Exponent & Logarithm

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

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 +
=

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

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= =

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=

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

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
−
− −
=

( ) ( )
( ) ( )
( )
( ){ } ( ){ }
( ) ( ){ }
( )
( )
( )
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
−
− −
− −
   
 ÷  ÷
   
−

( )
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
−
=
− = −
= =
− − = −
= = =

( ) ( )
( ) ( )
( )
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
=
=
=

( ) ( ){ }
( )
( )
( )
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
   
= = ÷  ÷
   
= = −
−−

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=

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−
=

( )
( ) ( )
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!!

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
− −
−
   −
 ÷  ÷
   

( )
( ) ( ) ( )
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
−
−
−
−
−
− −
−
− −
=
=
= =
=
= = =
− − = − =
−

( )
( )
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
−−
− − −
−
−
− − −− −
=
+ + +
− −
=
+
+
= −
−

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
− − − − − −
− − −
− − − −
− −
− − − −
− − −
   − −
= ÷  ÷
   
−
=
= −
= −
=
−

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=

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
 
=  ÷
=
=
 

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
−
−
= =

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
−−
−
 
 ÷
 ÷
 

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
−
−

( )
( )
2
3
4 33
3
4
3
42
3
. 27
1
.
2 .2
16
.
81
1
. 125
81
a
b
c
d
−
−
−
−
 
 ÷
 
 
+ ÷
 
( )
1
2 2 2
3 2 1
5 34
. 3 4
.16 32 125
e
f
−
+
+ −
3. Evaluate!

4. Find the length of diagonal of rectangle
if it has dimension length is cm and
width is cm!
8
2

( )
( ) ( )
( ) ( )
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
− − −+
− + −
−
− + −
= = =
= =
= = =
   
= ÷  ÷
   

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
− −−−
−−
−
−
−
= = = =
  
 ÷= = = = ÷
 ÷  ÷
   

( )
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
− − = − − −
= − −
= −

( )
( )
( ) ( )
22
3 233
6 23 64 2 33
33
4 3 344
4 3 3
3 3 324 42
3 3 23
4
1
3. . 27 3 3
9
1 1 1 1 1
.
2 422 .2 2
16 2 2 3 27
.
81 3 3 2 8
1 1 1 1 1
. 125 5 5 27 27
81 3 3 25 25
a
b
c
d
−− −
−− −
−
− − −−−
−
= = =
= = = =
  
= = = = ÷ ÷
   
     
+ = + = + = + = ÷  ÷  ÷
     

( ) ( ) ( ) ( )
( ) ( ) ( )
1 11 1
2 2 2 12 22 2
3 2 13 2 1
4 5 34 5 35 34
3 2
1
. 3 4 9 16 25 5 5
5
.16 32 125 2 2 5
2 2 5 8 4 5 7
e
f
− −− − −
+ = + = = = =
+ − = + −
= + − = + − =

( ) ( )
2 2
2
2
2
4. 8 2 .
8 2
10
10
x
x
x
x
= +
= +
=
=

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
−
=
 
=  ÷
 

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
−
−
⇔ =
⇔ =
⇔ − =
⇔ =
⇔ =
⇔ =

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
−
+ −
+ −
 
=  ÷
 
=
=

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
−
−
−
− −
−
−
⇔ =
⇔ =
⇔ = − −
⇔ − = − −
⇔ =
⇔ =

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
+ −
⇔ =
⇔ + = − +
⇔ + = −
⇔ =
⇔ =

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
+ +
− + =
+ =
− + =

)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

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 =

( ))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
⇔ − + =
⇔ = >
⇔ − + =
⇔ − − =
⇔ = =
⇔

(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

(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

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

( )
( )
( )
( )
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

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−

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

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
−

( ) ( )
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
− = − = −
= = =
= =

( )
( )
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− = + − ⇒ ≥

( ) ( )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

( )
( )
( )
) 8 2 12 6 2 2 6.2 6 2
) 15 2 90 8 7 2 56 8 7
) 9 80 9 2 20 4 5 2 5
) 14 6 5 14 2.3 5 14 2 45 9 5 2 9.5 9 5
) 15 10 2 15 2 50 10 5
5 2 5 1 5 1 1 1 1 1 1 1
) 4. 2 2 .
6 3 6 6 6 6 2 3 2 3 2 3
a
b
c
d
e
f
− = + − = −
+ = + + = +
− = − = − = −
+ = + = + = + + = +
− = − = −
   
+ = + = + = + + = + ÷  ÷
   

REMEMBER !!
1) .
2) .
3)
4)
a a a
a b ab
a b a b
a b a b
=
=
+ = +
− = −

1. Evaluate the value of the following root forms
( )
( )( )
( )( )
( ) ( )
2
2 2
)5 2 2 2 2
)20 3 80 45
) 150 3 54 5 96)
) 2 8
) 6 4 6 4
1 1
) 3. 27.10 48
2 3
) 4 8 3 4 8 3
) 1 5 10 1 5 10
a
b
c
d
e
f
g
h
+ −
− −
− +
+
− +
+ + − −
+ + − − −
ANSWER

( )
( )
( )
( )
( )( ) ( )
2
2
2
)5 2 2 2 2 5 3 1 2 7 2
)20 3 80 45 2 5 3.4 5 3 5 2 12 3 5 7 5
) 150 3 54 5 96) 5 6 3.3 6 5.4 6 5 9 20 6 16 6
) 2 8 2 8 2 16 18
) 6 4 6 4 6 4 10
1 1 1 1 1
) 3. 27.10 48 3. .3 3.10.4 3 .40 3 20 3
2 3 2 3 2
a
b
c
d
e
f
+ − = + − =
− − = − + = − + = −
− + = − + = − + =
+ = + + =
− + = − = −
= = =

( )( ) ( ){ } ( ){ }
( )
( )
2
2
) 4 8 3 4 8 3 4 8 3 4 8 3
4 8 3
16 8 3 2 8. 3
5 2 24
5 4 6
g + + − − = + + − −
= − +
= − + +
= −
= −
( ) ( ) ( ){ } ( ){ }
( )
( )
2 22 2
2
2
) 1 5 10 1 5 10 1 5 10 1 5 10
1 5 10
1 5 10 2 5. 10
16 2 50
16 10 2
h + + − − − = + + − − −
= + +
= + + +
= +
= +

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

1 ) 10 10 10 10...a
10 10 10 10...x =
0 10x x= ∨ =
10x∴ =
2
10 10 10 10...x⇔ =
2
10x x⇔ =
( )
2
10 0
10 0
x x
x x
⇔ − =
⇔ − =

1 ) 72 72 72 ...b + + +
72 72 72 ...x = + + +
9 8x x⇔ = ∨ = −
9x∴ =
( ) ( )
2
2
2
72 72 72 ...
72
72 0
9 8 0
x
x x
x x
x x
⇔ = + + +
⇔ = +
⇔ − − =
⇔ − + =

1 ) 56 56 56 56 ...c − − − −
56 56 56 56 ...x = − − − −
8 7x x⇔ = − ∨ =
7x∴ =
2
56 56 56 56 ...x⇔ = − − − −
2
56x x⇔ = −
( ) ( )
2
56 0
8 7 0
x x
x x
⇔ + − =
⇔ + − =

( ) ( )2 ) 2 2 2 2+ 5 2 2 5
4 2 5 4 2 5
8
a p q+ = + −
= + + −
=
( )( )
( )
2 ) 4 4 2+ 5 2 5
4 4 5
4
b pq = −
= −
= −
( ) ( )
( ) ( )
2 2
2 2
2 ) p +q = 2 5 2 5
4 4 5 5 4 4 5 5
4 4 5 5 4 4 5 5
18
c + + −
= + + + − +
= + + + − +
=

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
++
= =
+− − +

2 2 3 2 3
) .
33 3 3
a = =
5 5 12 5.2 3 15
) .
12 12 612 12
b = = =
4 4 3 2 4 3 4 2
) . 4 3 4 2
3 23 2 3 2 3 2
c
− − + − −
= = = − −
−− − +

( )
2
2
2 3 2 3 8 2 2 24 4 3
) .
8 2 8 2 8 2 8 2
4 6 4 3
6 3
4
d
− −
= =
+ + − −
−
= = −
( )
( ) ( )
2
2 2
6 26 2 6 2 6 2
) .
6 2 6 2 6 2 6 2
6 2 2 12 8 4 3
2 3
6 12 4
e
−− − −
= =
+ + − −
+ − −
= = = −
−

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=

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=

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 !!

2 2 4
) log 16 log 2
= 4
a =
5 5
2
5 -2
1 1
) log = log
25 5
log 5
2
b
=
= −
-3
) log 0.001 = log 10
-3
c
=
1
3 log 2 3 log 22
1
3 log 2 2
1
2
) ( 3) (3 )
= (3 )
2
2
d =
=
=
2 2
log 3 log 3 -2
-2
1
) ( ) = (2 )
4
= 3
1
9
e
=

3 3 ½ 3 3
3
3
)      log 2 log 4 = log 2 + log 2
=   log 2.2
log 4
f +
=
2 2 2
2
5
4
5 5
)    log 5 - log =   log
4
log 4
=  2
g
=
13 3
4 4
4 3
)    log log
3 4
=  -1
h
−
 
=  ÷
 

3 2 7 3 7 2
3
) log 7 . log 27 . log 2 = log 7 . log 2 . log 27
= log 27
= 3
i
2
1
2
9 3 3
3
3 2
3
3
1
2
log 8 log 2
) =
log 4
log 2
3
log 2
2
2
log 2
3 1
= .
2 4
3

8
j
=
=

2
2
25 5 5 2
5 5
5
5
1 1 1
) log5 =
log 4 log 2 log 2
1 1

log 2 log 2
1
log 2
log 2
1
k
x
− −
= −
=
=

2 3
4 2 8
1
2
2 2 2
13
2
3
2
2 2 2
3
3 2
1
3 3
1 2
2 2 22
3
3 2
7 7
) log21 log7 log .
24 3
7 7
log 21 log7 log .
2 .3
3
1 1 7
log 21 - log7 - log
2 3
2 .3
7
log 21 log7 - log
2 .3
l
 
− −  ÷ ÷
 
 
 ÷= − −
 ÷
 ÷
 
=
 
 ÷= −
 ÷
 ÷
 
{
1
222 2
1
2
1
2
2 2
1
2
2
2
7
2 3
7
log 21 log7 log
2.3
7
log 21 log7 .
2.3
7 3
log
7
3
log
14
}
x
x
 
 ÷= − +
 ÷
 ÷
 
 
 ÷= −
 ÷
 ÷
 
=
=
NEXT

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

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
− − − = + − − −
= + −
=

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

3. Given that :
log 2 = 0.301
log 3 = 0.477
log 4 = 0.845
EVALUATE !!
)log5
7
)log
2
6
)log
7
a
c
e
12
)log6
3
)log
2
) log 3
b
d
f
ANSWER

10
)log5 log
2
log10 log 2
1 0.301
0.699
a =
= −
= −
=
)log6 log 2 . log3
0.301 0.477
0.778
b =
= +
=
7
)log log7 log 2
2
0.845 0.301
0.544
c = −
= −
=
3 1 1
)log log3 log 2
2 2 2
1 1
. 0,477 . 0,301
2 2
0.2385 0,1505
0,088
d = −
= −
= −
=
NEXT

4
9
3
9
4. Given that :
) log 6 = A
     Express log 8 with A
b) log 5 = B
     Express log 375 with B
a
3
2
6
) log 2 = M
     log 7 = N
    Express log 98
c
ANSWER

( )
4
9
4
4 3
4 2
4
4
log8
) log 8 =
log9
log 2
log3
3 . log 2
2 . log3
13 .
2
12 .
2
3
22 .
2 1 2
3
4 2
a
A
A
A
=
=
=
−
=
−
=
−
:
4log 6 = A
4log 3 . 2 = A
4log 3 + 4log 2 = A
14log 3 + = A
2
14log 3 = A -
2
Note

3
9
3
3
3 3 3
3
log 375
) log 375 =
log 9
log 125 . 3
2
log 5 + log 3
2
3 log5 + 1
2
3 1
2
b
B
=
=
=
+
=

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

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

( )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
−
⇔ =
−
−
⇔ =
−
⇔ − = −
⇔ =

4 4 4 4 4
) log . log x - log log log16 = 2c { }
{ }
4 4
4 4 2 4 2
4 2
2 16
16
log 2 log x 2
log log x log4
log x 16
x 4
2
65536
x
x
⇔ =
⇔ =
⇔ =
⇔ =
⇔ =
⇔ =
( ) ( ){ }
( )
4 4 4 4 4
4
4
4 4
4
4
4
4
4
log log x - log log log16 2
log x
log 2
log log16
log x
log 2
log2
log x
log 2
1
2
⇔ =
  
⇔ = 
  
 
⇔ = 
 
  
⇔ = 
  

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

( ) ( )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
⇔ − + =
⇔ =
+
⇔ = +
⇔ − − =
⇔ + − =

( ){ } { })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
⇔ + =
⇔ + =
⇔ − − =
⇔ − + =

( )) 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

( ) ( )
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
⇔ + − + − =
⇔ + − + + =

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