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\begin{document}
\chapter{Exponents and Polynomials}
\section{Definition of Exponents}
When we study Geometry course, we take a square to find an area,
and take a cube to find a volumn.
The square and cube are special cases of \emph{exponents}.
In this section, let us introduce the formal definition
of integer (positive , zero and negative) exponents,
and discuss their properties.
\bigskip
\begin{tabular}{|rcl|}\hline
\multicolumn{3}{|l|}{} \\
\multicolumn{3}{|l|}{\large \bf Definition} \\
\multicolumn{3}{|l|}
{If $x$ is nonzero real number and $n$ is a positive integer,
then} \\
\multicolumn{3}{|l|}{}\\
\multicolumn{1}{|r} {$x^n$} & \multicolumn{1}{c}{=} &
\multicolumn{1}{l|} {$ \underbrace{x \cdot x \cdots x}_{n \hbox{ times}} $ } \\
\multicolumn{1}{|r} {} & \multicolumn{1}{c}{} &
\multicolumn{1}{l|} {} \\
$x^{-n}$ & = & $\frac{1}{x^n}$ \\
\multicolumn{1}{|r} {} & \multicolumn{1}{c}{} &
\multicolumn{1}{l|} {} \\
$x^0 $ & = & $1 $ \\
\hline
\end{tabular}
\bigskip
In the exponent form $x^n$, $x$ is called the {\it base}
and $n$ is called the {\it power index} or {\it exponent}
In this section, we will be concerned with the simplification
of expressions with exponents. We start from some simple
examples and make some generalizations in order to obtain
important properties.
\begin{exa}
{\rm Write the product $x^3 \cdot x^5$ with a single exponent.}
\end{exa}
\\
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
x^3 \cdot x^5 & = & (x \cdot x \cdot x) \cdot
(x \cdot x \cdot x \cdot x \cdot x) \\
& = & x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\
& = & x^8 .
\end{eqnarray*}
\begin{exa}
{\rm Write the product $x^3 \cdot x^{-5}$ with a single exponent.}
\end{exa}
\\
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
x^3 \cdot x^{-5} & = & x^3 \cdot \frac{1}{x^5} =
\frac{x \cdot x \cdot x} {x \cdot x \cdot x \cdot x \cdot x} \\
& = & \frac{1}{ x \cdot x } = \frac{1} {x^2} = x^{-2}.
\end{eqnarray*}
Therefore, we will have the first property for exponent:
$x^m \cdot x^n = x ^{m+n} $ for any nonzero $x$ and any interges
$m$ and $n$.
\begin{exa}
{\rm Write the quotient $\frac{x^3}{x^5}$ with a single exponent.}
\end{exa}
\\
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
\frac{x^3}{x^5} & = & \frac{x \cdot x \cdot x}
{x \cdot x \cdot x \cdot x \cdot x} = \frac{1} {x^2}\\
& = & x^{-2}
\end{eqnarray*}
\begin{exa}
{\rm Write the quotient $\frac{x^3}{x^{-5}}$ with a single exponent.}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
\frac{x^3}{x^{-5}} & = & \frac{x^3}{\frac{1} {x^5} } =
(x^3) \cdot (x^5) \\
& = & x^8
\end{eqnarray*}
Therefore, we will have the second property for exponent:
$ \frac{x^m} {x^n} = x ^{m-n} $ for any nonzero $x$ and any interges
$m$ and $n$.
\begin{exa}
{\rm Write the product $(x^3)^4$ with a single exponent.}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
(x^3)^4 & = & (x \cdot x \cdot x) \cdot (x \cdot x \cdot x) \cdot
(x \cdot x \cdot x) \cdot (x \cdot x \cdot x) \\
& = & \underbrace{ x \cdot x \cdots x}_{12 \hbox{ times}} \\
& = & x^{12}.
\end{eqnarray*}
Therefore, we will have the third property for exponent:
$ (x^m)^n = x ^{mn} $ for any nonzero $x$ and any interges
$m$ and $n$.
\begin{exa}
{\rm Expand $ (3x)^4$ and then multiply.}
\end{exa}
\\
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
(3x)^4 & = & (3x) \cdot (3x) \cdot (3x) \cdot (3x) \\
& = & ( 3 \cdot 3 \cdot 3 \cdot 3) \cdot
(x \cdot x \cdot x \cdot x) = 3^4 \cdot x^4 \\
& = & 81 x^4.
\end{eqnarray*}
Therefore, we will have the forth property for exponent:
$ (x \cdot y)^n = x^n \cdot y^n $ for any nonzero $x$ and $y$,
and any interge $n$.
\begin{exa}
{\rm Simplify $ \left ( \frac{3}{4} \right ) ^{-2} $ and then
rewrite it as an exponent form with a positive power index}
\end{exa}
\\
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
\left ( \frac{3}{4} \right ) ^{-2}
& = & \frac{1}{\left ( \frac{3}{4} \right ) ^2} \\
& = & \frac{1}{ \frac{9}{16}} = \frac{16}{9} \\
& = & \left ( \frac{4}{3} \right ) ^2
\end{eqnarray*}
Therefore, we will have the fifth property for exponent:
$ \left ( \frac{x}{y} \right ) ^{-n} = \left ( \frac{y}{x} \right ) ^{n} $
for any nonzero $x$ and $y$,
and any interge $n$.
\begin{exa}
{\rm Expand and rewrite $ \left ( \frac{x}{y} \right ) ^3 $}
\end{exa}
\\
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
\left ( \frac{x}{y} \right ) ^3
& = & \left ( \frac{x}{y} \right ) \cdot
\left ( \frac{x}{y} \right ) \cdot
\left ( \frac{x}{y} \right ) \cdot \\
& = & \frac{ x \cdot x \cdot x }{ y \cdot y \cdot y} \\
& = & \frac{x^3}{y^3}
\end{eqnarray*}
Therefore, we will have the sixth property for exponent:
$ \left ( \frac{x}{y} \right ) ^{n} = \frac{x^n}{y^n} $
for any nonzero $x$ and $y$,
and any interge $n$.
\section{Equations With Absolute Values}
\section{Inequalities With Absolute Values}
\newpage
\section{The Absolute Value}
\begin{defn}
For any real number $x$, the {\bf absolute value} of $x$, written as $|x|$,
is defined by
$$ |x| = \left \{ \matrix
{ x & \hbox{ if } & x \ge 0 \cr
-x & \hbox{ if } & x<0 \cr} \right . $$
\end{defn}
We can try to remember the definition in the following way.
If the original number $x$ is positive or zero, then $|x|$ is $x$ itself.
If the original number $x$ is negative, then $|x|$ is its opposite which is
still positive.
No matter what $x$ is, $|x|$ is always nonnegative.
Therefore, it is easy to figure out $|x|$, if $x$ is a given real number.
The following are some examples:
a) $ | 5 | = 5 $
b) $ | \sqrt 3 | = \sqrt 3 $
c) $ | -4 | = -(-4) = 4 $
d) $ | - {2 \over 3} | = -(- {2 \over 3}) = {2 \over 3} $
However, we don't have a simple way to remove the absolute value symbol
from $|x|$, if $x$ is an expression with variables.
The following two properties are true for expressions with absolute values.
\begin{tabular}{|l|}\hline
\\
{\bf {\large The Properties for Absolute Values}}
\\
\\For any real numbers $a$ and $x$
\\
$ |ax| = |a| \cdot |x| $ \\
$ | {x \over a} | = {{|x|} \over {|a|} } $
as long as $x \neq 0$
\\ \hline \end{tabular}
\\
\\`
The following are some simple examples of using these properties:
a) $ | 5 x | = 5 |x| $
b) $ | - \sqrt 3 \cdot x | = |-\sqrt 3| \cdot |x| = \sqrt 3 \cdot |x| $
c) $ | {x \over {-3}} | = {{|x|} \over {|-3|}} = {{|x|} \over 3} $
But, if there are additions or subtractions inside absolute value symbol,
we cannot do anything to simplify them.
For example, the following properties are {\bf NOT} true:
1, $ |x+5| = |x| + 5 $
2, $ |x-5| = |x| - 5 $
After using graphing calculator to graph them, you will realize that, they are
totally different functions.
\begin{exa}
Which of the following rewriting equalities are correct or wrong? Why?
\end{exa}
a) $ 3 + | x - 3 | = |x| $
b) $ 5 \cdot | \frac{x-1}{5} | = |x-1| $
c) $ 5 \cdot | \frac{x-1}{-5} | = |x-1| $
d) $ (-5) \cdot | \frac{x-1}{-5} | = |x-1| $
e) $ 5 \cdot | \frac{x-1}{5} | = x-1 $
\newline
\newline
\noindent {\bf SOLUTION:}
Rewriting equality a) is wrong. Since we don't have any property to help us
remove the absolute value symbol, we cannot cancel $3$ which is outside
the absolute value symbol, and $-3$, which is inside
the absolute value symbol.
If we choose $x =-1$, then the left is equal to $7$, meanwhile the right is
equal to $1$.
Rewriting equality b) is correct, since we can use the above property to
move $5$ out from the absolute value symbol.
After that, we can cancel $5$.
Rewriting equality c) is also correct, Since we can use the above property to
seperate $|x-1|$ and $|-5|$, which is equal to $5$.
After that, we can cancel $5$.
Rewriting equality d) is wrong. Although we can use the above property to
seperate $|x-1|$ and $|-5|$, which is equal to $5$ and cancel $5$.
But $-1$ will be left. Therefore, the correct answer should be $ - | x-1|$.
Rewriting equality d) is wrong. Although we can use the above property to
seperate $|x-1|$ and $|5|$, which is equal to $5$ and cancel $5$.
But $|x-1|$ is not equal to $ x-1$. Therefore, the
correct answer should be $ | x-1|$. $\Box$
\newline
\section{The Equations With Absolute Value}
Lack of properties for absolute value makes solving
equations with absolute value and inequalities with absolute value much more
difficult than solving linear equations and linear inequalities.
In this section, we discuss the methods of solving the following four types
of equations with absolute values.
In next section, we handle inequalities with absolute value.
Type I: $ | \hbox{linear expression with one variable} | = \hbox{positive number} $.
Type II: $ | \hbox{linear expression with one variable} | = 0 $.
Type III: $ | \hbox{linear expression with one variable} | = \hbox{negative number} $.
Type IV: $ | \hbox{first linear expression with one variable} | =
\hbox{second linear expression with one variable} $.
We handle the easiest type III first. Since absolute value of any real number
is nonnegative, the equation has no solution.
\begin{exa}
Solve $| 3 -4(x-2) | = -5 $.
\end{exa}
\newline
\noindent {\bf SOLUTION:}
This is Type III, hence no solution.
You don't even have to simplify the expression. $\Box$
Type II is also easy. Since absolute value of any expression
is zero if and only if the expression itself is zero,
we can remove the absolute value symbol simply if it is case II.
After that, it can be solved by the methods in section 2.2.
If the expression is linear, there is only one solution.
\begin{exa}
Solve $| 3 -4(x-2) | = 0 $.
\end{exa}
\newline
\noindent {\bf SOLUTION:}
This is Type II,
$ 3 -4(x-2) = 0 $, $ 3- 4x + 8 = 0$, $ 4x = 11 $, $ x = {{11} \over 4} $.
$\Box$
Type I: $ | \hbox{linear expression with one variable} |
= \hbox{positive number} $.
If the linear expression with one variable is equal to the positive number
on the
right, the equation becomes a true statement. From here, we can find one
solution.
If the linear expression with one variable is equal to the opposite
of the positive number on the right,
the equation also becomes a true statement. From here, we can find another
solution.
Generally speaking, we simplify the expression first.
Then set the expression equal to the right positive number, solve it.
Then set the expression equal to the opposite of
the right positive number, solve it.
\begin{exa}
Solve $| 3 -4(x-2) | = 5 $.
\end{exa}
\newline
\noindent {\bf SOLUTION:}
This is Type I,
$ |3 -4x + 8 | = 5 $, $ | 11 - 4x| = 5 $,
$ 11 - 4x = 5 $,\hskip 1in $ 11 - 4x = -5 $
$ 4x = 6 $ \hskip 1in $ 4x = 16 $,
$ x = {3 \over 2} $ \hskip 1in $ x = 4$. $\Box$
Type IV: $ | \hbox{first linear expression with one variable} | =
\hbox{second linear expression with one variable} $.
If two linear expressions with one variable is equal to each other,
the equation becomes a true statement. From here, we can find one solution.
If one linear expression with one variable is equal to the opposite
of another linear expression with one variable,
the equation also becomes a true statement. From here, we can find another
solution.
Since We have mentioned in section 2.2 that the solution set of
a linear equation may have none, or one, or infinitely many solutions (all
reals).
The solution set of equations of Type IV may have one, or two, or infinitely
many solutions (all reals).
Generally speaking, we simplify both the expressions first.
Then set the first expression equal to the second expression, solve it.
Then set the first expression equal to the second expression, solve it.
Then set the expression equal to the opposite of the second expression,
solve it.
\begin{exa}
Solve $| 3 -4(x-2) | = | 3(x+1) - 6| $.
\end{exa}
\newline
\noindent {\bf SOLUTION:}
This is Type IV.
$ | 11 - 4x| = | 3x - 3| $,
$ 11 - 4x = 3x - 3 $ ,\hskip 1in or \hskip 1in $ 11 - 4x = 3 - 3x $
$ -7x = -14 $ \hskip 1in or \hskip 1in $ 11 - 3 = 4x - 3x $,
$ x = 2 $ \hskip 1in or \hskip 1in $ x = 8 $.
The equation has two solutions$ x = 2 $, $ x = 8 $. $\Box$
\begin{exa}
Solve $| 3 -4(x-2) | = | 2(2x+1) - 6| $.
\end{exa}
\newline
\noindent {\bf SOLUTION:}
This is Type IV,
$ | 11 - 4x| = | 4x - 4| $,
$ 11 - 4x = 4x - 4 $ ,\hskip 1in or \hskip 1in $ 11 - 4x = 4 - 4x $
$ -8x = -15 $ \hskip 1in or \hskip 1in $ 11 = 4 $,
$ x = {{15} \over 8} $ \hskip 1in or \hskip 1in no solution.
The equation has one solution $ x = {{15} \over 8} $. $\Box$
\begin{exa}
Solve $| 3 -4(x-2) | = | 2(2x+1) - 13| $.
\end{exa}
\newline
\noindent {\bf SOLUTION:}
This is Type IV,
$ | 11 - 4x| = | 4x - 11| $,
$ 11 - 4x = 4x - 11 $ ,\hskip 1in or \hskip 1in $ 11 - 4x = 11 - 4x $
$ -8x = -22 $ \hskip 1in or \hskip 1in $ 0 = 0 $,
$ x = {{11} \over 4} $ \hskip 1in or \hskip 1in all reals.
The solution set of the equation is the set of all reals. $\Box$
\section{Inequalities With Absolute Value}
When solving inequalities with absolute value algebraically,
there are several possibilities to be considered.
\end{document}
\end
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