\hfill \hfill } %} \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}{Definition} \newtheorem{example}{Example} \newtheorem{exercise}{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}} \input{tcilatex} \geometry{left=0.4in,right=0.6in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Exponents 1} \rhead{ page \ \thepage} \cfoot{} \lfoot{\small \copyright \; Hidegkuti, Powell, 2011} \rfoot{\small Last revised: June 4, 2015} \textwidth 7.6in \textheight 9.6in \setlength{\headheight}{25pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\Large Sample Problems}\bigskip \bigskip \end{center} \begin{enumerate} \item Simplify each of the following.% %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %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) $\ \dfrac{\left( -2x\right) ^{2}y^{3}}{2x^{3}y^{2}}$\bigskip h) $\ \dfrac{\left( 2ab\right) ^{3}\left( -3a^{2}b\right) ^{2}}{-b\left( 6ab^{2}\right) ^{2}}$\bigskip i) $\ \left( \dfrac{-2ab}{3b^{3}}\right) ^{3}\left( \dfrac{6ab^{4}}{4a^{3}b}% \right) ^{2}$\bigskip\ %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) \ $5^{2x-1}\cdot 25^{3-x}$\bigskip \item Let us denote $3^{100}$ by $M$. \ \ Express each of the following in terms of $M.$\medskip a) \ $3^{101}\medskip \qquad $b) \ $3^{100}-2\cdot 3^{101}+3^{102}\medskip \qquad $c) \ $3^{99}\medskip \qquad $d) \ $9^{100}\medskip $ \end{enumerate} \begin{center} {\Large Practice Problems}\bigskip \bigskip \end{center} \begin{enumerate} \item Simplify each of the following.\medskip %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion a) \ $-3^{2}$\bigskip b) \ $\left( -3\right) ^{2}$\bigskip c) \ $\left( -\dfrac{2}{3}\right) ^{3}$\bigskip d) \ $\left( -\dfrac{1}{2}\right) ^{4}$\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 k) \ $\dfrac{\left( -5s^{3}t\right) ^{4}\left( s^{2}t\right) ^{3}}{\left( 10st^{3}\right) ^{2}}$\bigskip l) $\ \left( -2xy^{3}\right) ^{2}xy^{5}x^{2}$\bigskip m) $\ \dfrac{\left( 3ab^{2}\right) ^{2}\left( -2a^{3}b\right) ^{4}}{\left( -2ab\right) ^{3}}$\bigskip n) $\ \dfrac{\left( -2x^{2}y^{3}\right) ^{4}xy^{3}\left( 2x^{2}y\right) ^{2}% }{\left( 2x\right) ^{2}y^{9}\left( 2x^{2}y\right) ^{4}}$\bigskip o) \ $\left( \dfrac{2a^{3}b}{-3ab^{2}}\right) ^{2}\left( \dfrac{3ab}{6b^{2}}% \right) ^{3}$\bigskip %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) \ $\dfrac{2^{2x-1}}{4^{x-2}}$\bigskip\ \ \ \ \ \ \ \ \ \ b) \ $\dfrac{% 100^{x+1}}{2^{2x+1}\cdot 5^{x-1}}$ \ \ \ \ \ \ \ c) \ $\dfrac{9^{x-1}\cdot 4^{x+2}}{6^{2x+1}}$ \ \ \ \ \ \ \ \ \ d) \ $\dfrac{4^{x-1}\cdot 5^{x+2}}{% 10^{x-1}}$ \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) \ $5^{2015}-3\cdot 5^{2016}+5^{2017}$ \end{enumerate} \pagebreak \begin{center} {\Large Sample Problems - Answers}\bigskip \bigskip \end{center} 1. \ a) $\ 2x^{9}$ \ \ \ b) $\ 32x^{9}$ \ \ \ c) $\ 16x^{20}$\ \ \ d) $\ -x^{5}y^{8}$ \ \ \ e) $\ -8a^{11}$ \ \ \ f) $\ -16a^{7}b^{8}$\ \ \ \ g) $\ \dfrac{2y}{x}$ \ \ \ h) $\ -2a^{5}$ \ \ i) $\ -\dfrac{2}{3a}$\bigskip 2. \ a) \ $27$ \ \ \ \ \ b) \ $2$ \ \ \ \ \ \ c) \ $3125$ \ \ \ \ \ \ \ 3. \ a) \ $3M\qquad $b) \ $4M\qquad $c) \ $\dfrac{M}{3}$ \ \ \ \ \ \ d) \ $M^{2}$ \bigskip \begin{center} {\Large Practice Problems - Answers}\bigskip \bigskip \end{center} 1. \ a) \ $-9$ \ \ \ \ \ b) \ $9$ \ \ \ \ \ c) \ $-\dfrac{8}{27}$ \ \ \ \ d) \ $\dfrac{1}{16}$ \ \ \ \ \ e) \ $8a^{6}b^{3}$ \ \ \ \ \ f) \ $64a^{6}b^{3}$ \ \ \ \ g) \ $4a^{2}b^{3}$ \ \ \ \ \ h) \ $m^{6}$ \bigskip i) \ $6p^{3}q^{8}$ \ \ \ \ \ j) \ $a^{9}$\ \ \ \ \ \ k) \ $\dfrac{25s^{16}t}{% 4}$ \ \ \ \ \ \ l) $\ 4x^{5}y^{11}$ \ \ \ \ m) $\ -18a^{11}b^{5}$ \ \ \ \ \ \ n) $\ x^{3}y^{4}$ \ \ \ \ \ o) \ $\dfrac{a^{7}}{18b^{5}}$\bigskip 2. \ a) \ $8$\bigskip\ \ \ \ \ \ b) \ $250\cdot 5^{x}$ \ \ \ \ \ \ c) \ $% \dfrac{8}{27}$ \ \ \ \ \ \ \ d) \ $\dfrac{125}{2}\cdot 2^{x}$ 3. \ a) \ $5P$ \ \ \ \ \ b) \ $25P$ \ \ \ \ \ \ \ c) \ $\dfrac{P}{5}$ \ \ \ \ \ \ \ \ d) \ $P^{2}$ \ \ \ \ \ \ e) \ $11P$\pagebreak \begin{center} {\Large Sample Problems - Solutions}\bigskip \end{center} Let us recall the rules of exponents.% \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 1. \ Simplify each of the following.\bigskip a) $\ \left( 2x^{5}\right) \left( x^{4}\right) $\newline Solution: \ $\left( 2x^{5}\right) \left( x^{4}\right) =2x^{5}x^{4}=2x^{5+4}=2x^{9}$ \ \ by rule 1.\bigskip b) $\ \left( 2x\right) ^{5}\left( x^{4}\right) $\newline 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} \\ &=&32x^{9} \end{eqnarray*} c) $\ \left( 2x^{5}\right) ^{4}$\newline Solution:% \begin{eqnarray*} \left( 2x^{5}\right) ^{4} &=&2^{4}\left( x^{5}\right) ^{4}\text{ \ \ by rule 4} \\ &=&16x^{20}\text{ \ \ \ \ \ by rule 3} \end{eqnarray*} d) $\ \left( -xy\right) ^{2}\left( -xy^{2}\right) ^{3}$\newline 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} \\ &=&-1x^{2+3}y^{2+6}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule 1} \\ &=&-x^{5}y^{8} \end{eqnarray*} e) $\ -2a^{3}\left( -2a^{4}\right) ^{2}$\newline Solution:% \begin{eqnarray*} -2a^{3}\left( -2a^{4}\right) ^{2} &=&-2a^{3}\left( -2a^{4}\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 4} \\ &=&-2a^{3}\left( -2\right) ^{2}\left( a^{4}\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ rule 3} \\ &=&-2a^{3}\left( 4\right) a^{8}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiplication is commutative} \\ &=&-2\left( 4\right) a^{3}a^{8}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 1} \\ &=&-8a^{3+8}=-8a^{11} \end{eqnarray*} f) $\ 2a^{3}\left( -2ab^{2}\right) ^{3}ab^{2}$\newline Solution:% \begin{eqnarray*} 2a^{3}\left( -2ab^{2}\right) ^{3}ab^{2} &=&2a^{3}\left( -2ab^{2}\right) ^{3}ab^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 4} \\ &=&2a^{3}\left( -2\right) ^{3}a^{3}\left( b^{2}\right) ^{3}ab^{2}\text{ \ \ \ \ \ \ \ \ \ \ rule 3} \\ &=&2a^{3}\left( -8\right) a^{3}b^{6}ab^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiplication is commutative} \\ &=&2\left( -8\right) a^{3}a^{3}ab^{6}b^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 1} \\ &=&-16a^{3+3+1}b^{6+2}=-16a^{7}b^{8} \end{eqnarray*} g) $\ \dfrac{\left( -2x\right) ^{2}y^{3}}{2x^{3}y^{2}}$\newline Solution:% \begin{eqnarray*} \dfrac{\left( -2x\right) ^{2}y^{3}}{2x^{3}y^{2}} &=&\dfrac{\left( -2\right) ^{2}x^{2}y^{3}}{2x^{3}y^{2}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 4} \\ &=&\dfrac{4x^{2}y^{3}}{2x^{3}y^{2}}=\dfrac{2x^{2}y^{3}}{x^{3}y^{2}}\text{ \ \ \ \ \ \ \ \ cancel out }x^{2}y^{2} \\ &=&\dfrac{2y}{x} \end{eqnarray*} h) $\ \dfrac{\left( 2ab\right) ^{3}\left( -3a^{2}b\right) ^{2}}{-b\left( 6ab^{2}\right) ^{2}}$\newline Solution: \begin{eqnarray*} \dfrac{\left( 2ab\right) ^{3}\left( -3a^{2}b\right) ^{2}}{-b\left( 6ab^{2}\right) ^{2}} &=&\dfrac{\left( 2ab\right) ^{3}\left( -3a^{2}b\right) ^{2}}{-1b\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 b\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 b\cdot 36\cdot a^{2}b^{4}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule 3} \\ &=&\dfrac{8\cdot 9\cdot a^{3}a^{4}b^{3}b^{2}}{-1\cdot 36\cdot a^{2}\cdot b\cdot b^{4}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiplication is commutative} \\ &=&\dfrac{72a^{7}b^{5}}{-36a^{2}b^{5}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by rule 1} \\ &=&\dfrac{-2a^{7}b^{5}}{a^{2}b^{5}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ simplify numbers: }\dfrac{72}{-36}=\dfrac{-72}{36}=\dfrac{-2}{1} \\ &=&\dfrac{-2a^{7}}{a^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ cancel out }a^{2}\text{ \ (or use rule 2)} \\ &=&-2a^{5} \end{eqnarray*}% \pagebreak i) $\ \left( \dfrac{-2ab}{3b^{3}}\right) ^{3}\left( \dfrac{6ab^{4}}{4a^{3}b}% \right) ^{2}$\newline Solution: \ We first simplify each expression within the parentheses.% \begin{eqnarray*} \left( \dfrac{-2ab}{3b^{3}}\right) ^{3}\left( \dfrac{6ab^{4}}{4a^{3}b}% \right) ^{2} &=&\left( \dfrac{-2a}{3b^{2}}\right) ^{3}\left( \dfrac{3b^{3}}{% 2a^{2}}\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 5} \\ &=&\dfrac{\left( -2a\right) ^{3}}{\left( 3b^{2}\right) ^{3}}\cdot \dfrac{% \left( 3b^{3}\right) ^{2}}{\left( 2a^{2}\right) ^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 4} \\ &=&\dfrac{\left( -2\right) ^{3}\left( a\right) ^{3}}{3^{3}\left( b^{2}\right) ^{3}}\cdot \dfrac{3^{2}\left( b^{3}\right) ^{2}}{2^{2}\left( a^{2}\right) ^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ rule 3} \\ &=&\dfrac{-8a^{3}}{27b^{6}}\cdot \dfrac{9b^{6}}{4a^{4}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiplication of fractions} \\ &=&\dfrac{-8a^{3}\left( 9b^{6}\right) }{27b^{6}\left( 4a^{4}\right) }=\dfrac{% -8\cdot 9a^{3}b^{6}}{4\cdot 27a^{4}b^{6}}\text{ \ \ \ \ \ \ \ \ \ \ cancel} \\ &=&\dfrac{-2a^{3}b^{6}}{3a^{4}b^{6}}=\dfrac{-2}{3a} \end{eqnarray*}% \bigskip 2. \ Write each of the following expressions in terms of a fixed number or a single exponential expression.\bigskip a) \ $\dfrac{3^{2x+1}}{9^{x-1}}$\newline Solution 1:% \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{Rules 1,2}}{~~~~~=~~~~}3^{2x+1-\left( 2x-2\right) }=3^{2x+1-2x+2}=3^{3}=27 \end{equation*}% Solution 2: \ \begin{equation*} \dfrac{3^{2x+1}}{9^{x-1}}\overset{\text{Rules 1,2}}{~~~~~=~~~~}\dfrac{% 3^{2x}\cdot 3^{1}}{\dfrac{9^{x}}{9^{1}}}=\dfrac{3^{2x}\cdot 3}{9^{x}\cdot \dfrac{1}{9}}=\dfrac{3^{2x}\cdot 3}{9^{x}}\cdot \dfrac{9}{1}=\dfrac{27\cdot 3^{2x}}{9^{x}}~~\overset{\text{Rule 3}}{=}~~\dfrac{27\cdot \left( 3^{2}\right) ^{x}}{9^{x}}=\dfrac{27\cdot 9^{x}}{9^{x}}=27 \end{equation*} b) \ $\dfrac{\left( 8^{b-2}\right) \left( 2^{b+1}\right) }{4^{2b-3}}$% \begin{eqnarray*} &=&\dfrac{\left( 8^{b-2}\right) \left( 2^{b+1}\right) }{4^{2b-3}}\overset{% \text{Rules 1,2}}{~~~=~~~}\dfrac{\dfrac{8^{b}}{8^{2}}\cdot 2^{b}\cdot 2^{1}}{% \dfrac{4^{2b}}{4^{3}}}\overset{\text{Rule 3}}{~~~=~~~}\dfrac{\dfrac{% 8^{b}\cdot 2^{b}\cdot 2}{64}}{\dfrac{\left( 4^{2}\right) ^{b}}{64}}\text{ \ \ \ \ to divide is to multiply by reciprocal} \\ &=&\dfrac{8^{b}\cdot 2^{b}\cdot 2}{64}\cdot \dfrac{64}{\left( 4^{2}\right) ^{b}}\overset{\text{Rule 4}}{~~~=~~~}\dfrac{\left( 8\cdot 2\right) ^{b}\cdot 2}{16^{b}}=\dfrac{16^{b}\cdot 2}{16^{b}}=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}=3125 \end{equation*} \bigskip \pagebreak 3. \ Let us denote $3^{100}$ by $M$. \ \ Express each of the following in terms of $M.$ \qquad a) \ $3^{101}\medskip $ \qquad Solution: \ Using rule 1, we write $3^{101}=3^{100+1}=3^{100}\cdot 3^{1}=M\cdot 3=3M$ \bigskip \qquad b) \ $3^{100}-2\cdot 3^{101}+3^{102}\medskip $ \qquad 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*}% \qquad \qquad 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=4M \end{equation*} \qquad c) \ $3^{99}\medskip $ \qquad Solution: \ Using rule 2, we write $3^{99}=3^{100-1}=\dfrac{3^{100}}{% 3^{1}}=\dfrac{M}{3}$\bigskip \qquad d) \ $9^{100}\medskip $ \qquad 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*} \qquad 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=M^{2}% \text{ } \end{equation*} \bigskip \vspace{4in} \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}