Zero and Negative Exponents
Zero as an Exponent:For every nonzero number a, a° =1. Examples:
r°
2
3L
-‘
3
Negative Exponent:
For every nonzero number a and integern, a =
Examples:
I
Example 1: Simplifying a Power Simplify.
a. 43 b. (—l.23)°
Example 2: Simplifying an Exponential Expression Simplify each expression.
b.-i
_
Example 3: Evaluating an Exponential Expression Evaluate 3m2t2 form = 2 andt=-3.
37
_
Multiplication Properties of Exponents Mu1tiplyin Powers With the Same Base Property:
For every nonzero number a and integers m and n, am a” = am.
Examples: 5)_
5
2TLi
51
2’
Exam pie 1: Multiplying Powers
Rewrite each expression using each base only once.
a. ii4 •ii3 b. 52 .52
(
2-2[1
‘
5
Example 2: Multiplying Powers in an Algebraic Expression Simplify each expression.
a. 2n5 3n2 b. 5x2y4 •3x8
4
°LN
Raising a Power to a Power Property:
For every nonzero number a and integers m and n, (atm)” =atm”.
Examples: 2..\)3
5-
(—2’)
Example 3: Simplifying a Power Raised to a Power Simplify (x3)6
Example4: Simplifying an Expression With Powers
Simplify c’(c3)2.
Example 5: Simplifying a Product Raised to a Power
Write the expression that represents the area of the square.
2x
tL
(axjz
2
Example 6: Simplifying a Product Raised to a Power Simplify (x2)2(3xy2)4.
N,
cHI1f
,J:71
Li
Raising a Product to a Power Property:
For every nonzero number a and b and integer n, (ab)’7 = a’.
Examples:
(3\)a.
32. 292
Division Properties of Exponents
Example 1: Simplifying an Algebraic Expression Simplify each expression.
a.
-2
c1d3ft.
L
Example 2: Raising a Quotient to a Power Dividing Powers With the Same Base Property:
For every nonzero number a and integers mand n, = a’3.
Example:
2
2
r
1
b
Cd
1
Raising a Ouotient to a Power Property:
For every
nonzero number a and
band integer n(J =Example:
(
/
(4N3
Which expression is equivalent to I —i- ?
x)
H
3
(L
-Example3: Simplifying an Exponential Expression
Simplify each expression.
a. H— —
Section 7.1 Roots and Radical Expressions Since 52 =25, 5 is a ____________ root of 25. Since 53 = 125, 5 is a ______________ root of 125.
Since 54 =625,5isaC(+k rooto
Since 55 =3125,5isa
_________
rootof
3j
Definition of thenth Root:
For any real numbers a
and
b,and any
positive integer n, if a’ =b, then a is an nthroot of b. -
9
oc
c*d
— O4-
rcc*s
)-I(hC5flOftoi
-‘khQ C
4Z
j
Summary of the possible real roots of a real number.
Numberof Rea’ nth
NLInbef’
of ReI ,,thType ofNimiber Roots Whenns Even ROGtSWeii IFJs Odd
-:JciF(\)
c
QOc
h
ñ
Example 1: Finding All Real Roots Find all the real roots.
a. The cube roots of 8, -1000, and
t\
I
-DC
hQ1\cc4
Iäo
0
\‘\
hQ\’Q_I
OO&
16 b. The fourth roots of 1, -0.0001, and —.
L
‘jiI.tc{th rooi-
od
-VH and
(-r)ZI
V1T)
hCJL QJJ
(TJ3S
1JTD1Ob
1O
V
t
&--- , -I 1 ___
5
Radical Sign:
\,cJ —ko
\r$&CCk(
Radicand:
—[-
cctr rdr
*
tijç oJ
Index:
Th°
-.
PrincipalRoot:
-JQ
n c
n’JU)cr
\CkS
+o oc*.
TYL i
nc
4 rc
O
6
ThQ
Lthrc’ct
Q$T
Example 2: Finding Roots Find each real-number root.
C. Jx’y
c@oJ iu&br
3
a
_codcO
b. J—ioo‘S
Sc)c\m
a.Vi-;
Example 3: Simplifying Radical Expressions Simplify each radical expression.
a.
b. Ja3b6
Section
7.2Multiplyin2 and Dividing Radical Expressions
MultipIyin Radical Expressions Property:If and ‘[ are real numbers, then
.
=Example
1:
Multiplying Radicals Multiply. Simplify if possible.-\
=
.
pap+j cbQS nc*
cppk scc
\
-S
rob-
a
QcJ
nuiDg.
Example 2: Simplifying Radical Expressions
Simplify each expression. Assume that all variables are positive. Then absolute value symbols are never needed in the simplified expression.
a.
\J
__
b. kI8On
J
10
z
)ion
Dividing Radical Expressions Property:
Example 4: Dividing Radicals
Divide and simplify. Assume that allvariables arepositive.
a.
b
3
___
Rationalize the Denominator
f)J
ç-
-\-\Q-
áQf1ofl
()Q
O
\Dd
cc.
Example 5: Rationalizing the Denominator
ncO1
Rationalize the denominator of each expression. Assume that all variablesarepositive.
a
•J
\{3
_1çI
-5X\/
b c. 3J_V
3x ‘I4D
—.L’E
-cL_
Section 7.3
-Binomial Radical Expressions
Like Radicals:
ea\ ftSS
[CC
Example!: Adig ubtingdial
-W’&... bxiQ
Add or subtract if possible.
a.
5k1—3&
b.4Ii+5J
rcC-Example 2: Simplifying Before Adding and Subtracting Simplify 6-lu +
4J
—3-In.
J9Y
-33
4l_
+
-L:
Example
3: Multiplying Binomial Radical Expressions Multiply (3+ 2J)(2+4J).
Pci :
t4
Example 4: Multiplying Conjugates Multiply (2+ J)(2—
Fo
IL
cc
Q
T
DrQnc c
- \(cbbcb2
()zExample 5: Rationalizing Binomial Radical Denominators
3+
Rationalize the denominator of
-(3
4I
-‘-5
-LjSection 7.4
Rational Exponents
Rational Exponent:
-
CO
1sScor.
VL
Example 1: Simplifying Expressions With Rational Exponents Simplify each expression.
a. 125
b. 5 5}
\)
5
1o.1oo =
C)
Definition of Rational Exponents: -
TL
(Th(If the nth root of a is a real number and m is an integer, then
a = and a =
(
‘J
OCrdk
Example
2: Converting to and From Radical Forma. Write the exponential expressions x and y2.5 in radical form.
b. Write the radical expressions
J1
and (J)2 in exponential form.b
2
Example 4: Simplifying Numbers With Rational Exponents
ExampleS: Writing Expressions in Simplest Form Write (16y8) in simplest form.
-Summary of Properties of Rational Exponents:
Let m
and
n represent rational numbers. Assume that no denominator equals 0.Example
]‘Lg
4
( )2.. L3 -5
Property a 11 ’ a am+n (am)n =am (ab) 11 =a11n11 _n a a” m = a 11 -1) L 9Vi f \11 (a a [L’\3 “ b 11ai
3
Simplify each number.
a. (—32)
_33/5
\j(33)
3
. = b. 435 L L4 N I1(s”I
\
1
J4/
1
\j
—_i-—
Ni 1 )3
.-Section
7.5Solve Square Root and Other Radical Equations
Radical Equation:
ç-
rn
\xI) @ ‘\Jc
QE*
0
c
Solving a Radical Equation.
sc tcU
+hp
rod
cc] cc+-
+
cjxthc
-
thSQ
boW’
Qa\ O
-Exampie
1: Solving Square Root EquationsSolve 2+iJ3x—2=6.
\3K-
H
CHY
-i-•
-Example
2: Solving Radical Equations With Rational Exponents____
X-
53
tb
x
-Solve 2(x—2) =50.(x
‘A
-or Sc&
c$
+i4
CbQC.
0cj9
r’33Q
::
50
50
:
50
• (
50
50
53DQ
5Q.
V
.‘-.
1
‘F I’ cNcx-.--.
.(ES
I x-\) 9Extraneous Solutions:
A tuEcn
d
cn
c± cn
•i
Qd
om
Example 3: Checking For Extraneous Solutions
32jj
f5
-4-3
_o
Example 4: Solving Equations With Two RationalExponents Solve (2x+ 1)0.5 —
(3x
+4)0.25
0. Check for extraneous solutions.
Q3
oc
/
“I) Solve -Jx —3 +5 = x. Check
for
extraneous solutions.5K
-5
A
Al)
(7S(i-’Xx
\rTE ÷31
J+
5
7
1 7 ,/
\J
÷5
7 .3—H
ChQch
(./)
+O05
/3/
.
17 ÷ \ -.(%)°
/ 5iEz --L (i:) .7 --(3(/)
/c -1- L)
(3KH)
c
(x
i)a
(±i)
z(*t)q
4
q+(
3x4
4xx
-3
D(So
1
.
--,q
--So
I
(3
(E:)0
0
-1
f\Q.5
v-I)
2±0
rai