\hfill \thepage} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \cfoot{} \renewcommand{\headrulewidth}{0.8pt} \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 } \newcommand{\matek}{ \setstretch{1.04} \normalsize \setlength{\parskip}{0.04in} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} } \setstretch{1.06} \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 Rules of Exponents } \rhead{\small page \thepage} \textwidth 7.7in \textheight 9.7in \setlength{\headheight}{27pt} \lfoot{\footnotesize \copyright \; Hidegkuti, Emmett, Powell, 2008} \rfoot{\footnotesize Last revised: August 26, 2018} \setlength{\parindent}{0in} \setlength{\parskip}{0.05in} \input{tcilatex} \begin{document} \begin{center} {\Large Part 1 - The Definition} \end{center} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Exponential notation expresses repeated multiplication. \ \vspace{0.04in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Definition}: \ $\ $We define $2^{7}$ to denote the factor $2$ multiplied by itself repeatedly $7$ times, 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*}% The new operation defined is called \textbf{exponentiation.} \ The factor (in this case $2$) is called the \textbf{base}. \ The number written above the base, in smaller font (in this case, $7$) is called the \textbf{exponent}% . %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Since the definition does not elegantly fit the case when the exponent is one, we also define $5^{1}$ to be $5$. \ One factor, so technically, no multiplication. When we enlarge our mathematical notation by the inclusion of exponential expressions, a few things might become problematic. \ For example, is there a difference between $-3^{2}$ \ and $\left( -3\right) ^{2}$? Recall that a negative sign in front of anything can be interpreted as '% \textit{the opposite of}', which is the same as mutliplication by $-1$. \ We can interpret $-3$ as $-1\cdot 3,$ and so we can re-interpret the original question from comparing \ $-3^{2}$ and $\left( -3\right) ^{2}$ to a question comparing $-1\cdot 3^{2}$ and $\left( -1\cdot 3\right) ^{2}$.\ \ The rest is really just an order of operations problem. Recall that in our order of operations agreement, \FRAME{itbpFX}{0.3563in}{% 0.5146in}{0.2413in}{}{}{pemdas.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.3563in;height 0.5146in;depth 0.2413in;original-width 0.7195in;original-height 1.0533in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pemdas.bmp';file-properties "XNPEU";}}, exponentiation\vspace{0.04in} superseeds multiplication. \ So, when presented by multiplication and exponentiation, we first execute the exponentiation and then the multiplication. \ And so, if there is no parentheses, we have \begin{eqnarray*} &&-1\cdot 3^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ perform exponentiation} \\ &&-1\cdot 9\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ perform multiplication} \\ &&-9 \end{eqnarray*}% And if we have a parentheses, that serves to overwrite the usual order of operations:% \begin{eqnarray*} &&\left( -1\cdot 3\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ whatever is in the parentheses first} \\ &&\left( -3\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square the number }-3 \\ &&9 \end{eqnarray*}% The difference between $-3^{2}$ and $\left( -3\right) ^{2}$ is truly an order of operations thing: we are talking about taking the opposite and squaring, but in different orders.% \begin{eqnarray*} &&-3^{2}\text{ is square 3 and then take the result's opposite \ \ \ \ ----\ \ \ \ \ or the opposite of the square of }3 \\ && \\ &&\left( -3\right) ^{2}\text{ is take the opposite of }3\text{ and then square \ \ \ \ \ \ \ ------\ \ \ \ \ \ \ or the square of the opposite of }3 \end{eqnarray*}% In algebra, it is important to correctly read notation. \ Confusing $-3^{2}$ \ and $\left( -3\right) ^{2}$ is an error that commonly occurs and messes up computations.\vspace{0.1in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion Caution! \ \ \ In the expression\textbf{\ }$\left( -3\right) ^{2}$, the\ base of the exponentiation is $-3$. \ In the expression $-3^{2}$, the\ base of the exponentiation is $3$. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 1. \ }Simplify each of the given expressions. \ %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $-2^{4}$ \ \qquad b) \ $\left( -2\right) ^{4}$ \ \qquad c) \ $-1^{3}$ \ \qquad d) \ $\left( -1\right) ^{3}$ \ $\qquad $e)\ \ $-\left( -2\right) ^{2}$ \ \qquad f) \ $-\left( -2^{2}\right) $ %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution: }\ a) \ The base of the exponentiation is $2$.% \begin{equation*} -2^{4}=-1\cdot 2^{4}=-1\cdot (2\cdot 2\cdot 2\cdot 2)=-1\cdot 16=\fbox{$-16$} \end{equation*}% $-2^{4}$ \ can be read as the opposite of $2^{4}$. %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ The base of the exponentiation is $-2$.% \begin{equation*} \left( -2\right) ^{4}=\left( -1\cdot 2\right) ^{4}=\left( -1\cdot 2\right) \left( -1\cdot 2\right) \left( -1\cdot 2\right) \left( -1\cdot 2\right) =\left( -2\right) \left( -2\right) \left( -2\right) \left( -2\right) =\fbox{$% 16$} \end{equation*}% $\left( -2\right) ^{4}$ \ can be read as the fourth power of $-2$. %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ The base of the exponentiation is $1$.% \begin{equation*} -1^{3}=-1\cdot 1^{3}=-1\cdot 1\cdot 1\cdot 1=\fbox{$-1$} \end{equation*} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion d) \ The base of the exponentiation is $-1$.% \begin{equation*} \left( -1\right) ^{3}=\left( -1\right) \left( -1\right) \left( -1\right) =% \fbox{$-1$} \end{equation*} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion e) \ The base of the exponentiation is $-2$.% \begin{equation*} -\left( -2\right) ^{2}=-\left( \left( -2\right) \left( -2\right) \right) =-\left( 4\right) =\fbox{$-4$} \end{equation*} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion f) \ Careful! \ \ The base of the exponentiation is $2$. \ This is NOT squaring $-2$. \ This is squaring $2$ and then taking the opposite of the result twice.% \begin{equation*} -\left( -2^{2}\right) =-\left( -1\cdot 2\cdot 2\right) =-\left( -4\right) =% \fbox{$4$} \end{equation*} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{0.9in}}{\begin{minipage}{0.9in}}}% %BeginExpansion \begin{minipage}{0.9in}% %EndExpansion \FRAME{itbpF}{0.7048in}{0.7117in}{0.2006in}{}{}{question.bmp}{\special% {language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file "F";width 0.7048in;height 0.7117in;depth 0.2006in;original-width 0.4869in;original-height 0.6797in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'question.bmp';file-properties "XNPEU";}}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{5.9in}}{\begin{minipage}{5.9in} %\setlength{\hangindent}{63pt}}}% %BeginExpansion \begin{minipage}{5.9in} \setlength{\hangindent}{63pt}% %EndExpansion \textbf{Discussion}: \ Explain why in the expression $-\left( -5\right) ^{2}$% , the two negatives do not cancel out to a positive.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion $\vspace{0.09in}$ %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \begin{center} {\Large Part 2 - Rules of Exponents} \end{center} When mathematicians agreed to define $3^{5}$ as \ $3\cdot 3\cdot 3\cdot 3\cdot 3$, 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. \ The following statements are straightforward consequences of the definition - and the mathematics we already have. \ Consider the expression $2^{3}\cdot 2^{4}$. \ If we re-write the expression using the definition of exponents, we quickly get that% \begin{equation*} 2^{3}\cdot 2^{4}~~\overset{\text{def}}{=}~~\left( 2\cdot 2\cdot 2\right) \cdot \left( 2\cdot 2\cdot 2\cdot 2\right) ~~~~\overset{\text{mult is }}{% \underset{\text{associative}}{=}}~~~2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2~~\overset{\text{def}}{=}~~2^{7} \end{equation*}% The computation above illustrates why the following theorem is true. %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem} \textbf{1.} \ $\ $If $a$ is any number and $m$ and $n$ are any two positive integers, then% \begin{equation*} a^{n}\cdot a^{m}=a^{n+m} \end{equation*}% %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion We can see that this rule follows from the definition of exponents and from the fact that multiplication is associative. Consider now the expression $\dfrac{2^{5}}{2^{3}}$. \ If we re-write the expression using the definition of exponents, we quickly get that% \begin{equation*} \dfrac{2^{5}}{2^{3}}~\overset{\text{def}}{=}~\dfrac{2\cdot 2\cdot 2\cdot 2\cdot 2}{2\cdot 2\cdot 2}~~\overset{\text{cancellation }}{=}~~\dfrac{\NEG% {2}\cdot \NEG{2}\cdot \NEG{2}\cdot 2\cdot 2}{\NEG{2}\cdot \NEG{2}\cdot \NEG% {2}}~\overset{\text{def}}{=}~\dfrac{2^{2}}{1}=2^{2} \end{equation*} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem 2.} \ $\ $If $a$ is any number and $m$ and $n$ are any two positive integers, then% \begin{equation*} \dfrac{a^{n}}{a^{m}}=a^{n-m} \end{equation*}% %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion As our computation shows, this property is a consequence of the definition of exponentials and our rules of cancellation. Consider now $\left( 2^{3}\right) ^{5}$. \ If we re-write the expression using the definition of exponents, we quickly get that% \begin{eqnarray*} &&\left( 2^{3}\right) ^{5}\overset{\text{def}}{=}\left( 2^{3}\right) \cdot \left( 2^{3}\right) \cdot \left( 2^{3}\right) \cdot \left( 2^{3}\right) \cdot \left( 2^{3}\right) \overset{\text{def}}{=}\left( 2\cdot 2\cdot 2\right) \cdot \left( 2\cdot 2\cdot 2\right) \cdot \left( 2\cdot 2\cdot 2\right) \cdot \left( 2\cdot 2\cdot 2\right) \cdot \left( 2\cdot 2\cdot 2\right) \\ &&\overset{\text{mult is}}{\underset{\text{associative}}{=}}2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\overset{\text{def}}{=}2^{15} \end{eqnarray*}% Clearly, we have five groups of three two-factors. %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem 3.} \ $\ $If $a$ is any number and $m$ and $n$ are any two positive integers, then% \begin{equation*} \left( a^{n}\right) ^{m}=a^{nm} \end{equation*}% %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion As our computation shows, this property is a consequence of the definition of exponentials and the fact that multiplication is associative. Consider now $\left( 2\cdot 3\right) ^{4}$. \ If we re-write the expression using the definition of exponents, we quickly get that% \begin{eqnarray*} &&\left( 2\cdot 3\right) ^{4}\overset{\text{def}}{=}\left( 2\cdot 3\right) \cdot \left( 2\cdot 3\right) \cdot \left( 2\cdot 3\right) \cdot \left( 2\cdot 3\right) \overset{\text{mult is}}{\underset{\text{associative}}{=}}% 2\cdot 3\cdot 2\cdot 3\cdot 2\cdot 3\cdot 2\cdot 3 \\ &&\overset{\text{mult is}}{\underset{\text{commutative}}{=}}2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 3\cdot 3\overset{\text{mult is}}{\underset{\text{% associative}}{=}}\left( 2\cdot 2\cdot 2\cdot 2\right) \cdot \left( 3\cdot 3\cdot 3\cdot 3\right) \overset{\text{def}}{=}2^{4}\cdot 3^{4} \end{eqnarray*} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem 4.} \ $\ $If $a$ and $b$ are any numbers and $n$ is any positive integer, then% \begin{equation*} \left( ab\right) ^{n}=a^{n}\cdot b^{n} \end{equation*}% %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion As our computation shows, this property is a consequence of the definition of exponentials and the fact that multiplication is commutative and associative. \pagebreak Caution! \ Exponentiation denotes repeated multiplication, so it is a fundamentally multiplicative concept. \ Exponents will exhibit nice behvior with respect to multiplication and division, but NOT with respect to addition and subtraction. Similar-looking statements fail to be true if addition or subtraction is involved. \ For example, $\left( 2\cdot 5\right) ^{2}=2^{2}\cdot 5^{2},$ but $\left( 2+5\right) ^{2}\not=2^{2}+5^{2}$. Caution! \ Another common mistake is to confuse $\left( ab\right) ^{n}$ with $ab^{n}$. \ (It is an order of operations thing.) \ The base of exponentiation is \ $ab$ in $\left( ab\right) ^{n}$ but only $b$ in $ab^{n}$. \begin{equation*} ab^{n}=a\cdot \underset{n\text{ times}}{\underbrace{~b\cdot b\cdot ...\cdot b% }}\text{ \ \ \ \ and \ \ }\left( ab\right) ^{n}=\underset{n\text{ times}}{% \underbrace{~\left( ab\right) \cdot \left( ab\right) \cdot ...\cdot \left( ab\right) }} \end{equation*} Consider now the expression $\left( \dfrac{3}{4}\right) ^{5}$. \ If we re-write the expression using the definition of exponents, we quickly get that% \begin{equation*} \left( \dfrac{3}{4}\right) ^{5}~\overset{\text{def}}{=}~\left( \dfrac{3}{4}% \right) \cdot \left( \dfrac{3}{4}\right) \cdot \left( \dfrac{3}{4}\right) \cdot \left( \dfrac{3}{4}\right) \cdot \left( \dfrac{3}{4}\right) ~\overset{% \text{rules of multipying}}{\underset{\text{fractions}}{=}}~\dfrac{3\cdot 3\cdot 3\cdot 3\cdot 3}{4\cdot 4\cdot 4\cdot 4\cdot 4}~\overset{\text{def}}{=% }~\dfrac{3^{5}}{4^{5}} \end{equation*} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem 5.} \ $\ $If $a$ and $b$ are any numbers and $n$ is any positive integer, then% \begin{equation*} \left( \dfrac{a}{b}\right) ^{n}=\dfrac{a^{n}}{b^{n}} \end{equation*}% %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion As our computation shows, this property is a consequence of the definition of exponentials and the rule of how we multiply fractions.\vspace{0.1in} Caution! \ Exponentiation denotes repeated multiplication, so it is a fundamentally multiplicative concept. \ Exponents will exhibit nice behavior with respect to multiplication and division, but NOT with respect to addition and subtraction. Similar-looking statements fail to be true if addition or subtraction is involved. \ For example, $\left( \dfrac{2}{5}% \right) ^{2}=\dfrac{2^{2}}{5^{2}}$, but $\left( 5-2\right) ^{2}\not=5^{2}-2^{2}$. To summarize what just happened: \ once we defined exponentiation as repeated multiplication, certain properties immediately followed from the definition. \ These properties are as follows.\vspace{0.08in} %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion \textbf{Theorem:} \ $\ $If $a,b$ are any numbers and $m$, $n$ are any positive integers, then% \begin{eqnarray*} \text{1.\ \ }a^{n}\cdot a^{m} &=&a^{n+m}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.\ \ \ \ }\left( ab\right) ^{n}=a^{n}b^{n} \\ && \\ \text{2. \ \ \ \ \ \ \ }\dfrac{a^{n}}{a^{m}} &=&a^{n-m}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5.\ \ }\left( \dfrac{a}{b}% \right) ^{n}=\dfrac{a^{n}}{b^{n}} \\ && \\ \text{3. \ \ \ }\left( a^{n}\right) ^{m} &=&a^{nm} \end{eqnarray*}% %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \pagebreak %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 2. }\ Simplify each of the following expressions. %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $x^{5}\cdot x^{3}$ \ \ \ \ \ \ \ \ b) \ $\left( x^{5}\right) ^{3}$ \ \ \ \ \ \ \ \ \ c) \ $\left( 2x^{5}\right) ^{3}$\ \ \ \ \ \ \ \ \ \ \ d) \ $% \dfrac{\left( 2x\right) ^{5}}{2x^{3}}$\ \ \ \ \ \ \ \ \ \ \ \ e) \ $\left( -x^{2}\right) ^{3}$ \ \ \ \ \ \ \ f) \ $\left( -x^{3}\right) ^{2}\vspace{% 0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ In case of $x^{5}\cdot x^{3}$, we can apply the definition or our first rule. \ Either way, we will end up adding the exponents.$\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion \ \ \ \ \ \ \ \ \ \ \ $x^{5}\cdot x^{3}=\fbox{$x^{8}$}\vspace{0.04in}\vspace{% 0.04in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ In case of $\left( x^{5}\right) ^{3}$, we have repeated exponentiation. Applying the definition, we see three groups of factors, each group with five factors in it. \ So, we multiply the exponents.$\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion \ \ \ \ \ \ \ \ \ \ \ $\left( x^{5}\right) ^{3}=\fbox{$x^{15}$}\vspace{0.04in% }\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ In case of $\left( 2x^{5}\right) ^{3}$, we will$\vspace{0.04in}$ have to apply several rules. \ First, when a product is exponentiated, we exponentiate each factor, i.e. \ $\left( ab\right) ^{n}=a^{n}b^{n}$.$\vspace{% 0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion \ \ \ \ \ \ \ \ \ \ \ $\left( 2x^{5}\right) ^{3}=2^{3}\cdot \left( x^{5}\right) ^{3}=8\cdot x^{15}=\fbox{$8x^{15}$}\vspace{0.04in}\vspace{0.04in% }\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion d) \ In the expression \ $\dfrac{\left( 2x\right) ^{5}}{2x^{3}}$, \ the number $2$ is part of the base in the exponentiation in the numerator, but not in the exponentiation in the denominator. \ Once we got rid of the parentheses in the numerator, we subtract the exponents corresponding to cancellation.$\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion \ \ \ \ \ \ \ \ \ \ \ $\dfrac{\left( 2x\right) ^{5}}{2x^{3}}=\dfrac{% 2^{5}\cdot x^{5}}{2x^{3}}=\dfrac{32x^{5}}{2x^{3}}=16x^{5-3}=\fbox{$16x^{2}$}$% .$\vspace{0.04in}\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion e) \ In the expression $\left( -x^{2}\right) ^{3}$, the leading negative sign brings some algebraic complications. \ One way to handle this, either mentally or also in writing, to interpret $-x^{2}$ as the opposite of $% x^{2}, $ or $-1\cdot x^{2}$. \ Notice that the $-1$ is not getting raised to second power, only to the third power.$\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion \ \ \ \ \ \ \ \ \ \ \ $\left( -x^{2}\right) ^{3}=\left( -1\cdot x^{2}\right) ^{3}=\left( -1\right) ^{3}\left( x^{2}\right) ^{3}=-1\cdot x^{6}=\fbox{$% -x^{6}$}$.$\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion f) \ This expression is very similar to the previous one. \ Here the negative sign gets squared only but not cubed.$\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solcont}{\solcont} }% %BeginExpansion \solcont %EndExpansion \ \ \ \ \ \ \ \ \ \ \ $\left( -x^{3}\right) ^{2}=\left( -1\cdot x^{3}\right) ^{2}=\left( -1\right) ^{2}\left( x^{3}\right) ^{2}=1\cdot x^{6}=\fbox{$x^{6}$% }$.$\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 3. }\ Simplify the expression \ \ \ $\dfrac{\left( -2a\right) ^{4}\left( -ab^{4}\right) ^{3}ab^{2}}{\left( -2ab^{2}\right) ^{3}ba^{2}}$. $\ \ $Assume that $a$ and $b$ represent non-zero numbers.$% \vspace{0.04in}\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ There are several ways to solve this problem. \ We will take our time and apply one rule at the time. \ We will start with the rule $% \left( ab\right) ^{n}=a^{n}b^{n}$. \ Also, standalone negative signs will be replaced with a $-1$ multiplyer.$\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ \ \ $\dfrac{\left( -2a\right) ^{4}\left( -ab^{4}\right) ^{3}ab^{2}}{\left( -2ab^{2}\right) ^{3}ba^{2}}=\dfrac{\left( -2a\right) ^{4}\left( -1ab^{4}\right) ^{3}ab^{2}}{\left( -2ab^{2}\right) ^{3}ba^{2}}=% \dfrac{\left( -2\right) ^{4}a^{4}\left( -1\right) ^{3}a^{3}\left( b^{4}\right) ^{3}ab^{2}}{\left( -2\right) ^{3}a^{3}\left( b^{2}\right) ^{3}ba^{2}}\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Next, we perform the exponentiation on the numbers and bring them forward, re-ordered the variables alphabetically, $\vspace{0.04in}$and simplify expressions contining repeated exponentiation using the rule $\left( a^{n}\right) ^{m}=a^{nm}$.$\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ \ $\dfrac{\left( -2\right) ^{4}a^{4}\left( -1\right) ^{3}a^{3}\left( b^{4}\right) ^{3}ab^{2}}{\left( -2\right) ^{3}a^{3}\left( b^{2}\right) ^{3}ba^{2}}=\dfrac{16a^{4}\left( -1\right) a^{3}b^{12}ab^{2}}{% -8a^{3}b^{6}ba^{2}}=\dfrac{-16a^{4}a^{3}ab^{12}b^{2}}{-8a^{3}a^{2}b^{6}b}% \vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Next, we apply our first rule, $a^{n}a^{m}=a^{n+m}$ in both numerator and denominator.$\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ \ $\dfrac{-16a^{4}a^{3}ab^{12}b^{2}}{-8a^{3}a^{2}b^{6}b}=% \dfrac{-16a^{4+3+1}b^{12+2}}{-8a^{3+2}b^{6+1}}=\dfrac{-16a^{8}b^{14}}{% -8a^{5}b^{7}}\vspace{0.04in}\vspace{0.04in}$ \pagebreak %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion At this point, both numerator and denominator are completely simplified. \ We can tell because both expressions have no repetition of sign, number, or any of the variables. \ What is left for us to do, is to consolidate numerator and denominator. \ We will $\vspace{0.04in}$simplify (or perform the division) between $-16$ and $-8$, and perform cancellation between the variables, using the rule $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.$\vspace{0.04in}% \vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ $\dfrac{-16a^{8}b^{14}}{-8a^{5}b^{7}}=\dfrac{-2a^{8-5}b^{14-7}% }{1}=2a^{3}b^{7}\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion So the simplified expression is \fbox{$2a^{3}b^{7}$}.$\vspace{0.04in}\vspace{% 0.04in}\vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{0.9in}}{\begin{minipage}{0.9in}}}% %BeginExpansion \begin{minipage}{0.9in}% %EndExpansion \FRAME{itbpF}{0.9686in}{0.8968in}{0in}{}{}{question.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.9686in;height 0.8968in;depth 0in;original-width 1.0533in;original-height 0.9729in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'question.bmp';file-properties "XNPEU";}}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion $\vspace{0.04in}\vspace{0.04in}\vspace{0.04in}\vspace{0.04in}$% %TCIMACRO{% %\TeXButton{\begin{minipage}{5.9in}}{\begin{minipage}{5.9in} %\setlength{\hangindent}{61pt} %\setlength{\leftskip}{25pt}}}% %BeginExpansion \begin{minipage}{5.9in} \setlength{\hangindent}{61pt} \setlength{\leftskip}{25pt}% %EndExpansion \textbf{Discussion:} \ Why was it necessary in the previous example to assume that $a$ and $b$ represent non-negative numbers?$\vspace{0.04in}$% \newline Explain why this step is incorrect: \ $2\cdot 5^{x}=10^{x}$.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 4. \ \ }Consider the expression $3^{x}\cdot 3^{x}$.$\vspace{% 0.04in}$ %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ Simplify the expression using our first rule, $a^{n}\cdot a^{m}=a^{n+m}$ only.$\vspace{0.04in}$ %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion b) \ Simplify the expression using our fourth rule, $\left( ab\right) ^{n}=a^{n}b^{n}$ only.$\vspace{0.04in}$ %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion c) \ Is there a rule of exponentiation that can be used to verify that the results from part a) and part b) are the same?$\vspace{0.04in}\vspace{0.04in} $ %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ The base is the same, so $\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ $3^{x}\cdot 3^{x}=3^{x+x}=$\fbox{$3^{2x}$}.$\vspace{0.04in}% \vspace{0.04in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ Now we will use the fact that the exponents of factors are the same, and so we will re-write $a^{n}\cdot b^{n}$ as $\left( ab\right) ^{n}$.$% \vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ $3^{x}\cdot 3^{x}=\left( 3\cdot 3\right) ^{x}=\fbox{$9^{x}$}% \vspace{0.04in}\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ The expressions $3^{2x}$ and $9^{x}$ are the same. \ We can use our third rule, $\left( a^{n}\right) ^{m}=a^{nm}$.$\vspace{0.04in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ $9^{x}=\left( 3^{2}\right) ^{x}=3^{2\cdot x}=3^{2x}$. %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \vspace{0.05in}\vspace{0.05in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 5. \ }Find the prime-factorization of $24^{100}$. %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion Solution: \ If we tried to enter $24^{100}$ in our calculator, we will probably find that the number is too great for it to handle. \ So it would be futile to compute the gigantic number and start the prime-factorization from scratch. \ Instead, we will use the prime-factorization of $24$ and rules of exponents. \ The prime-factorization of $24=2^{3}\cdot 3$.\vspace{% 0.05in} %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ $24^{100}=\left( 2^{3}\cdot 3\right) ^{100}=\left( 2^{3}\right) ^{100}\cdot 3^{100}=2^{300}\cdot 3^{100}$.\vspace{0.05in}% \vspace{0.05in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion So the prime-factorization of $24^{100}$ is \fbox{$2^{300}\cdot 3^{100}$}. %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \pagebreak \begin{center} {\Large Part 3 - Scientific Notation} \end{center} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion In physics, we use units that are internationally established by scientists. \ The abbreviation SI stands for (Syst\`{e}me international d'unit\'{e}s). \ In natural sciences such as chemistry and physics, we often have to communicate numbers so large that it is uncomfortable for both of our imagination and even for our notation. \ Consider for example, the radius of the sun. \ The SI unit of measuring length is meters. The radius of the sun is $695\,700\,000$ meters. \ The distance between our planet Earth and the Sun is approximately$\allowbreak $\newline $149\,\allowbreak 600\,000\,000$ meters. \ The basic particles that make up material are so tiny that a handful of some material includes a very large number of particles. \ In chemistry, the basic measurement of how how much material we have (i.e. how many particles) is $1$ mole. \ One mole of a substance contains approximately $602\,214\,\allowbreak 076\,000\,000\,\allowbreak 000\,000\,000$ particles. \ Scientific notation was developed to handle such uncomfortably large numbers, using properties of exponentiation. \ Scientific notation expresses a single number as a product, where the first number \ (kind of, sort of) expresses the number, and the second part expresses the number of zeroes to place after the number. \ Naturally, the definition is going to be a bit more rigorous.\vspace{0.08in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 6. \ }Re-write $4\cdot 10^{7}$ using regular notation. \ \vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ If we look at ten-powers, we will notice some very nice properties. \ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $10^{1}=10$ \ and multiplying an integer by $10$ results in adding a zero as its last digit. \ $4\cdot 10^{1}=40$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $10^{2}=100$ \ and multiplying an integer by $100$ results in adding two zeroes as its last digits. \ $4\cdot 10^{2}=400$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $10^{3}=1000$ \ and multiplying an integer by $1000$ results in adding three zeroes as its last digits. \ $4\cdot 10^{3}=4000$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $10^{4}=10\,000$ \ and multiplying an integer by $10\,000$ results in adding four zeroes as its last digits. \ $4\cdot 10^{4}=40\,000$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion This is indeed a very nice pattern. \ The exponent on $10$ is the same as the number of zeroes to be added at the end. \ Therefore, we can interpret $% 4\cdot 10^{7}$ as placing $7$ zeroes after the digit $4$. %TCIMACRO{\TeXButton{\solinside}{\solinside} }% %BeginExpansion \solinside %EndExpansion \ \ \ \ \ \ \ \ \ \ \ $4\cdot 10^{7}=\,$\fbox{$40\,000\,000$}.\vspace{0.08in}% \vspace{0.08in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 7. \ }Re-write $3.215\cdot 10^{12}$ using regular notation. \ \vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: \ \ }When we are dealing with decimals, a multiplication by a $10-$power means moving the decimal point. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $3.215\cdot 10=\allowbreak 32.\,15$, \ \ $3.215\cdot 100=\allowbreak 321.\,\allowbreak 5,$ \ \ $3.215\cdot 1000=\allowbreak 3215,$ and $% 3.125\cdot 10000=\allowbreak 31250$ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Once we reached the end of the decimal, i.e. multiplied it by a ten-power large enough to create an integer, multiplication by the remaining $10-$% powers is again a matter of placing zeroes at the end. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $3.215\cdot 10^{12}=3.215\cdot 10^{3+9}=3.215\cdot 10^{3}\cdot 10^{9}=3215\cdot 10^{9}=\,$\fbox{$3215\,000\,000\,000$}.\vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %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 Please note that scientific notation uses the cross notation for multiplication. \ We will break with tradition and simply use the dot notation for scientific notation. \ Part of the problem with the cross notation is that in hand-written computations it looks too much like the letter $x$.\vspace{0.08in} \pagebreak %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 8. \ }Re-write $602\,200\,\allowbreak 000\,000\,000\,\allowbreak 000\,000\,000$ using scientific notation. \ \vspace{0.04in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: \ }Let us first count the trailing zeroes at the end. There% \vspace{0.04in} are six groups of three zeroes and then two more, so that is $20$. \ So now we can re-write this giant number as $602\,2\cdot 10^{20}$.% \vspace{0.04in} \ \ But remember that the first factor in scientific notation must be between $1$ and $10$.\vspace{0.04in} \ So we will extract more ten-powers by moving the decimal point. \ When in doubt, check with the calculator.\vspace{0.04in} $6022=602.2\cdot 10=60.22\cdot 100=6.022\cdot 1000.$ \ Thus our number is \vspace{0.04in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $602\,200\,\allowbreak 000\,000\,000\,\allowbreak 000\,000\,000=602\,2\cdot 10^{20}=\left( 6.022\cdot 1000\right) \cdot 10^{20}=6.022\cdot 10^{3+20}=$ \fbox{$6.022\cdot 10^{23}$}\vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion This number is famous in chemistry, it is called Avogadro's number. \ 1 mole of a substance contains $6.022\cdot 10^{23}$ particles. \ Mole is the SI unit for the amount of a substance. \ \vspace{0.08in}\vspace{0.04in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 9. \ }Suppose that two numbers, $A$ and $B$ are given in scientfici notation as follows. \ $A=3\cdot 10^{8}$ and $B=6\cdot 10^{3}$. \ Compute each of the following. \ Present your answer in scientific notation.% \vspace{0.04in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $A+B$ \ \ \ \ \ \ \ \ b) \ $A-B$ \ \ \ \ \ \ \ c) \ $AB$ \ \ \ \ \ \ \ \ d) \ $\dfrac{A}{B}$\vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ Surprisingly, $A+B$ will look a whole lot like $A$% . \ Scientific notation will serve us very well in multiplications and divisions, but we must be careful when adding or subtracting. \ To see what happens, we will return to regular notation. \ Notice also that there is no rule for how to add exponential expressions in the rules we just learned. \ \vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $A=3\cdot 10^{8}=300\,000\,000$ and $B=6\cdot 10^{3}=6000$. \ So $% A+B=300\,000\,000+6000=300\,006\,000$. \ \vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion This can be written in scientific notation as $3.00006\cdot 10^{3}$ or simply $3\cdot 10^{8}$. \ Basically, $A$ is so much larger than $B$, that the addition of $B$ is almost negligible compared to the size of $A$. \ (Imagine that we\vspace{0.04in} added 1 inch to 1 mile or a penny to a million dollars) \ Indeed, in science, there will be agreements about the precision of results, and in most agreements, $3\cdot 10^{8}+6\cdot 10^{3}$% \vspace{0.04in} is simply $3\cdot 10^{8}$, as we round $3.00006$ down to $3$% . \ So the answer is \fbox{$300\,006\,000$} or \fbox{$3.00006\cdot 10^{3}\approx 3\cdot 10^{8}$}.\vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ Very similarly, $A-B=300\,000\,000-6000=\,$\fbox{$299\,994\,000$}.% \vspace{0.04in} \ Using scientific notation, the result is %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion \fbox{$2.99994\cdot 10^{8}\approx 3\cdot 10^{8}$}.\vspace{0.04in} \ We can imagine that we subtracted an inch from a mile or a penny from a million dollars. %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ Scientific notation will be awesome for mutiplication and division!% \vspace{0.04in} \ Consider $AB=\left( 3\cdot 10^{8}\right) \left( 6\cdot 10^{3}\right) $. \ \vspace{0.04in} \ We will group the first factors and the ten-powers. \ $3\cdot 6=\allowbreak 18$ and $10^{8}\cdot 10^{3}=10^{3+8}=10^{11}$.\vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $AB=\left( 3\cdot 10^{8}\right) \left( 6\cdot 10^{3}\right) =\left( 3\cdot 6\right) \cdot \left( 10^{8}\cdot 10^{3}\right) =18\cdot 10^{11}$\vspace{% 0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion This is not in scientific notation. \ Recall that in scientific notation, the first factor must be less than $10$. \ So we have to re-write $18$ as $% 1.8\cdot 10$. \ \vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $AB=18\cdot 10^{11}=\left( 1.8\cdot 10\right) \cdot 10^{11}=1.8\cdot 10^{12}$% . \ So the answer is \ \fbox{$1.8\cdot 10^{12}=1800\,000\,000\,000$}.\vspace{% 0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion d) \ In order to help the division, we will re-write $A$ from $3\cdot 10^{8}$ to $30\cdot 10^{7}$. \ Then the rules of exponents will work with the notation very nicely.\vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $\dfrac{A}{B}=\dfrac{3\cdot 10^{8}}{6\cdot 10^{3}}=\dfrac{30\cdot 10^{7}}{% 6\cdot 10^{3}}$\vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion We simply divide $30$ by $6.$ \ As for the ten-powers, then cancellation can be expressed via the rule\vspace{0.04in} $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $\dfrac{A}{B}=\dfrac{30\cdot 10^{7}}{6\cdot 10^{3}}=\dfrac{5\cdot 10^{7-3}}{1% }=\,$\fbox{$5\cdot 10^{4}$ \ or \ $50\,000$}.\vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \pagebreak \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}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item Simplify each of the following.% %TCIMACRO{\TeXButton{4col begin}{\begin{multicols}{4}}}% %BeginExpansion \begin{multicols}{4}% %EndExpansion \medskip a) $\ \left( 2x^{5}\right) \left( x^{4}\right) $\bigskip b) $\ \left( 2x\right) ^{5}\left( x^{4}\right) $\bigskip c) $\ \left( 2x^{5}\right) ^{4}$\bigskip d) $\ \left( -xy\right) ^{2}\left( -xy^{2}\right) ^{3}$\bigskip e) $\ -2a^{3}\left( -2a^{4}\right) ^{2}$\bigskip f) $\ 2a^{3}\left( -2ab^{2}\right) ^{3}ab^{2}$\bigskip g) $\ \dfrac{\left( -2x^{5}\right) ^{2}y^{3}}{2x^{3}y^{2}}$ h) $\ \dfrac{\left( 2ab\right) ^{3}\left( -3a^{2}b\right) ^{2}}{-a\left( 6ab^{2}\right) ^{2}}$\ %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \item Write each of the following expressions in terms of a fixed number or a single exponential expression.\medskip a) \ $\dfrac{3^{2x+1}}{9^{x-1}}$ \ \ \ \ \ \ \ \ \ \ \ \ \ b) \ $\dfrac{% \left( 8^{b-2}\right) \left( 2^{b+1}\right) }{4^{2b-3}}$ \ \ \ \ \ \ \ \ \ \ \ \ c) \ $5^{2x-1}\cdot 25^{3-x}$\vspace{0.09in} \item Let us denote $3^{100}$ by $M$. \ \ Express each of the following in terms of $M.$\medskip a) \ $3^{101}\qquad $b) \ $3^{100}-2\cdot 3^{101}+3^{102}\qquad $c) \ $% 3^{99}\qquad $d) \ $9^{100}$ \item Find the prime-factorization for each of the following numbers. a) \ $10^{2018}$ \ \ \ \ \ \ b) \ $18^{1000}$ \ \ \ \ \ \ c) \ $360^{50}$ \ \ \ \ \ \ d) \ $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$ \item Re-write each of the given numbers using scientific notation. a) \ $3800\,000\,000$ \ \ \ \ \ \ b) \ $6250\,\allowbreak 000\,000\,000$ \item Suppose that $A=5\cdot 10^{18}$, and \ $B=8\cdot 10^{7}.$ \ \ Compute each of the following. \ Present your answer using scientific notation. a) \ $AB$ \ \ \ \ \ \ b) \ $A^{2}$ \ \ \ \ c) \ $2A$ \ \ \ \ \ d) \ $\dfrac{% 4A}{B}$ \end{enumerate} \pagebreak \FRAME{itbpF}{0.7074in}{0.6797in}{0.2508in}{}{}{work.jpg}{\special{language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file "F";width 0.7074in;height 0.6797in;depth 0.2508in;original-width 4.6977in;original-height 5.2607in;cropleft "0";croptop "1.003173";cropright "1";cropbottom "0.003173";filename 'work.jpg';file-properties "XNPEU";}}\ \ {\LARGE Practice Problems }% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item Simplify each of the following.\medskip %TCIMACRO{\TeXButton{5col begin}{\begin{multicols}{5}}}% %BeginExpansion \begin{multicols}{5}% %EndExpansion a) \ $-3^{2}$\bigskip b) \ $\left( -3\right) ^{2}$\bigskip c) \ $\left( -2a^{3}\right) ^{4}$\bigskip d) \ $\left( -2a^{4}\right) ^{3}$\bigskip e) \ $\left( 2a^{2}b\right) ^{3}$\bigskip f) \ $\left( \left( 2a\right) ^{2}b\right) ^{3}$\bigskip g) \ $\left( 2a\right) ^{2}b^{3}$\bigskip h) \ $\dfrac{m^{4}m^{5}}{m^{3}}$\bigskip i) \ $\left( 3p^{2}q^{5}\right) \left( 2pq^{3}\right) $\bigskip j) \ $\dfrac{\left( a^{2}\right) ^{6}a^{3}}{\left( -a^{3}\right) ^{2}}$% \bigskip %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \qquad k) \ $\dfrac{\left( -5ts^{3}t\right) ^{3}\left( 4s^{2}t\right) ^{2}}{% \left( 10st^{3}\right) ^{2}}$\bigskip \qquad l) $\ \left( -2xy^{3}\right) ^{2}xy^{5}x^{2}$ m) $\ \dfrac{\left( 3ab^{2}\right) ^{2}\left( -2a^{3}b\right) ^{4}}{\left( -2ab\right) ^{3}}$\vspace{0.1in} n) $\ \dfrac{\left( -2x^{2}y^{3}\right) ^{4}xy^{3}\left( 2x^{2}y\right) ^{2}% }{\left( 2x\right) ^{2}y^{9}\left( 2x^{2}y\right) ^{4}}$\vspace{0.1in} o) \ $\left( \dfrac{6a^{3}b^{5}}{-3ab^{2}}\right) ^{2}\left( \dfrac{12ab^{3}% }{-6b^{2}}\right) ^{3}$% %TCIMACRO{\TeXButton{multcol end}{\end{multicols}} }% %BeginExpansion \end{multicols} %EndExpansion \ \ \item Write each of the following expressions in terms of a fixed number or a single exponential expression.\medskip a) \ $\dfrac{2^{2x-1}}{4^{x-2}}$\ \ \ \ \ \ \ \ \ \ b) \ $\dfrac{100^{x+1}}{% 2^{2x+1}\cdot 5^{x-1}}$ \ \ \ \ \ \ \ c) \ $\dfrac{9^{x}\cdot 4^{x+2}}{% 6^{2x-1}}$ \item Let $P$ denote $5^{2015}$. \ \ Express each of the following in terms of $P$.\medskip a) \ $5^{2016}$ \ \ \ \ \ \ \ b) \ $5^{2017}$ \ \ \ \ \ \ c) \ $5^{2014}$ \ \ \ \ \ \ \ d) \ $25^{2015}$ \ \ \ \ \ \ e) \ $5^{2015}-3\cdot 5^{2016}+5^{2017}$ \item Find the prime-factorization for each of the given numbers. a) \ $20^{100}$ \ \ \ \ \ \ \ b) \ $18^{2000}$ \ \ \ \ \ \ \ c) \ $120^{120}$ \ \ \ \ \ \ d) \ $\left( 5\cdot 4\cdot 3\cdot 2\cdot 1\right) ^{3}$ \item Re-write each of the given numbers in scientific notation. a) \ $21\,\allowbreak 000\,000\,000$ \ \ \ \ b) \ $300\,\allowbreak 000\,000\,000$ \ \ \ \ \ \ c) \ $325\,000\,000$ \item Suppose that $X=3\cdot 10^{10}$, \ and \ $Y=6\cdot 10^{4}$. \ \ Compute each of the following. \ Present your answer using scientific notation. a) \ $X^{2}$ \ \ \ \ \ \ b) \ $XY$ \ \ \ \ \ c) \ $\dfrac{X}{Y}$ \ \ \ \ d) \ $\dfrac{3X}{Y^{2}}$ \end{enumerate} \pagebreak {\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 Discussion}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item Two minus signs still cancel out each other, it is just that there are really three negative signs in the \newline expression $-\left( -5\right) ^{2}=-\left( -5\right) \left( -5\right) $. \item $a$ and $b$ appear in the denominator and division by zero is not allowed. \item $2\cdot 5^{x}=2\cdot \underset{x\text{ times}}{\underbrace{5\cdot 5\cdot \ldots \cdot 5}}$ \ \ Imagine that $x$ is very large. \ Then we have lots of $5$-factors but just one $2$-factor. In order to be able to simplify to $10^{x}$, we would need to see $% 2^{x}\cdot 5^{x}$.\vspace{0.1in}\vspace{0.2in} \end{enumerate} {\Large Sample Problems}{\large \ }% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item a) $\ 2x^{9}$ \ \ \ b) $\ 32x^{9}$ \ \ \ c) $\ 16x^{20}$\ \ \ d) $\ -x^{5}y^{8}$ \ \ \ e) $\ -8a^{11}$ \ \ \ f) $\ -16a^{7}b^{8}$\ \ \ \ g) $\ 2x^{7}y$ \ \ \ h) $\ -2a^{4}b$ \vspace{0.04in} \item a) \ $27$ \ \ \ \ \ b) \ $2$ \ \ \ \ \ \ c) \ $3125$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3. \ a) \ $3M\qquad $b) \ $4M\qquad $c) \ $\dfrac{M}{3} $ \ \ \ \ \ \ d) \ $M^{2}$\vspace{0.04in} \item[4.] a) \ $2^{2018}\cdot 5^{2018}$ \ \ \ \ \ \ b) \ $2^{1000}\cdot 3^{2000}$ \ \ \ \ \ c) \ $2^{150}\cdot 3^{100}\cdot 5^{50}$ \ \ \ \ \ d) \ $% 2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7$\vspace{0.04in} \item[5.] a) \ $3.8\cdot 10^{9}$ \ \ \ \ \ \ \ b) \ $6.25\cdot 10^{12}$% \vspace{0.04in} \ \ \ \ \ \ \ \ 6. \ a) \ $4\cdot 10^{26}$\ \ \ \ \ \ \ b) \ $2.5\cdot 10^{37}$\ \ \ \ \ \ c) \ $1\cdot 10^{19}$\ \ \ \ \ d) \ $2.5\cdot 10^{11}$ \item[6.] a) \ $9\cdot 10^{20}$ \ \ \ \ \ b) \ $1.8\cdot 10^{15}$ \ \ \ \ \ c) \ $5\cdot 10^{5}$ \ \ \ \ \ \ d) $2.5\cdot 10$ \end{enumerate} \vspace{0.2in} {\Large Practice Problems}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item a) \ $-9$ \ \ \ \ b) \ $9$ \ \ \ \ c) \ $16a^{12}$ \ \ \ d) \ $% -8a^{12} $\ \ \ \ \ e) \ $8a^{6}b^{3}$ \ \ \ \ \ f) \ $64a^{6}b^{3}$ \ \ \ \ g) \ $4a^{2}b^{3}$ \ \ \ \ \ h) \ $m^{6}$ \ \ \ \ i) \ $6p^{3}q^{8}$ \ \ \ \ \ j) \ $a^{9}$ k) \ $-20s^{11}t^{2}$\ \ \ \ \ l) $\ 4x^{5}y^{11}$ \ \ \ \ m) $\ -18a^{11}b^{5}$ \ \ \ \ \ \ n) $\ x^{3}y^{4}$ \ \ \ \ \ o) \ $-32a^{7}b^{9}$% \ \ \ \ \ \ \ \ 2. \ a) \ $8$\ \ \ \ \ \ b) \ $250\cdot 5^{x}$ \ \ \ \ \ \ c) \ $96$ \item[3.] a) \ $5P$ \ \ \ \ \ b) \ $25P$ \ \ \ \ \ \ \ c) \ $\dfrac{P}{5}$ \ \ \ \ \ \ \ \ d) \ $P^{2}$ \ \ \ \ \ \ e) \ $11P$ \ \ \ \ \item[4.] a) \ $2^{200}5^{100}$ \ \ \ \ \ \ \ b) \ $2^{2000}\cdot 3^{4000}$ \ \ \ \ \ \ \ c) \ $2^{360}\cdot 3^{120}\cdot 5^{120}$ \ \ \ \ \ \ d) \ $% 2^{9}\cdot 3^{3}\cdot 5^{3}$ \item[5.] a) \ $2.1\cdot 10^{10}$ \ \ \ \ \ \ b) \ $3\cdot 10^{14}$ \ \ \ \ \ c) \ $3.25\cdot 10^{8}$ \item[6.] a) \ $9\cdot 10^{20}$ \ \ \ \ \ b) \ $1.8\cdot 10^{15}$ \ \ \ \ \ \ \ c) \ $5\cdot 10^{5}$ \ \ \ \ \ \ \ d) \ $2.5\cdot 10$ \pagebreak \end{enumerate} \begin{center} \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 \end{center} Let us recall the rules of exponents.% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{eqnarray*} \text{1) \ \ \ }a^{n}\cdot a^{m} &=&a^{n+m} \\ \text{2) \ \ \ \ \ \ \ }\dfrac{a^{n}}{a^{m}} &=&a^{n-m} \\ \text{3) \ \ \ \ }\left( a^{n}\right) ^{m} &=&a^{nm} \\ \text{4) \ \ \ \ \ }\left( ab\right) ^{n} &=&a^{n}b^{n} \\ \text{5) \ \ \ \ }\left( \dfrac{a}{b}\right) ^{n} &=&\dfrac{a^{n}}{b^{n}} \end{eqnarray*}% \bigskip \begin{enumerate} \item Simplify each of the following. a) \ $\left( 2x^{5}\right) \left( x^{4}\right) $ Solution: \ $\left( 2x^{5}\right) \left( x^{4}\right) =2x^{5}x^{4}=2x^{5+4}=\,$\fbox{$2x^{9}$} \ \ by rule 1.\vspace{0.08in} b) $\ \left( 2x\right) ^{5}\left( x^{4}\right) $ Solution: \ \begin{eqnarray*} \left( 2x\right) ^{5}\left( x^{4}\right) &=&2^{5}x^{5}x^{4}\text{ \ \ \ \ \ \ \ \ \ by rule 4} \\ &=&32x^{5+4}\text{ \ \ \ \ \ \ \ \ \ by rule 1} \\ &=&\fbox{$32x^{9}$} \end{eqnarray*} c) $\ \left( 2x^{5}\right) ^{4}$ Solution:% \begin{eqnarray*} \left( 2x^{5}\right) ^{4} &=&2^{4}\left( x^{5}\right) ^{4}\text{ \ \ \ \ \ \ \ \ \ \ \ by rule 4} \\ &=&\fbox{$16x^{20}$}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule 3} \end{eqnarray*} d) $\ \left( -xy\right) ^{2}\left( -xy^{2}\right) ^{3}$ Solution:% \begin{eqnarray*} \left( -xy\right) ^{2}\left( -xy^{2}\right) ^{3} &=&\left( -1xy\right) ^{2}\left( -1xy^{2}\right) ^{3}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ the }1\text{s \ will help with signs} \\ &=&\left( -1\right) ^{2}x^{2}y^{2}\left( -1\right) ^{3}x^{3}\left( y^{2}\right) ^{3}\text{ \ \ \ \ \ \ \ \ \ \ \ \ by rule 4} \\ &=&1\cdot x^{2}y^{2}\left( -1\right) x^{3}y^{6}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule 3} \\ &=&1\left( -1\right) x^{2}x^{3}y^{2}y^{6}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiplication is commutative} \\ &=&-1~x^{2+3}y^{2+6}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule 1} \\ &=&\fbox{$-x^{5}y^{8}$} \end{eqnarray*} \pagebreak e) $\ -2a^{3}\left( -2a^{4}\right) ^{2}$ Solution:% \begin{eqnarray*} -2a^{3}\left( -2a^{4}\right) ^{2} &=&-2a^{3}\left( -2\right) ^{2}\left( a^{4}\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 4} \\ &=&-2a^{3}\left( 4\right) a^{8}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 3 } \\ &=&-2\left( 4\right) a^{3}a^{8}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiplication is commutative} \\ &=&-8a^{3+8}=\fbox{$-8a^{11}$}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 1% } \end{eqnarray*} f) $\ 2a^{3}\left( -2ab^{2}\right) ^{3}ab^{2}$ Solution:% \begin{eqnarray*} 2a^{3}\left( -2ab^{2}\right) ^{3}ab^{2} &=&2a^{3}\left( -2\right) ^{3}a^{3}\left( b^{2}\right) ^{3}ab^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 4} \\ &=&2a^{3}\left( -8\right) a^{3}b^{6}ab^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 3} \\ &=&2\left( -8\right) a^{3}a^{3}ab^{6}b^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiplication is commutative} \\ &=&-16a^{3+3+1}b^{6+2}=\fbox{$-16a^{7}b^{8}$}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 1} \end{eqnarray*} g) $\ \dfrac{\left( -2x^{5}\right) ^{2}y^{3}}{2x^{3}y^{2}}$ Solution:% \begin{eqnarray*} \dfrac{\left( -2x^{5}\right) ^{2}y^{3}}{2x^{3}y^{2}} &=&\dfrac{\left( -2\right) ^{2}\left( x^{5}\right) ^{2}y^{3}}{2x^{3}y^{2}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 4} \\ &=&\dfrac{4x^{10}y^{3}}{2x^{3}y^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 3} \\ &=&\dfrac{4x^{10-3}y^{3-2}}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 2} \\ &=&\dfrac{4x^{7}y^{1}}{2}=\fbox{$2x^{7}y$} \end{eqnarray*} h) $\ \dfrac{\left( 2ab\right) ^{3}\left( -3a^{2}b\right) ^{2}}{-a\left( 6ab^{2}\right) ^{2}}$ Solution: \begin{eqnarray*} \dfrac{\left( 2ab\right) ^{3}\left( -3a^{2}b\right) ^{2}}{-a\left( 6ab^{2}\right) ^{2}} &=&\dfrac{\left( 2ab\right) ^{3}\left( -3a^{2}b\right) ^{2}}{-1a\left( 6ab^{2}\right) ^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ the }1\text{\ \ will help with signs} \\ &=&\dfrac{2^{3}a^{3}b^{3}\left( -3\right) ^{2}\left( a^{2}\right) ^{2}b^{2}}{% -1\cdot a\cdot 6^{2}\cdot a^{2}\left( b^{2}\right) ^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule\ 4} \\ &=&\dfrac{8a^{3}b^{3}\cdot 9\cdot a^{4}b^{2}}{-1\cdot a\cdot 36\cdot a^{2}b^{4}}=\dfrac{8\cdot 9\cdot a^{3}a^{4}b^{3}b^{2}}{-1\cdot 36\cdot a\cdot a^{2}\cdot b^{4}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule 3} \\ &=&\dfrac{72a^{3+4}b^{3+2}}{-36a^{1+2}b^{4}}=\dfrac{72a^{7}b^{5}}{% -36a^{3}b^{4}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule 1} \\ &=&\dfrac{-2a^{7}b^{5}}{a^{3}b^{4}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ simplify numbers: }\dfrac{72}{-36}=% \dfrac{-72}{36}=\dfrac{-2}{1} \\ &=&\dfrac{-2a^{7-3}b^{5-4}}{1}=-2a^{4}b^{1}=\fbox{$-2a^{4}b$}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 2\ } \end{eqnarray*} \item Write each of the following expressions in terms of a fixed number or a single exponential expression.\vspace{0.04in} a) \ $\dfrac{3^{2x+1}}{9^{x-1}}$ Solution: \ We will re-write the denominator in terms of base $3.$ After that, we can apply $\dfrac{a^{n}}{a^{m}}=a^{n-m}$.% \begin{equation*} \dfrac{3^{2x+1}}{9^{x-1}}=\dfrac{3^{2x+1}}{\left( 3^{2}\right) ^{x-1}}~~% \overset{\text{Rule 3}}{=}~~\dfrac{3^{2x+1}}{3^{2\left( x-1\right) }}=\dfrac{% 3^{2x+1}}{3^{2x-2}}\overset{\text{Rule 2}}{~~~~~=~~~~}3^{2x+1-\left( 2x-2\right) }=3^{2x+1-2x+2}=3^{3}=\fbox{$27$} \end{equation*} b) \ $\dfrac{\left( 8^{b-2}\right) \left( 2^{b+1}\right) }{4^{2b-3}}$ Solution: \ We will re-write each exponential expressions in terms of base $% 2.$ \begin{eqnarray*} \dfrac{\left( 8^{b-2}\right) \left( 2^{b+1}\right) }{4^{2b-3}} &=&\dfrac{% 8^{b-2}\cdot 2^{b+1}}{4^{2b-3}}=\dfrac{\left( 2^{3}\right) ^{b-2}\cdot 2^{b+1}}{\left( 2^{2}\right) ^{2b-3}}\overset{\text{Rule 3}}{~~~=~~~}\dfrac{% 2^{3\left( b-2\right) }\left( 2^{b+1}\right) }{2^{2\left( 2b-3\right) }}=% \dfrac{2^{3b-6}\cdot 2^{b+1}}{2^{4b-6}}\overset{\text{Rule 1}}{~~~=~~~}% \dfrac{2^{3b-6+\left( b+1\right) }}{2^{4b-6}} \\ &=&\dfrac{2^{4b-5}}{2^{4b-6}}\overset{\text{Rule 2}}{~~~=~~~}2^{\left( 4b-5\right) -\left( 4b-6\right) }=2^{4b-5-4b+6}=2^{1}=\fbox{$2$} \end{eqnarray*} c) \ $5^{2x-1}\cdot 25^{3-x}$% \begin{equation*} 5^{2x-1}\cdot 25^{3-x}=5^{2x-1}\cdot \left( 5^{2}\right) ^{3-x}\overset{% \text{Rule 3}}{~~~=~~~}5^{2x-1}\cdot 5^{2\left( 3-x\right) }=5^{2x-1}\cdot 5^{6-2x}=\overset{\text{Rule 1}}{~~~=~~~}5^{2x-1+6-2x}=5^{5}=\fbox{$3125$} \end{equation*} \item Let us denote $3^{100}$ by $M$. \ \ Express each of the following in terms of $M.$ a) \ $3^{101}$ Solution: \ Using rule 1, we write $3^{101}=3^{100+1}=3^{100}\cdot 3^{1}=M\cdot 3=\,$\fbox{$3M$}\vspace{0.09in} b) \ $3^{100}-2\cdot 3^{101}+3^{102}$ Solution: \ Using rule 1, we re-write $3^{101}$ and $3^{102}$% \begin{eqnarray*} 3^{101} &=&3^{100+1}=3^{100}\cdot 3^{1}=M\cdot 3=3M \\ 3^{102} &=&3^{100+2}=3^{100}\cdot 3^{2}=M\cdot 9=9M \end{eqnarray*}% Then our expression becomes% \begin{equation*} 3^{100}-2\cdot 3^{101}+3^{102}=M-2\cdot \left( 3M\right) +9M=M-6M+9M=-5M+9M=% \fbox{$4M$} \end{equation*} c) \ $3^{99}$ Solution: \ Using rule 2, we write $3^{99}=3^{100-1}=\dfrac{3^{100}}{3^{1}}% =\,$\fbox{$\dfrac{M}{3}$} d) \ $9^{100}$ Solution: \ This time we will use rule 3 in a novel way: $\left( a^{n}\right) ^{m}=\left( a^{m}\right) ^{n}$\ \begin{equation*} 9^{100}=\left( 3^{2}\right) ^{100}=\left( 3^{100}\right) ^{2}=M^{2}\text{ } \end{equation*}% We can also solve this problem using rule 4% \begin{equation*} 9^{100}=\left( 3\cdot 3\right) ^{100}=3^{100}\cdot 3^{100}=M\cdot M=\fbox{$% M^{2}$}\text{ } \end{equation*} \item[4.] Find the prime-factorization for each of the following numbers. a) \ $10^{2018}$ Solution: \ We will find the prime-factorization of the base and then apply rules of exponents.\vspace{0.04in} $10^{2018}=\left( 2\cdot 5\right) ^{2018}=\fbox{$2^{2018}\cdot 5^{2018}$}% \vspace{0.09in}$ b) \ $18^{1000}$ Solution: \ We will find the prime-factorization of the base and then apply rules of exponents.\vspace{0.04in} $18^{1000}=\left( 2\cdot 9\right) ^{1000}=\left( 2\cdot 3^{2}\right) ^{1000}=2^{1000}\cdot \left( 3^{2}\right) ^{1000}=2^{1000}\cdot 3^{2\cdot 1000}=\,$\fbox{$2^{1000}\cdot 3^{2000}$}$\vspace{0.09in}$ c) \ $360^{50}$ Solution: \ We will find the prime-factorization of the base and then apply rules of exponents.\vspace{0.04in} $360=36\cdot 10=\left( 4\cdot 9\right) \cdot \left( 2\cdot 5\right) =2^{3}\cdot 3^{2}\cdot 5$. \ \ So the prime factorization of $360$ is $% 2^{3}\cdot 3^{2}\cdot 5$. $360^{50}=\left( 2^{3}\cdot 3^{2}\cdot 5\right) ^{50}=\left( 2^{3}\right) ^{50}\left( 3^{2}\right) ^{50}\left( 5\right) ^{50}=2^{3\cdot 50}\cdot 3^{2\cdot 50}\cdot 5^{50}=\,$\fbox{$2^{150}\cdot 3^{100}\cdot 5^{50}$}$% \vspace{0.09in}$ d) \ The product $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$ comes up a lot in mathematics, so there is notation for it. \ $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=10!$ \ We pronounce $10!$ as ten factorial. Now we just find the prime-factorization of each factor and collect the prime-factorization that way.% \begin{eqnarray*} 10! &=&10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1% \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ we drop }1\text{ at the end} \\ &=&\left( 2\cdot 5\right) \left( 3^{2}\right) \left( 2^{3}\right) \cdot 7\cdot \left( 2\cdot 3\right) \cdot 5\cdot 2^{2}\cdot 3\cdot 2 \end{eqnarray*}% We collect the two-factors: one from $10,$ three from $8,$ one from $6$, two from $4,$ and one from two. \ Similarly, the three-factors: two from $9,$ one from $6$ and one from $3.$ \ Five factors come out only from $10$ and $% 5, $ one from each, and the greatest prime factor is $7$.% \begin{eqnarray*} 10! &=&\left( 2\cdot 5\right) \left( 3^{2}\right) \left( 2^{3}\right) \cdot 7\cdot \left( 2\cdot 3\right) \cdot 5\cdot 2^{2}\cdot 3\cdot 2 \\ &=&2^{1+3+1+2+1}\cdot 3^{2+1+1}\cdot 5^{1+1}\cdot 7=\fbox{$2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7$} \end{eqnarray*} \pagebreak \item[5.] Re-write each of the given numbers using scientific notation. a) \ $3800\,000\,000$ Solution: \ $3800\,000\,000$ ends in eight zeroes. \ This can be translated as $3800\,000\,000=38\cdot 10^{8}$. \ \ We are not done yet: the first factor is too big, it must be between $1$ and $10$. So we re-write $38$ as $% 3.8\cdot 10$ and use the rules of exponents: $3800\,000\,000=38\cdot 10^{8}=\left( 3.8\cdot 10\right) \cdot 10^{8}=\,$% \fbox{$3.8\cdot 10^{9}$} b) \ $6250\,\allowbreak 000\,000\,000$ Solution: \ We count ten trailing zeroes, so $6250\,\allowbreak 000\,000\,000=625\cdot 10^{10}$. \ We re-write $625$ as $% 6.25\cdot 10^{2}$. \ So $6250\,\allowbreak 000\,000\,000=625\cdot 10^{10}=\left( 6.25\cdot 10^{2}\right) \cdot 10^{10}=\,$\fbox{$6.25\cdot 10^{12}$} \item[6.] Suppose that $A=5\cdot 10^{18}$, and \ $B=8\cdot 10^{7}.$ \ \ Compute each of the following. \ Present your answer using scientific notation. a) \ $AB$ \ \ \ \ \ \ b) \ $A^{2}$ \ \ \ \ c) \ $2A$ \ \ \ \ \ d) \ $\dfrac{% 4A}{B}$ Solution: \ $AB=\left( 5\cdot 10^{18}\right) \left( 8\cdot 10^{7}\right) =\left( 5\cdot 8\right) \left( 10^{18}\cdot 10^{7}\right) =40\cdot 10^{25}=\, $\fbox{$4\cdot 10^{26}$} b) \ $A^{2}$ Solution: \ $A^{2}=\left( 5\cdot 10^{18}\right) ^{2}=5^{2}\cdot \left( 10^{18}\right) ^{2}=25\cdot 10^{18\cdot 2}=25\cdot 10^{36}=2.5\cdot 10\cdot 10^{36}=\,$\fbox{$2.5\cdot 10^{37}$} c) \ $2A$ Solution: \ $2A=2\left( 5\cdot 10^{18}\right) =\left( 2\cdot 5\right) \cdot 10^{18}=10\cdot 10^{18}=\,$\fbox{$1\cdot 10^{19}$} d) \ $\dfrac{4A}{B}\vspace{0.04in}$ Solution: \ $\dfrac{4A}{B}=\dfrac{4\left( 5\cdot 10^{18}\right) }{8\cdot 10^{7}}=\dfrac{20\cdot 10^{18}}{8\cdot 10^{7}}\vspace{0.04in}$ Unfortunately, $8$ is not a divisor of $20$, but it is a divisor of $200$. \ So, we borrow a ten from the ten-power.$\vspace{0.04in}$ \ \ $\dfrac{20\cdot 10^{18}}{8\cdot 10^{7}}=\dfrac{200\cdot 10^{17}}{8\cdot 10^{7}}=25\cdot 10^{10}=\,$\fbox{$2.5\cdot 10^{11}$}\vspace{1in}\vspace{1in}$% \vspace{0.2in}\vspace{0.2in}\vspace{0.2in}\vspace{0.2in}\vspace{0.2in}$ \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}