2. Consider the following the problem
What’s the product of 24
and 25
?
Solutions:
am
= a.a.a…..a
m times
24
= 2.2.2.2 =16
4 times
25
=2.2.2.2.2 =32
So that the answer of the problem above is:
24
.25
= 16.32
= 512
We know :
512 = 29
=16.32
=24
.25
= 2 4+5
3. If we replace the base a by a, the exponents 4 and 5 by
positive integers m and n :
We will get:
am
= a.a.a….a
m times
an
= a.a.a….a
n times
am
.an
= (a.a….a).(a.a….a)
m times n times
= a.a.a……..a
(m+n) times
FORMULA I
with condition a € R, a ≠ 0
a is called the base number and m,n is called exponent
.m n m n
a a a +
=
4. if m>n, them use formula I, we shall obtain:
am-n
.an
= am-n+n
am-n
.an
= am
Formula II
m
m n
n
a
a
a
−
=
an
5. pay attention the following problem :
(am
)n
= am
.am
…..am
n times
= am+m+m….m
= am.n
Conclusion
Formula III
.
( ) ( )m n n m m n
a a a= =
6. If besides a we also use the base of b, then the
form (ab)m
. Can be defined :
(ab)m
= ab.ab……ab
m times
= (a.a.a.a).(b.b.b.b)
m times m times
Formula IV
( ) .m m m
ab a b=
7. So that by the same way we will get
Conclusion :
b≠ 0 Formula V
Because : is not defined
m m
m
a a
b b
= ÷
0
number
8. Example : Simplify!
( )
3 5 2
5 2 4
2 3
54
11 8 19
6 5 11
1.5 .5 .5
2.( ) . . .
3.2 .(2 ) .(2 )
4.( 3) . 3
3 3 3
5.
3 3 3
x x x x
x x x
−
− −
=
9. ( ) ( )
( ) ( )
( )
( ){ } ( ){ }
( ) ( ){ }
( )
( )
( )
8 9
7 2
85
2 53 4
32 3
5
3 42 2
3 4
42 3
32
3 . 3
6)
3 . 3
7)
8) 2 . 2
3 . 3
9)
3
10) .
.
11)
.
a a
a a
p
y y
y y
y
x x
y y
x y
x y
−
− −
− −
÷ ÷
−
10. ( )
3 5 2 10
5 2 4 12
2 3 2 2 3 3 6 6
54 9
11 8 19
19 11 8
6 5 11
1.5 .5 .5 5
2.( ) . . .
3.2 .(2 ) .(2 ) 2 .2 .2 2
4.( 3) . 3 3
3 3 3
5. 3 3
3 3 3
x x x x x
x x x x x x x
−
=
− = −
= =
− − = −
= = =
11. ( ) ( )
( ) ( )
( )
8 9 17 17
8 8
7 2 9 9
85 5.8 40
3 . 3 3 .
6) 3 .
3 .3 . 3
7)
a a a
a
aa a
p p p
−
= = −
−− −
= =
( ){ } ( ){ } ( ) ( )
( ) ( )
2 5 2 53 4 3 3 4 4
6 6 20 20
26 26
8) 2 . 2 2 . . 2 .
2 . . 2 .
2 .
y y y y
y y
y
=
=
=
12. ( ) ( ){ }
( )
( )
( )
32 3 32 2 3 3
5 5 5
2 2 9 9
5 5
11 11
5 5
6 6
3 . 3 3 . . 3 .
9)
3 .3
3 . . 3 .
3 .
3 .
3 .
3 .
y y y y
yy
y y
y
y
y
y
− − −
=
−
=
−
=
= −
( )
( )
3 42 2 6 8 14
3 4 9 16 25
42 3 8 12
5 6
3 3 62
.
10) .
.
. .
11) .
..
x x x x x
y y y y y
x y x y
x y
x yx y
= = ÷ ÷
= = −
−−
13. In this section we shall see that formula I-V hold for integers
exponents, either positive, negative and zero
If we substitute m = n use formula II we shall obtain
Formula VI
a € R, a ≠ 0
m n
n n
n n
a a
a
a a
−
= =
0
1 a=
14. If we subsituty m=0 use formula II we will get
Formula VII
a € R, a ≠ 0
0
1m
n n n
a a
a a a
= =
0 1n
n
a
a
−
=
1n
n
a
a
−
=
1 n
n
a
a−
=
15. ( )
( ) ( )
5 2
7
5
2 2
3
4
25
3
3 42 2
1)3 .3
4
2)
4
3) .
4)
5)
6) .
x x
x
x
x
x
x y x y
−
−
−
−
−
−
−
− −
SIMPLIFY!!
16. 3 3
22
2 3
7) .
8)
x y
ab
b c
− −
−
− −
÷
( ) ( )
( ) ( )
4 3
1 2
2 1 . 1 2
9)
3 . 3
x x
x x
−
− −
− −
+ +
3 13 4 4
2 3 5
10) :
a b a b
c b c
− −
−
−
÷ ÷
17. ( )
( ) ( ) ( )
5 2 3
7
12
5
2 2 0
3
1
4
25 10
13
3 3 13
3 4 12 2 2
2
1)3 .3 3
4
2) 4
4
3) . 1
4)
1
5)
1
6) .
x x x
x
x
x
x x
x
x x x
x y x y x y
x y
−
−
−
−
−
− −
−
− −
=
=
= =
=
= = =
− − = − =
−
18. ( )
( )
33 3
3
22 2 4
2 3 4 6 2 8 6
1
7) .
1
8)
x y xy
xy
ab a b
b c b c a b c
−− −
− − −
− −
= =
= = ÷
( ) ( )
( ) ( )
( ) ( ){ }
( )
( )
( )
( )
( )
344 3
1 2 3
1
3
3
2 1 2 12 1 . 1 2
9)
3 . 3 3
2 1
3
3
2 1
x xx x
x x x
x
x
x
x
−−
− − −
−
−
− − −− −
=
+ + +
− −
=
+
+
= −
−
19. 3 13 4 4 9 12 4 1
2 3 5 6 3 5
9 12 3 5
6 4 1
9 15 6 4 1 5
5 14 11
5 14 11
10) : :
:
. . . . .
. .
1
a b a b a b a b
c b c c b c
a b b c
c a b
a b c a b c
a b c
a b c
− − − − − −
− − −
− − − −
− −
− − − −
− − −
− −
= ÷ ÷
−
=
= −
= −
=
−
20. By the Formula I we obtain am
.an
= am+n
If we substituting n=m, then we will get
am
.am
= a2m
so that a2m
= a 2m =1
m =1/2
form a2m
=a ,can be changed
(am
) 2
= a
am
= √a
a1/2
= √a
In general can written
1
nn
a a=
21. In the fractional exponents:
If is changed ,then we will get :
n,m is positive integers
Formula VIII
1
n
m
n
( )
1 mm
n n
m
m
nn
m
n mn
aa a
a a
a
= ÷
=
=
22. By the formula VIII:
,then if ,we obtain
Formula IX
Furthermore the formula I - IX,also hold for fractional
exponents
1
n=n
n
p
a
a q
−
=
1 1
orp
q
p p
q q
p
q
a a
a a
−
−
= =
1 1
orm
n
m m
n n
m
n
a a
aa
−
−
= =
23. Example
1. Simplify and write down in positive exponents!
( )
( ) ( )
131
3 62 2
2 1
3 6
31
4 4
2
3
. 2 .2 . 2
. . .
3 . 3
.
3
.
a
b x x x
x x
c
x
a
d
b
−
−
−
÷
5 35
6
1 5
2
3
34
3 .2
.
.
.
x x
e
x x
x
f
x
−−
−
÷
÷
24. 2. Simplify and write down in the roots form !
( ) ( )
1 2
34
. 1 3 . 3 1c x x− −
( )
1
3 3
1
1 6
2
5 45
1
52
.
.
1 .
2.
4 .
x x
a
x
x x
b
x x
−
−
26. 4. Find the length of diagonal of rectangle
if it has dimension length is cm and
width is cm!
8
2
27. ( )
( ) ( )
( ) ( )
13 31 1 1 1
3 62 2 2 2 2 2
1
2
2 1 2 1 41
3 6 3 3 3
31
4 4 31 1 0
4 4 2
22
3
3
1
1. . 2 .2 . 2 2 2
2
. . .
3 . 3
. 3 3 1
3
.
a
b x x x x x
x x
c x x
x
a a
d
b b
− − −+
− + −
−
− + −
= = =
= =
= = =
= ÷ ÷
28. 31 4
5 35 5 5 5 15
35
6 11 11
1 5 55
22 3
3 32 3
2
3 3 334 4 2 2
3 .2 6 . 6
. 6 6
.
1
.
x x x x x
e x x
x xx x
x x x
f x
x x x x
− −−−
−−
−
−
−
= = = =
÷= = = = ÷
÷ ÷
29. ( )
1 3 1
3 3 32 3 1 1 12
2 43 6 62
1 1
1 6 6
62 2 4
5 45 5 5 5
1 1 61 1
52 52 52
. .
2. .
1 1 1. .
2 2 2. 1
4 . 4 . . 4 .
x x x x
a x x x x
xx
x x x x x
b
x x x x x
+ +
−−
− − −
= = = = =
= = =
( ) ( ) ( ){ } ( )
( )
( )
11 22 434 3
11
12
11
12
. 1 3 . 3 1 3 1 . 3 1
3 1
1 3
c x x x x
x
x
− − = − − −
= − −
= −
32. ( ) ( )
2 2
2
2
2
4. 8 2 .
8 2
10
10
x
x
x
x
= +
= +
=
=
33. f(x) p
I. a =a f(x)=p⇒
Example :
1. Find the solution set :
3
1
1
)2 8
2
1 1
) 27
3 243
x
x
a
b
−
=
= ÷
34. 3 1
)2 8
2
x
a =
1
1 1
) 27
3 243
x
b
−
= ÷
3
3 1 2
1
3 2
2 2 .2
2 2
1
3
2
1
6
x
x
x
x
−
⇔ =
⇔ =
⇔ =
⇔ =
3
1 52
3
1
52
3 .3 3
3 3
3
1 5
2
3
6
2
3 12
4
x
x
x
x
x
x
−
−
⇔ =
⇔ =
⇔ − =
⇔ =
⇔ =
⇔ =
35. f(x) ( )
II. a =a f(x)=g(x)g x
⇒
Example :
1. Find the solution set :
( )
3 1
4
3 5( 1)
2
3 2 3 4 8
16 1
)
1 162
2
)27 3 .81
) 3 27
xx
x x
x x
a
b
c
−
+ −
+ −
= ÷
=
=
36. 3 1
4
16 1
)
1 162
2
xx
a
−
= ÷
3 4 4
4 4
1
1 2
4 4 4 1
3 4 2
4 4
3
2
2
2 .2
2 2 .2
1
2
3
4 4 3
2
5
7
2
5
14
x
x
x x
x
x
x x
x
x
−
−
−
− −
−
−
⇔ =
⇔ =
⇔ = − −
⇔ − = − −
⇔ =
⇔ =
37. 3 5( 1)
)27 3 .81x x
b + −
=
( )
2
3 2 3 4 8
) 3 27x x
c + −
=
( ) ( )2 2 3 3 4 8
3 2
3 3
4 6 12 24
3 2
8 12 36 72
84 28
3
x x
x x
x x
x
x
+ −
⇔ =
+ −
⇔ =
⇔ + = −
⇔ =
⇔ =
3( 3) 5( 1) 4
3 3 .3
3 9 5 5 4
3 9 5 1
10 2
5
x x
x x
x x
x
x
+ −
⇔ =
⇔ + = − +
⇔ + = −
⇔ =
⇔ =
38. III. Exponent equation is changed to quadratic equation
Example :
1. Determine the solution set :
( )
1 2 3
)3.9 10.3 3 0
)2 2 36
)5 6 5 5 0
x x
x x
x
x
a
b
c
+ +
− + =
+ =
− + =
39. )3.9 10.3 3 0x x
a − + =
( )
( ) ( )
2
2
3. 3 10.3 3 0
3 , 0
3 10 3 0
3 1 3 0
x x
x
a a
a a
a a
⇔ − + =
⇔ = >
⇔ − + =
⇔ − − =
1
3
1
3
3
1
x
a
x
⇔ =
⇔ =
⇔ = −
3
3 3
1
x
a
x
=
=
=
V
40. 1 2 3
) 2 2 36x x
b + +
+ =
( ) ( )
2 3
2
2
2 , 0
2 .2 2 .2 36 0
2 8 36 0
4 18 0
4 9 2 0
x
x x
p p
p p
p p
p p
⇔ = >
⇔ + − =
⇔ + − =
⇔ + − =
⇔ + − =
: 2
9
4
p⇔ = −
2 2
1
x
x
⇔ =
⇔ =
V 2p =
41. ( ))5 6 5 5 0
x
x
c − + =
( ) ( )
2
5 5
2
x
x
=
=
( ) ( )
0
5 5
0
x
x
=
=
V
V
( ) ( )
( )
( ) ( )
2
2
5 6 5 5 0
5 , 0
6 5 0
5 1 0
5 1
x x
x
p p
p p
p p
p p
⇔ − + =
⇔ = >
⇔ − + =
⇔ − − =
⇔ = =
⇔
42. (1) Rational Number
Rational Number are numbers which can be put in a/b
form with a and b are integers numbers and b ≠ 0.
Example:
1. 3 coz 3 can be stated in etc
2. 0.444…= 0.4 can be stated in
3. 1.12 can be stated in
Notes !
The lines above 12(no.3)shows that 12 is repeated
unlimitedly
Generally !
Rational number are repeated decimals
6 9
,
2 3
2
5111
97
43. (2) Irrational Numbers
Irrational number are number which can’t be stated in
Generally :
Irrational numbers are unrepeated decimal
Example!
1. =1.4142135….
2. = 5.1461524….
3. Log 2 = 0.3010….
4. e =2.71828182…..
coz can’t be put in
2
a
b
3
3
a
b
44. 1.Please check whether the following form
are rational or irrational.
3
3
a.-2 5
1
.log
10
. log 2
. 12
b
c
d −
. 0.1
2
.
9
.1.2
.8.34
e
f
g
h
45. ( )
( )
( )
( )
3
3
a.-2 5
1
.log
10
. log 2
. 12
I
b R
c I
d I−
( )
( )
( )
( )
. 0.1
2
.
9
.1.2
.8.34
e I
f I
g R
h R
46. Root number are the rational number which the result are irrational
EXAMPLE :
3
3 6
1) 3,2 12, 8 are root forms
2) 4, 27, 64, , are not root form because the result are rationalπ−
47. 3
5
)2 7
) log81
) 0,05
9
)
25
) 32
a
b
c
d
e −
1. Please check whether the following forms are root forms or not
5
)1, 2
) 81
1
)
3
)
)ln 2
f
g
h
i e
j
Note :
Ln = Log number with base e
Ln x = e
log x
48. 2. :Simplify the following root form
4
5
5 44
) 32
)4 250
) 243
) 64
1
) 288
2
) 405
a
b
c
d
e
f x y
−
49. ( ) ( )
511
54 4444
) 32 16.2 4 2
)4 250 4 25.10 20 10
) 243 243 3 3 3 3
a
b
c
= =
= =
= = = =
( )
5 55 5
5 4 4 4 54 4
) 64 2 .2 2 2
1 1 1
) 288 144.2 .12 2 6 2
2 2 2
) 405 81.5 . . 3 5
d
e
f x y x x y xy x
− = − = −
= = =
= =
50. ( )
( )
2
2
Notes !
1) 2
2) 2
a b a b ab
a b a b ab
+ = + +
− = + −
( ) 2a b a b ab+ = + +
( ) 2a b a b ab a b− = + − ⇒ ≥
51. ( ) ( )3. inSimplify a b or a b form+ −
) 8 2 12
) 15 2 90
) 9 80
) 14 6 5
) 15 10 2
5 2
)
6 3
a
b
c
d
e
f
−
+
−
+
−
+
ANSWER
57. 1. Find the value of the following problem !!
) 10 10 10 10...
) 72 72 72 ...
) 56 56 56 56 ...
a
b
c
+ + +
− − − −
2+ 5
2 5
if p
q
=
= −
2. Find 2p+2q, 4pq and p2
+q2
ANSWER
62. A fractional of root on its denominator such as :
1
, ,
c c
a a b a b+ −
Can be simplified by rationalizing
1 1 1
.
a
a
a a a a
= =
( ).
c a bc c a b
a ba b a b a b
−−
= =
−+ + −
( ).
c a bc c a b
a ba b a b a b
++
= =
+− − +
66. a c
1. log b c a b= ⇔ =
a
2. log a 1=
a
3. log 1 0=
a a a
4. log x . y log x log y= +
a a a
5. log log x log y
x
y
= −
a n a
6. log x n . log x=
67. x log
9. log y
log
a
a
x
y
=
a y a
10. log b log b
x y
x
=
x
a y
8. log a
y
x
=
x 1
7. log y
logy
x
=
BACK
11. log . log . log loga x y a
x y b b=
log
12.
a
x
a x=
68. 2
2
2
3 log 2
log 3
3 3 ½
2 2
) log 16
1
) log
25
) log 0.001
) ( 3)
1
) ( )
4
) log 2 log 4
5
) log 5 - log
4
a
b
c
d
e
f
g
+
3
4
3 2 7
9
3
2
25
4 2 8
4
) log
3
) log 7 . log 27 . log 2
log 8
)
log 4
1
) log3
log 4
7 7
) log 21 log7 log .
24 3
h
i
j
k
l
−
− − ÷ ÷
ANSWER
1. EVALUATE !!
74. 2. SIMPLIFY !!
( ) ( )
2
2 2
23
1 1
) log x + log - log
) log log
) log log logx x x
a
x x
b x y x y
c x x x
− − −
+ −
ANSWER
75. 2
2
2
2
1 1 1
) log x + log - log = log x . log . log x
1
log x . . x
log x
2 log x
a
x x x
x
= ÷
=
=
( ) ( ) ( ) ( ) ( )2 2
) log log log log
log log . log
log .
{ }b x y x y x y x y x y
x x
x y
y y
x y
− − − = + − − −
= + −
=
76. 1 1
3 2
23
x
log . log
) log log log
2 logx
1 1
.
3 2
2
1
12
x x
x x x x x
c x x x+ − =
=
=
NEXT
82. 3
6
3
3
3
3 2 3
3 3
3
3 2
log98
) log 98 =
log6
log 49.2
log3.2
log7 log 2
log3 log 2
2 log7
1
2 log 2 log7
1
2
1
c
M
M
M
M
MN M
M
=
+
=
+
+
=
+
+
=
+
+
=
+
NEXT
83. 5. Find the value of x that fulfill of the following equation!
( )
( ) ( )
3 3 3
3 3 3
4 4 4 4 4
) log log 1 log 2
) log 2 1 log 3 log7
) log . log x - log log log16 = 2
a x x
b x x
c
+ + =
− − − =
ANSWER
84. ( )3 3 3
) log log 1 log 2a x x+ + =
2 1x x= − ∨ =
TM { }1HP =
( ) ( )3 3 3
) log 2 1 log 3 log7b x x− − − =
{ }4HP =
( )
( )
( ) ( )
3 3
log 1 log 2
1 2
2 1 0
x x
x x
x x
⇔ + =
⇔ + =
⇔ + − =
( )
( )
( )
( )
3 32 1
log log7
3
2 1
7
3
2 1 7 21
4
x
x
x
x
x x
x
−
⇔ =
−
−
⇔ =
−
⇔ − = −
⇔ =
86. 1. Find the value of x that fulfill of the following equation !!
( ) ( )
( )
( ){ } { }
( )
2 2
5 2 5
) log 2 3 4 log 2 8
) log 2 log 5 1
)log log 3 log 2 log log16
) log 2 2x
a x x
b x x
c x x
d x
− + = −
− + =
+ + =
+ =
( )
( ) ( )
5 2 5 3
32 2 2
) log 12 3. log 4 1 0
) log log 2 0
) log 1 log 1 10 0
x x
e x
f x x
g x x
+ − + =
− + =
+ − + − =
ANSWER
87. ( ) ( )2 2
) log 2 3 4 log 2 8a x x− + = −
( )5 2 5
) log 2 log 5 1b x x− + =
5
10
2
x x⇔ = − ∨ =
HP =
5
,10
2
HP
= −
Ø
( ) ( )
( ) ( )
2 2 2
2 2
log 2 3 log16 log 2 8
log16 2 3 log 2 8
32 48 2 8
1
1
3
x x
x x
x x
x
⇔ − + = −
⇔ − = −
⇔ − = −
⇔ =
( )
( ) ( )
5 2 5 5
2
2
2
log 2 log 5 log5
2
5
5
2 5 25
2 5 25 0
2 5 5 0
x x
x
x
x x
x x
x x
⇔ − + =
⇔ =
+
⇔ = +
⇔ − − =
⇔ + − =
88. ( ){ } { })log log 3 log 2 log log16c x x+ + =
1 9x x⇔ = ∨ =
( )) log 2 2x
d x + =
2 1x x⇔ = ∨ = − { }2HP =
{ }1,9HP =
( ){ } { }
( ){ }
( )
( ) ( )
2
2
2
log log 3 log 2 log log16
2.log 3 log16
log 3 log16
6 9 16
10 9 0
1 9 0
x x
x x
x x
x x x
x x
x x
⇔ + + =
⇔ + =
⇔ + =
⇔ + + =
⇔ − + =
⇔ − − = ( )
( ) ( )
2
2
2
log 2 log
2
2 0
2 1 0
x x
x x
x x
x x
x x
⇔ + =
⇔ + =
⇔ − − =
⇔ − + =
89. ( )) log 12 3. log 4 1 0x x
e x + − + =
5 2 5 3
) log log 2 0f x x− + =
16 4x x⇔ = − ∨ =
5
log 1 0x⇔ − = 5
5
2
log 2 0
log 2
5
25
x
x
x
x
− =
=
=
=
{ }5,25HP =
{ }4HP =
( )
( )
( ) ( )
3
3 1
2
log 12 log 4 1
log 12 log 4 log
12 1
64
12 64 0
16 4 0
x x
x x x
x
x x
x
x
x x
x x
−
⇔ + − = −
⇔ + − =
+
⇔ =
⇔ + − =
⇔ + − =
( )( )
5 2 5
5 5
log 3. log 2 0
log 1 log 2 0
x x
x x
⇔ − + =
⇔ − − =
5
1
log 1
5
5
x
x
x
⇔ =
⇔ =
⇔ =
V
90. ( ) ( )
32 2 2
) log 1 log 1 10 0g x x+ − + − =
( )2
log 1 5 0x⇔ + − = ( )
( )
2
2
2
log 1 2 0
log 1 2
1 2
1
1
4
3
4
x
x
x
x
x
−
⇔ + + =
⇔ + = −
⇔ + =
⇔ = −
⇔ = −
3
,31
4
HP
= −
V
( )2
5
log 1 5
1 2
32 1
31
x
x
x
x
⇔ + =
⇔ + =
⇔ = −
⇔ =
( ) ( )
( ){ } ( ){ }
2 2 2
2 2
log 1 3. log 1 10 0
log 1 5 log 1 2 0
x x
x x
⇔ + − + − =
⇔ + − + + =