}}{\left( 2q^{0}p^{3}\right) ^{-1}}% \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply \ }% q^{0}=1\text{ \ and \ }\dfrac{1}{a^{-n}}=a^{n} \\ &=&\left( -3p^{3}q^{5}\right) ^{2}\left( 2\cdot 1p^{3}\right) ^{1} \\ &=&\left( -3p^{3}q^{5}\right) ^{2}\cdot 2p^{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply \ }\left( ab\right) ^{n}=a^{n}b^{n} \\ &=&\left( -3\right) ^{2}\left( p^{3}\right) ^{2}\left( q^{5}\right) ^{2}\cdot 2p^{3}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ apply \ }\left( a^{n}\right) ^{m}=a^{nm} \\ &=&9p^{3\cdot 2}q^{5\cdot 2}\cdot 2p^{3} \\ &=&18p^{6}q^{10}p^{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply \ }a^{n}\cdot a^{m}=a^{n+m} \\ &=&18p^{6+3}q^{10}=18p^{9}q^{10} \end{eqnarray*} \item $\left( \dfrac{2a^{-2}b^{3}}{-2^{2}\left( a^{-1}b\right) ^{-3}}\right) ^{-2}$\medskip Solution:% \begin{eqnarray*} E &=&\left( \dfrac{2a^{-2}b^{3}}{-2^{2}\left( a^{-1}b\right) ^{-3}}\right) ^{-2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( \dfrac{% a}{b}\right) ^{-n}=\left( \dfrac{b}{a}\right) ^{n} \\ && \\ &=&\left( \dfrac{-2^{2}\left( a^{-1}b\right) ^{-3}}{2a^{-2}b^{3}}\right) ^{2}% \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( ab\right) ^{n}=a^{n}b^{n} \\ && \\ &=&\left( \dfrac{-4\left( a^{-1}\right) ^{-3}b^{-3}}{2a^{-2}b^{3}}\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( a^{n}\right) ^{m}=a^{nm} \end{eqnarray*} \ \begin{eqnarray*} &=&\left( \dfrac{-4a^{-1\left( -3\right) }b^{-3}}{2a^{-2}b^{3}}\right) ^{2} \\ &=&\left( \dfrac{-2a^{3}b^{-3}}{a^{-2}b^{3}}\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\dfrac{a^{n}}{a^{m}}=a^{n-m} \\ &=&\left( -2a^{3-\left( -2\right) }b^{-3-3}\right) ^{2} \\ &=&\left( -2a^{3+2}b^{-3-3}\right) ^{2} \\ &=&\left( -2a^{5}b^{-6}\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply \ }\left( ab\right) ^{n}=a^{n}b^{n} \\ &=&\left( -2\right) ^{2}\left( a^{5}\right) ^{2}\left( b^{-6}\right) ^{2}% \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( a^{n}\right) ^{m}=a^{nm} \\ &=&4a^{5\cdot 2}b^{-6\cdot 2}=4a^{10}b^{-12}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }a^{-n}=\dfrac{1}{a^{n}} \\ &=&4a^{10}\cdot \dfrac{1}{b^{12}} \\ &=&\dfrac{4a^{10}}{1}\cdot \dfrac{1}{b^{12}}=\dfrac{4a^{10}}{b^{12}} \end{eqnarray*} \item $\left( -\dfrac{x^{3}y^{0}x^{-5}}{y^{-3}}\right) ^{-2}$\medskip Solution: \ \begin{eqnarray*} E &=&\left( -\dfrac{x^{3}y^{0}x^{-5}}{y^{-3}}\right) ^{-2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }% y^{0}=1\text{ and }a^{n}\cdot a^{m}=a^{n+m} \\ &=&\left( -\dfrac{x^{3+\left( -5\right) }}{y^{-3}}\right) ^{-2}=\left( \dfrac{-1x^{-2}}{y^{-3}}\right) ^{-2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply\ \ }\left( \dfrac{a}{b}\right) ^{n}=\dfrac{a^{n}}{b^{n}} \\ &=&\dfrac{\left( -1x^{-2}\right) ^{-2}}{\left( y^{-3}\right) ^{-2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply \ }\left( ab\right) ^{n}=a^{n}b^{n} \\ &=&\dfrac{\left( -1\right) ^{-2}\left( x^{-2}\right) ^{-2}}{\left( y^{-3}\right) ^{-2}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( a^{n}\right) ^{m}=a^{nm}% \text{ \ and \ }a^{-n}=\dfrac{1}{a^{n}} \\ &=&\dfrac{x^{-2\left( -2\right) }}{\left( -1\right) ^{2}y^{-3\left( -2\right) }}=\dfrac{x^{4}}{1y^{6}}=\dfrac{x^{4}}{y^{6}} \end{eqnarray*} \item $\left( -\dfrac{x^{3}y^{7}x^{-5}}{y^{-3}}\right) ^{0}$\medskip Solution: \ Any non-zero quantity raised to the power zero is $1.$ \ So the answer is $1$. \pagebreak \item $\dfrac{x^{-1}+y^{-1}}{x^{-2}-y^{-2}}$\medskip Solution: \ This problem is very different because there are addition and subtraction involved. \ Because of that, we can not simply move the expressions with negative exponents. \ Instead, this will be a problem involving complex fractions.% \begin{eqnarray*} E &=&\dfrac{x^{-1}+y^{-1}}{x^{-2}-y^{-2}}=\dfrac{\dfrac{1}{x^{1}}+\dfrac{1}{% y^{1}}}{\dfrac{1}{x^{2}}-\dfrac{1}{y^{2}}}=\dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{% \dfrac{1}{x^{2}}-\dfrac{1}{y^{2}}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ bring fractions to the common denominator} \\ && \\ &=&\dfrac{~~\dfrac{1\cdot y}{x\cdot y}+\dfrac{1\cdot x}{y\cdot x}~~}{~~% \dfrac{1\cdot y^{2}}{x^{2}\cdot y^{2}}-\dfrac{1\cdot x^{2}}{y^{2}\cdot x^{2}}% ~~}=\dfrac{~~\dfrac{y}{xy}+\dfrac{x}{xy}~~}{~~\dfrac{y^{2}}{x^{2}y^{2}}-% \dfrac{x^{2}}{x^{2}y^{2}}~~}=\dfrac{~~\dfrac{y+x}{xy}~~}{~~\dfrac{y^{2}-x^{2}% }{x^{2}y^{2}}~~}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ to divide is to multiply by the reciprocal} \\ && \\ &=&\dfrac{y+x}{xy}\cdot \dfrac{x^{2}y^{2}}{y^{2}-x^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ cancel out }xy \\ && \\ &=&\dfrac{y+x}{1}\cdot \dfrac{xy}{y^{2}-x^{2}}=\dfrac{xy\left( x+y\right) }{% y^{2}-x^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor }% y^{2}-x^{2}\text{ via the difference of squares theorem, cancel out }x+y \\ && \\ &=&\dfrac{xy\left( x+y\right) }{\left( y-x\right) \left( y+x\right) }=\dfrac{% xy}{y-x} \end{eqnarray*} \item $\dfrac{\left( -2a^{-2}\right) ^{-2}b^{3}a^{0}\left( -aba^{-2}b^{-2}\right) ^{-3}}{2a^{2}\left( -2a^{-2}b\right) ^{-2}ab^{0}}$% \medskip Solution: \ \begin{eqnarray*} E &=&\dfrac{\left( -2a^{-2}\right) ^{-2}b^{3}a^{0}\left( -aba^{-2}b^{-2}\right) ^{-3}}{2a^{2}\left( -2a^{-2}b\right) ^{-2}ab^{0}}% \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }a^{0}=b^{0}=1\text{ and }x^{n}x^{m}=x^{n+m} \\ && \\ &=&\dfrac{\left( -2a^{-2}\right) ^{-2}b^{3}\left( -a^{1+\left( -2\right) }b^{1+\left( -2\right) }\right) ^{-3}}{2a^{2+1}\left( -2a^{-2}b\right) ^{-2}}% =\dfrac{\left( -2a^{-2}\right) ^{-2}b^{3}\left( -1a^{-1}b^{-1}\right) ^{-3}}{% 2a^{3}\left( -2a^{-2}b\right) ^{-2}}\text{ \ \ \ \ \ \ \ \ \ apply \ }\left( xy\right) ^{n}=x^{n}y^{n} \\ && \\ &=&\dfrac{\left( -2\right) ^{-2}\left( a^{-2}\right) ^{-2}b^{3}\left( -1\right) ^{-3}\left( a^{-1}\right) ^{-3}\left( b^{-1}\right) ^{-3}}{% 2a^{3}\left( -2\right) ^{-2}\left( a^{-2}\right) ^{-2}b^{-2}}\text{ \ \ \ \ \ \ \ \ \ apply }\left( x^{n}\right) ^{m}=x^{nm} \\ && \\ &=&\dfrac{\left( -2\right) ^{-2}a^{-2\left( -2\right) }b^{3}\left( -1\right) ^{-3}a^{-1\left( -3\right) }b^{-1\left( -3\right) }}{2a^{3}\left( -2\right) ^{-2}a^{-2\left( -2\right) }b^{-2}}=\dfrac{\left( -2\right) ^{-2}a^{4}b^{3}\left( -1\right) ^{-3}a^{3}b^{3}}{2a^{3}\left( -2\right) ^{-2}a^{4}b^{-2}}\text{ \ \ \ \ \ \ cancel out }a^{4}\text{ and }a^{3}\text{ and }\left( -2\right) ^{-2} \\ && \\ &=&\dfrac{b^{3}\left( -1\right) ^{-3}b^{3}}{2b^{-2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }x^{n}x^{m}=x^{n+m} \\ && \\ &=&\dfrac{\left( -1\right) ^{-3}b^{3+3}}{2b^{-2}}=\dfrac{\left( -1\right) ^{-3}b^{6}}{2b^{-2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }x^{-n}=\dfrac{1}{x^{n}} \\ && \\ &=&\dfrac{b^{6}b^{2}}{\left( -1\right) ^{3}2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }x^{n}x^{m}=x^{n+m} \\ && \\ &=&\dfrac{b^{6+2}}{-1\cdot 2}=\dfrac{b^{8}}{-2}=-\dfrac{b^{8}}{2} \end{eqnarray*} \item $\left( \dfrac{-a^{2}\left( b^{-1}a\right) ^{-5}}{b^{7}\left( -ab^{2}\right) ^{-3}}\right) ^{-2}$\medskip Solution: \ \begin{eqnarray*} E &=&\left( \dfrac{-a^{2}\left( b^{-1}a\right) ^{-5}}{b^{7}\left( -ab^{2}\right) ^{-3}}\right) ^{-2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( xy\right) ^{n}=x^{n}y^{n} \\ &=&\left( \dfrac{-a^{2}\left( b^{-1}\right) ^{-5}a^{-5}}{b^{7}\left( -1\right) ^{-3}a^{-3}\left( b^{2}\right) ^{-3}}\right) ^{-2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( x^{n}\right) ^{m}=x^{nm} \\ &=&\left( \dfrac{-a^{2}b^{-1\left( -5\right) }a^{-5}}{b^{7}\left( -1\right) ^{-3}a^{-3}b^{2\left( -3\right) }}\right) ^{-2}=\left( \dfrac{% -a^{2}b^{5}a^{-5}}{b^{7}\left( -1\right) ^{-3}a^{-3}b^{-6}}\right) ^{-2}% \text{ \ \ \ \ \ \ \ \ \ \ \ apply \ }x^{n}x^{m}=x^{n+m} \\ &=&\left( \dfrac{-a^{2+\left( -5\right) }b^{5}}{\left( -1\right) ^{-3}b^{7+\left( -6\right) }a^{-3}}\right) ^{-2}=\left( \dfrac{-1\cdot a^{-3}b^{5}}{\left( -1\right) ^{-3}b^{1}a^{-3}}\right) ^{-2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ cancel out }a^{-3} \\ &=&\left( \dfrac{-1b^{5}}{\left( -1\right) ^{-3}b^{1}}\right) ^{-2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }a^{-n}=\dfrac{1}{a^{n}} \\ &=&\left( \dfrac{-1\left( -1\right) ^{3}b^{5}}{b^{1}}\right) ^{-2}=\left( \dfrac{-1\left( -1\right) b^{5}}{b^{1}}\right) ^{-2}=\left( \dfrac{1b^{5}}{% b^{1}}\right) ^{-2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ apply }\dfrac{x^{n}}{x^{m}}% =x^{n-m} \\ &=&\left( b^{5-1}\right) ^{-2}=\left( b^{4}\right) ^{-2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( x^{n}\right) ^{m}=x^{nm} \\ &=&b^{4\left( -2\right) }=b^{-8}=\dfrac{1}{b^{8}} \end{eqnarray*} \item $\dfrac{\left( x^{-2}\right) ^{-2}y^{3}x^{0}\left( -2yx^{0}y^{-2}x^{-2}\right) ^{0}}{yx^{5}\left( y^{-2}x\right) ^{-3}\left( 2x^{-1}yx^{3}\right) ^{-1}}$\ \ Solution: \ \begin{eqnarray*} E &=&\dfrac{\left( x^{-2}\right) ^{-2}y^{3}x^{0}\left( -2yx^{0}y^{-2}x^{-2}\right) ^{0}}{yx^{5}\left( y^{-2}x\right) ^{-3}\left( 2x^{-1}yx^{3}\right) ^{-1}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }a^{0}=1\text{ \ and }a^{n}a^{m}=a^{n+m} \\ &=&\dfrac{\left( x^{-2}\right) ^{-2}y^{3}}{yx^{5}\left( y^{-2}x\right) ^{-3}\left( 2x^{-1+3}y\right) ^{-1}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( ab\right) ^{n}=a^{n}b^{n} \\ &=&\dfrac{\left( x^{-2}\right) ^{-2}y^{3}}{yx^{5}\left( y^{-2}\right) ^{-3}x^{-3}\left( 2x^{2}y\right) ^{-1}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( ab\right) ^{n}=a^{n}b^{n}\text{ \ and }a^{n}a^{m}=a^{n+m} \\ &=&\dfrac{\left( x^{-2}\right) ^{-2}y^{3}}{yx^{5+\left( -3\right) }\left( y^{-2}\right) ^{-3}\left( 2\right) ^{-1}\left( x^{2}\right) ^{-1}y^{-1}}% \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }\left( a^{n}\right) ^{m}=a^{nm} \\ &=&\dfrac{x^{-2\left( -2\right) }y^{3}}{yx^{2}y^{-2\left( -3\right) }2^{-1}x^{2\left( -1\right) }y^{-1}}=\dfrac{x^{4}y^{3}}{% 2^{-1}yx^{2}y^{6}x^{-2}y^{-1}}\text{ \ \ \ \ \ \ \ \ \ \ apply }% a^{n}a^{m}=a^{n+m} \\ &=&\dfrac{x^{4}y^{3}}{2^{-1}y^{1+6+\left( -1\right) }x^{2+\left( -2\right) }}% =\dfrac{x^{4}y^{3}}{2^{-1}y^{6}x^{0}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x^{0}=1\text{ and }\dfrac{a^{n}}{a^{m}}=a^{n-m} \\ &=&\dfrac{x^{4}y^{3-6}}{2^{-1}}=\dfrac{x^{4}y^{-3}}{2^{-1}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ apply }a^{-n}=\dfrac{1}{a^{n}} \\ &=&\dfrac{2^{1}x^{4}}{y^{3}}=\dfrac{2x^{4}}{y^{3}} \end{eqnarray*} \end{enumerate} {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}