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LOGARITHM
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Basic Mathematics
Historical Development of Number System
Logarithm
Principal Properties of Logarithm
Basic Changing theorem
Logarithmic equations
Common & Natural Logarithm
Characteristic Mantissa
Absolute value Function
Solved examples
Exercise
Answer Key

13
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If p(x) is divided by (x – a), then the remainder is equal to p(a)
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Conversely, if (x – a) is a factor of p(x), then p(a) = 0
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If leading coefficient of p(x) is 1
then p(x) is called monic
...
)

SOME IMPORTANT IDENTITIES :
(1)

(a + b) 2 = a 2 + 2ab + b2 = (a – b)2 + 4ab

(2)

(a – b)2 = a2 – 2ab + b2 = (a + b)2 – 4ab

(3)

a2 – b2 = (a + b) (a – b)

(4)

(a + b)3 = a3 + b3 + 3ab (a + b)

(5)

(a – b)3 = a3 – b3 – 3ab (a – b)

(6)

a3 + b3 = (a + b)3 – 3ab (a + b) = (a + b) (a2 + b2 – ab)

(7)

a3 – b3 = (a – b)3 + 3ab (a – b) = (a – b) (a2 + b2 + ab)

(8)

1 1 1
(a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca) = a2 + b2 + c2 + 2abc    
...

(11)

a4 – b4 = (a2 + b2) (a2 – b2) = (a2 + b2) (a – b) (a + b)

(12)

If a, b  0 then (a – b) =

(13)

a4 + a2 + 1 = (a4 + 2a2 + 1) – a2 = (a2 + 1)2 – a2 = (a2 + a + 1) (a2 – a + 1)



a b



a b



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ACC-MT- LOGARITHM

Definition of Indices :
The product of m factors each equal to a is represented by am
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a ( m times)
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Law of Indices :
(1)

am + n = am · an, where m and n are rational numbers
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am
a0 = 1, provided a  0
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an
(am)n = amn
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q

Intervals :
Intervals are basically subsets of R (the set of all real numbers) and are commonly used in solving
inequaltities
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{x : a < x < b} i
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, end points are not included
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e
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This is possible only when both a and b are finite
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e
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Closed-open interval

[a, b)

{x : a  x < b} i
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, a is included and b is excluded
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e
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(3)

If their is no value of x, then we say x (i
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, null set or void set or empty set)
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A
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If

a c
 , then it is written as a : b = c : d or a : b : : c : d
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Product of extremes = product of means
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b d
b
d
b
d

(7)

If

a c
a b cd
 
=
(Componendo and dividendo)
b d
a b cd

(8)

If

a b
 then b2 = ac
...

b c

Histrorical Development of Number System :
I
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{1, 2, 3, 4,
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Whole number’s
Including zero (0) | cypher | ’kwU; | duck |love| knot along with natural numbers called as whole numbers
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}
i
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N W
0 is neither positive nor negative

III

Integer’s
Integer’s given by
I = {
...
}
i
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N W I
Type of Integer’s
(a) None negative integers
(b) Negative integers (I–)
(c) Non positive integers
(d) Positve integers (I+)

{ 0, 1, 2, 3,
...
–3, –2, –1}
{
...
}

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ACC-MT- LOGARITHM

IV
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Rational numbers are also represented by recurring & terminating or repeating decimal’s
e
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1
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333
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3333
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33
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The decimal representation of these number is non-terminating and non repeating
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414
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Real  Rational + Irrational
NWIQRZ

VII
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g
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}

VIII

Composite Number’s
Number’s which are multiples of prime are called composite number’s
{4, 6, 8, 9
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(2, 9),
(16, 25
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g
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1 is niether a prime nor a composite number
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A
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(1)
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49 = 72
where 'a' is also a positive real different than unity and is called the base and 'x' is called the exponent
...
(2)
Hence the two relations
ax  N 
and
log a N  x 

are identical where N > 0, a > 0, a  1

Hence logarithm of a number to some base is the exponent by which the base must be raised in order to
get that number
...

i
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(i) Let log8127 = x

27 = 81x

33 = 34x
(ii)
Let
log10100 = x

100 = 10x

102 = 10x
(iii)

(iii) log1/3 9 3

gives x = 3/4

gives x = 2

Let log1/3 9 3 = x



1
9 3 = 
3



–x
35 / 2 = 3

x

gives x = – 5/2

Note that :
(a)

Unity has been excluded from the base of the logarithm as in this case
log1N will not be possible and if N = 1
then log11 will have infinitelymanysolutions and will not be unique
which is necessaryin the functional notation
...
g
...
Hence
If log2512 is 9 then antilog29 is equal to 29 = 512

log2 5

=5

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ACC-MT- LOGARITHM

(d)

Using the basic definition of log we have 3 important deductions :
(i)
(ii)
(iii)



log 1 N = – 1 

N


loga1 = 0

logNN = 1

i
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logarithm of a number to the same base is 1
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e
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i
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logarithm of unity to any base is zero
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)
(iv)
(e)

a log a n  n is an identify for all N > 0 and a> 0 ; a  1 e
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2 log 2 5 = 5

Whenever the number and base are on the same side of unity then logarithm of that number to the base
is (+ve), however if the number and base are located an diffrent side of unity then logarithm of that
number to the base is (–ve)
e
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(i) log10100 = 2
(ii) log1/10100 = –2

For a non negative number ‘a’ & n  2, n  N

(f)

n

a  a1/ n

Illustration :
(i)

logsin 30° cos 60º = 1

(iv)
Sol
...
  1



(ii) log3/4 1
...
  x
5x  x



x2 = 5x



x=5

 log 5 5 = 1

(v)
Sol
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(log tan 89º) = 0
Since tan 45° = 1 thus log tan 45° = 0

(vi)
Sol
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x –3 + 2x – 6 – 12 = 0
3x = 21  x = 7

(viii)

log 2 (x  3)  4

Sol
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A
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1

Q
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3

Find the logarithms of the following numbers to the base 2:
1
1
(i) 3 8
(ii) 2 2
(iii) 5
(iv) 7
8
2
1
Find the logarithms of the following numbers to the base
3
1
1
(i) 81
(ii) 3 3
(iii) 7
(iv) 9 3
(v) 4
9 3
3
Find all number a for which each of the following equalities hold true?
(i)
(iii)

Q
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5

log2a = 2
log1/3(a2 – 1) = – 1
log 2 x 2 = 1
log1/2(2x + 1) = log1/2(x + 1)



If 2 3  5  13  48  =





(ii)
(v)

log3x = log3(2 – x)
log1/3(x2 + 8) = – 2

(iii)

log 4 x 2 = log4x

b where a and b are natural number find (a + b)
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1 (i) 1, (ii) 3/2, (iii) – 1/5, (iv) – 3/7

Q
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3 (i) 4, (ii) –5, 2, (iii) –2, 2, (iv) –3, 3

Q
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A
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5 8

17

ACC-MT- LOGARITHM

Solved Examples
Q
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Find the value of x satisfying log10 (2x + x – 41) = x (1 – log105)
...
Ans
...
2

If the product of the roots of the equation,

Sol
...

Take log on both the sides with base 2

1
a

b

5
3
1
2
 log 2 x   log 2 x   log 2 x =
4
4
2

log2x = y
3y3 + 4y2 – 5y – 2 = 0 
3y2(y – 1) + 7y(y – 1) + 2(y – 1) = 0

(y – 1)(3y2 + 7y + 2) = 0

(y – 1)(3y + 1)(y + 2) = 0
1

y = 1 or y = – 2 or y =
3
1
1
1

x = 2; ; 1 3  x1x2x3 = 3
 a + b = 19
4
2
16

Q
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a

Sol
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(2)

x+y=±a

 a 2a 
 and (– a, 2a)
possible ordered pairs =  ,
3 3 

Q
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Find (y1 + y2)
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From (1),
3 + log10(2xy) – log10x · log10y = 4
or
log10(xy) – log10x · log10y = 1 – log10(2)


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(1)

1
and
2

23

ACC-MT- LOGARITHM

EXERCISE-1 (Exercise for JEE Main)
[SINGLE CORRECT CHOICE TYPE]

Q
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2


3

(C) sec


3

(D) sin


3

[3010110650]

1
1
1
1
·
·
·
simplifies to
log 2 N log N 8 log32 N log N 128
3
(C) 5 ln 2

If p is the smallest value of x satisfying the equation 2x +
(B) 16

(C) 25

(D)

5
21

[3010110244]

15
p
x = 8 then the value of 4 is equal to
2
(D) 1
[3010110950]

The sum of two numbers a and b is 18 and their difference is 14
...
5


3

3
(B) 7 ln 2

3
7

(A) 9
Q
...
3

5
3


4
2

(B) 2

(C) 1

(D)

1
2

[3010112439]

The value of the expression (log102)3 + log108 · log105 + (log105)3 is
(A) rational which is less than 1
(B) rational which is greater than 1
(C) equal to 1

(D) an irrational number







[3010111646]



2

3 log 2  2 log log 103  log  log106 


N = 10

Q
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The characteristic of the
logarithm of N to the base 3, is equal to
(A) 2
(B) 3
(C) 4
(D) 5
[3010112388]

Q
...
8

10  2
10  2
and y =
, then the value of log2(x2 + xy + y2), is equal to
2
2
(B) 2
(C) 3
(D) 4
[3010112337]

Suppose that x < 0
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A
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1 to 3
A denotes the product xyz where x, y and z satisfy
log3x = log5 – log7
log5y = log7 – log3
log7z = log3 – log5
B denotes the sum of square of solution of the equation
log2 (log2x6 – 3) – log2 (log2x4 – 5) = log23
C denotes characterstic of logarithm
log2 (log23) – log2 (log43) + log2 (log45) – log2 (log65) + log2 (log67) – log2(log87)
Q
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2
Q
...
4

Let N =

log 3 135
log3 5


...
5



If a logb x



2

[3010112387]

 5 x logb a + 6 = 0, where a > 0, b > 0 & ab  1, then the value of x can be equal to

(A) 2logb a
Q
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A
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1

2
 ab  (ab) 2  4(a  b) 



 + log  ab  (ab)  4(a  b) 
Let A denotes the value of log10
10 



2
2




when a = 43 and b = 57







and B denotes the value of the expression 2log6 18 · 3log6 3
...

Q
...
2) in terms of x and y
...


[3010110921]

Q
...
2 to 0
...

[3010110177]

Q
...

Q
...

Find the value of

yz

...
6

Find the value of x satisfying log10 (2x + x – 41) = x (1 – log105)
...
7

Positive numbers x, y and z satisfy xyz = 1081 and (log10x)(log10yz) + (log10y)(log10z) = 468
...
8

Find the number of integral solution of the equation log

x

x  | x  2 | = logx(5x – 6 + 5 | x – 2|)
...
9

Suppose p, q, r and s  N satisfying the relation p 

1
q

=

1
r

89
, then find the value of (pq + rs)
...
10

If 'x' and 'y' are real numbers such that, 2 log(2y – 3x) = log x + log y, find

x

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P
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A
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1

The least value of the expression 2 log10x – logx (0
...
2

(B)2

(C) –0
...
3

The equation x 4

+ 23x + 21) = 4 –

(log 2 x ) 2 log 2 x –

5
4

log(3x + 7)(4x2

+ 12x + 9)
[3010110279]

= 2 has :

(A) at least one real solution
(C) exactly one irrational

[IIT 1989, 2M]
(B) exactlythree real solution
(D) Complex roots
[3010110651]

Q
...
5

Let (x0, y0) be the solution of the following equations
(2 x )ln 2  (3y)ln 3
3ln x = 2ln y
...
6



1
1
1
1

The value of 6  log 3 
4
4
4

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A
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2
Q
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12

D
C
A

EXERCISE-1
Q
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8
D

Q
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6
Q
...
1
Q
...
9
Q
...
2
A
Q
...
7
A, B, C
Q
...
4
Q
...
5
Q
...
4

A, C

Q
...
3

449

Q
...
8

1

Q
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7
5625

Q
...
2

Q
...
6

Q
...
12
Q
...
11 (a) 0
...
5386 ; 3
...
3522 (d) 3
(a) 140 (b) 12 (c) 47
Q
...
14 2
x  [1/3, 3] – {1}
Q
...
1
Q
...
2
Q
...
3

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B

Q
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19 y = 6

Q
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5
3


4
2

Let x =


5

4

5
1
3
5
25 3
 =  2· = 3
 x2 =  2
2
4
2
2
16 2


3 = tan 3
...

3
...


Ans
...
Ans
...


log102 = a and log105 = b

a + b = 1; a3 + 3ab + b3 = ?
3 = 1  a3 + b3 + 3ab = 1  (C)
Now (a + b)

6
...

7
...


52 6  52 6
=
2

5
24

=
4
4



log2 ( x  y) 2  xy
but

Ans
...

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3 2

Ans
...
Ans
...


y = 2x  | x  2 | = 2x  (2  x ) = | 3x – 2 | as x < 0 hence y = 2 – 3x

9
...


10
...


As,




=p



log p log q r

Ans
...
 , y > 0  y =

and let y =

But




log q log q r
log q p

6  y  y2 = 6 + y

y2 – y – 6 = 0  (y – 3) (y + 2) = 0
y > 0, so y = 3
...




1

log a 2  3

= log 2



3





1
 3 1 

log b 
 3 1



a  log 2

Now, 2  3



log

2 3

3

= log 2

Ans
...

Hence , (a + b) = 7
...




12
...


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2
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