Logarithm

Introduction We are familiar with a simple exponential identity a x

= b. Here, 'a' is called the base, 'x ' the exponent and 'b' the result.

Now, just as we can say J4 = 2 , which is basically

another way of saying 2 * 2 = 4, we can say

log a b = x

It is another way of saying a x = b

Thus a log or logarithm is an equivalent way of ex­pressing an exponential identity and the following two ex­pressions are completely equivalent

a x = b o log a b = x

log a b is generally expressed as log of b to the base a. Generally, the base is taken as 10 in which case the subscript for the base is not written.

Hence log b means log 1 0 b . Thus, i f no base is given assume that the base is 10.

Rule 1 I f log a b = x then, a = b

Illustrative Examples Ex. 1: I f log 3 a = 4 . Find the value of a.

Soln: log 3 a = 4=>3 4 = a :. a = 81

Ex.2: I f log 3 [ log 4 ( log 2 x)] = 0, find the value of x.

Soln : log 3 [log 4 (log 2 x)] = log 31 (As, log 31 = 0)

or, log 4 ( log 2 x) = \

or, 4 1 = log 2 x or, log 2 x = 4 or, 2 4 = x :, x = 16

Exercise 1. I f log x = 2, find the value of.x.

a) 4 b) 10

c)100 d) Can't be determined

2. I f log 8 64 = x, find the value of*.

a) 4 b ) l c)2 d)3

3. What is the value of l o g / ~ ^ ~ ] ?

a) 4 b)2 c)-2 d)~4

4. I f log6(O.OOOl) = - 4 , findb.

a) 10 b) io°

c) 10 2 d) Can't be determined

5. I f ' ° 8 2 N / 2 ^ ^ ^ j find the value of x.

16 16 a) j b)4 c)-4 d)

6. I f \oghV2= —, find the value of b.

a) 16 b)32 c)64 d)4

7. Find the value of x, i f log, [log 5 (log 3 JC)] = 0 .

a) 81 b)243 c) 128 d)256

8. I f log f l V3 find the value of a. 6

a) 9 b)27 c)18 d)3

Answers 1. c

2. c; Hint: x = log 8 64 or, 64 = 8X or, 8 2 = g *

.-. x=2.

3. d; Hint: log3( — j =x

or, — = 3V or, - ^ = 3* 0^3^**^ .-. x « - 4 81 3 4

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696 PRACTICE BOOK ON QUICKER MATHS

4. a; Hint: log A (0.000l)=-4 or, b - 4 = o.OOOl = 10~4

.-. b= 10

5. d; Hint: ^ - = (2V2")r=2" 1 _'/x - = 2 / 2 256

or, 2 " 8 = 2Ax o r , - x = -8 .* .*• •

6. b; Hint: logA V2 = - | . ^ V " = 2]/i

•16

or, b^s=(25) / l 5

b= 25 =32

7.b; Hint: log, [log 5(log 3)x] = 0

l o g 5 ( l o g 3 x ) = l or, log, x = 5 .-. * = 3 5 = 243

8.b; Hint: log a V3 = -

or, a ] ' 6 = V3 A a = ( V 3 f = 3 3 = 27

Rule 2 a , o g ' n = n

Proof: L e t x = a ' ° g a n

Presenting the above exponential identity in loga­rithm

log a O) = log a («) , . \B or, a ' ° B . » = „

Illustrative Examples Ex. 1: Find the value of 2 l o g 2 5 • Soln: Detail Method:

Let 2 L O 8 2 5 =x

.-. log2(x) = log 2(5) => x = 5 .-. 2 ' ° B = 5 = 5 Quicker Method: Applying the above formula, we can directly get the answer.

2 L O S 2 5 =5

Ex.2: Find the value of 3 2 - > ° g 3 5

Soln: 3 2 - i » K 3 5 _ 32 x 3 - i o G , 5

= 3 2 x 3 l o G J 5 ' - " = 3 2 x 5 - l = 9

5

Exercise 1. Find the value of Q 1 o 8 J 4 .

a)8 b)9 c)6 d) 16

2. Find the value of 3 2 + 1 ° g 3 9 - l 0 8 « i 9

a) 3 3 b) 3 5 / 2 c) 3 7 / 2 d) 3~ 7 / 2

3. Find the value of 55-iog5 25

a)25 b)5 c)125 d) 5 4

4. Find the value of 43+'og48-iog,f,2

a) 256V2 b) 128V2 c) 156^3 d) 256V3

5. Find the value of I<5 I oS-I 5.

a) 5 b) 16 c)25 d)36

6. The value of 2 2 + L ° S 2 5 is .

a)2 b)4 c) 10 d)20

7. Find the value of 3 2 + ^ , 5

a) 20 b)30 c)45 d)50

Answers

1. d; Hint: gios>4 _3 2iog,4 _ 3 iog ,4 2 = 16

2. c; Hint: 3 2+iog,9-iog 8 19 _ 3 2 x , iog 3 9 x 3 - i o g s i 9

= 3 2 x 9 x 3"* [Wc suppose x = log g l 9 ]

Now find the value of x, x = log g l 9

_ log 3 9 = 21og33 = I

~ l o g 3 8 1 41og33 2

[See Rule 7]

.-. required answer = 3 x3 2 = 3 2 = 3 2

3. c; Hint: s ^ ^ a s 2 ? = 5 5 x 5 - i o g 5 2 5

= 5 5 x25"' = 5 3 =125

4. a; Hint: 43+iog48-iogK,2 x 4 i o g j 8 x 4 - i o g » , 2

= 4 3 x 8 , x 4 _ l 0 8 ' 2 2 = 4 3 x 8 , x 4 < - 1 / 2 ^ 2

• = 4 3 x 8 , x ( 4 l 0 ^ 2 ) ~ 1 / 2 = 4 3 x 8 I x 2 - , / 2

= (6+3-1] 17 2^ 2> =22 = 256V2

5. c; Hint: I 6 1 0 8 4 5 = 4 2 l o ? 4 5 = 4 i ° g 4 2 5 = 2 5

6. d; Hint: 2 2 + 1 ° ^ 5 = 2 2 x 2 l oB> 5 = 2 2 x 5 = 20

7. c; Hint: 3 2 + 1 °g j 5 = 3 2 x 3 I o S J 5 « 9 x 5 = 45

Rule 3

logAr a* = - ( l o g f t a) y

I f b = a = n, then

Proof: Let log n , ( « T ) = .

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Logarithm 697

\ny) =nx ^>ny2=nx yz = x

Illustrative Example Ex.: Find the value of log 2 5 125 — log 8 4

Soln: log 2 5 ( l25)- log 8 (4)

= log 5 2 (53 ) - l o g 2 l ( 2 2 )

3 2 5 = —~— (from the above formula) = *r

2 3 6

Exercise 1. Find the value of log 9 81 - log 4 32 .

1 3 1 a ) " b ) - - c ) - : 2 ' 2 ' 2

2. Find the value of log 4 9 16807 - log 9 27

3 a)0 b ) l c)

d)2

d)-

3. Find the value of !og 3 2 2 8 + l o g 2 4 3 3 7 - log 3 6 1296

a)3 b)2 c ) l d)0

4. Find the value of log 0 1 2 5 64

a)-2 b)2 c)0 d) Can't be determined

5. I f 1 0 0 3 0 1 0 = 2 , then find the value of log 0 1 2 5 125 .

a) 699 3 0 l

699 d)^2

6. If log 8 x + log 4 x + log 2 x = 11, then the value of x is

0)4 a) 2

Answers

c)8 d)64

1. c; Hint: log 3 2 3 4 - log 2 2 2 = 2 ~ 2 = ~ 2

2. b 3.c 4. a; Hint:

log 0 , 6 4 = l o g 2 - ' 2 6 = — T l 0 § 2 2 = ~ 2 [ V ' ° g 2 2 = ! ]

2 5. b; Hint: log 0 . 1 2 5 125 = l o g r , 5 3 = - - l o g , 5 = - l o g 2 5

v 1 0 0 3 0 1 0 = 2 = > l o g 1 0 2 = 0.3010

10

or,

= 1-0.3010 = 0.6990

log 1 0 5 _ 0.6990 _ 699 log, 0 2 " 0.3010 ~ 301

[See Rule 71

- l o g 2 5 = - -

6. d; Hint: Iog 2 , x] + log 22 r + log 2 x = 11

or, i l o g 2 x + ^- log 2 x + l o g 2 x = l l

I x + | l og 2 x = l l 3 2

11 1x6 or, — log 2 x = l l or, log 2 x = = 6

x = 2 = 64

Rule 4

•"• log I 0 5 = log, = l og , 0 10 - l og 1 0 2

\ogab" =n\ogab

Illustrative Example 1

Ex.: I f logx = log5+21og3 - — log25, find the value ofx.

1 Soln : logx = Iog5 + 2 log3 - - log25

= log5+ log 3 2 -log(25)^ = log5 + log9-log5 =log9

.-. x = 9

Note: 1. \oga(b^)=-\ogab n

2. l o g a ( A - " ) = - n l o g 0 *

Exercise 1. I f s5'x = 2X~5, find the value ofx.

a) 5 b)0 c) 1 d) Can't be determined

2. I f 2!og 4 x = 1 + l o g 4 ( x - l ) , find the value ofx.

a) 2 b ) l c)4 d)3

3. I f log 3 = 0.477 and (l Q00)x = 3 , then x equals to

a) 0.159 b)10 c)0.0477 d)0.0159 SSC Graduate Level PT Exam - 2000

4. Find the value of log 1 0 (0.0001).

a) 4 b ) l c)0 d ) ^ l

5. Find the value of log 5 5{log 5 25 + log 5125}.

a) 2 b ) l c)5 d)4

6. Find the value of [log 1 0(51og l 0100)] 2 .

a)0 b ) l c)2 d)4

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698 PRACTICE BOOK ON QUICKER MATHS

7. The logarithm of 144 to the base 2^3 ' s •

a) 2 b)4 c)6 d)8 8. I f log 8 = 0.9031 an log 9 = 0.9542 then find the value of

log 6.

a) 0.3010 b) 0.4771 c) 0.7781 d) None of these

Answers

l.a; Hint: s5'x = 2 r " 5 or, ss~* Jjfi'^

or, ( 5 - x ) log 5 = - ( 5 - x ) log 2

or. (5-x)log5 + (5-x) log2 = 0

or, (5-x){log5 + log2}=0

( 5 - x ) | l o g ^ + l o g 2 } = 0 or,

or, (5 -x ){ log l0 - log2 + log2} = 0

or, 5 - x = 0 x = 5

2. a; Hint: 21og4 x = 1 + log 4 (x -1 )

l o g 4 x 2 = l o g 4 4 + l o g 4 ( x - l )

or, x 2 = 4 ( x - l ) or, x 2 _ 4^ + 4 = 0

or, ( x - 2 ) 2 =0

.-. x = 2

3. a; Hint: (lOOO)* =3

or, xlog, 0 3 = log3

log 3 0.477 n i c n

or, 3x = log3 .-. j r = - | - = - y - = 0.159

4. d; Hint: log 1 0(0.000l)= log 1 0 ( l0" 4 )= -41og, 010 = -4

5. a; Hint:

log 5 5{log5 25 + log 5125} = log 5 5{log5(25x125)}

= log 5 5 2 = 21og55 = 2 6. b; Hint:

[log l 0(5log l 0100)P = [ log l 0 5 log,„ 1 0 2 f = [log l o(l0)p = ( l ) 2 = I 7. b; Hint: 144

= 2 x 2 x 2 x 2 x X X X

Now, l o g 2 V 3 144 = ^ , ^ ( 2 / 3 ) * = 4 1 o g 2 V T 2V3 =4x1 = 4

8. c; Hint: log8 = log(2)3 = 3 log 2

0 9031 .-. 3 log2 = 0.9031 .-. log2 = — — = 0.3010

Now, log 9= log(3)2 = 2 log 3 .-. 2 log 3 = 0.9542 .-. log 3 = 0.4771

N O W , log6 = log(2x3)= log2 + log3 = 0 3010 + 0.477U 0.7781

Rule 5 log (xy) = log (x) + log (y)

Illustrative Examples

Ex.1: I f log wm = b- log| 0 n , find the value of m.

Soln: We have, log 1 0 in = b- log 1 0 n

=> log l 0 /n + l o g 1 0 « = 6

=> l o g , 0 ( « « ) = 6

10" 10° =mn m -

Ex. 2: I f log8 = 0.9031 and log9 = 0.9542 then find the value of log6.

Soln : Iog8 = log(2)3 = 3 log2

.-. 3 log2 = 0.9031

0.9031 .-. log2= — - — =0.3010

Now,log9= | 0 g(3) 2 =21og3

.-. 2 log3 = 0.9542

.-. log.3 = 0.4771 Now, log6 = log(2 x 3) = log2 + log3 =0.3010 + 0.4771 = 0.7781

Note: 1. log (x) + log (y) ^ log (x + y) 2. log(xy) * log(x) x log(y)

Exercise

1. I f log 2 = x , log 3 = y and log 7 = r , then the value of

log(4xV63") is . [Assistants' Grade Exam, 1998|

a) ~2x + -y + -z

c) 2x + -y--z

-, 2 1 b) 2x + -y + -z

- 2 1 d)2x--y + -z

j 3 3 3 I f log(0.57) = 1.756, then the value of

log57 + log(0.57) 3+logV0J7 is .

[SSC Graduate Level (PT) Exam, 1999]

a)0.902 b) 1.902 c) T . 1 4 6 d) 2.146

I f log90 = 1.9542 then log3 equals to . |SSC Graduate Level (PT) Exam, 2000|

a) 0.9771 b) 0.6514 c) 0.4771 d)0.3181

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Logarithm 699

4. Find the value of — log25-21og| 0 3 + log, 0 18

a)0 b)l c)2 d)

5. Find the value of log.v + log

a)0 b ) l c ) - l d)

6. The equation log a x + loga (l + x) = 0 can be written as

a) x- +x-l = 0

c) x" +x-e = 0

7. Find the value of log 8 + log

b) x2 +x + \ 0

d) x2 +x + e = 0

1

a)0 b ) l c)2 d) log(64)

8. Find the value of log KbcJ

( L2\ + log

\J + log

yabJ

a)0

Answers

b ) l c)abc d) a 2 b 2 c 2

1. b; Hint: ^ x fe)= log 2 2 x (3 x 3 x 7 ^

= log2 2 +log(3x3x7)i

= 21og2 + i l o g ( 3 2 x 7 )

= 21og2 + l [ l o g 3 2 + l o g 7 ]

= 21og2 + j l o g 3 + y log7

= 2x + — y + —z 3 3

.a;Hint: log 57x100

100 ) 3Iog(0.57)+|log(0.57)

= log(0.57)+logl0 2 +31og(0.57)+~log(0.57)

= [ l + 3 + i j l o g ( 0 . 5 7 ) + 2 [ v l o g l o 2 = 2 ]

= (4.5 x T .756)+ 2 = 4.5 x (-1 + 0.756)+ 2 = 0.902

3. c; Hint: Iog90 = 1.9542

or, log(3 2x 10)= 1.9542

or, 2log3 + log 10 = 1.9542

0 954? or, log3 = - ^ y ^ = 0.4771

4. b; Hint: ~ log 1 0 25 - 2 log l 0 3 + log, 0 18

= I o g , 0 ( 2 5 ) " 2 - l o g l 0 ( 3 ) 2 + l o g , 0 I 8

= l o g , 0 5 - l o g , 0 9 + log| 018

f 5x18^1 = log, log, 0 10 = 1

V * J

1 5. a;Hint: l o g x + l o g - = log;t+logl - log.r

vJt

= log,v- log.v+ 0 = 0

6. a; Hint: log f l x + log f l(l + .v) = 0

or, log,, x(x +1) = log„ 1 (Since log 1 = 0)

or, x(.r + l ) = l or, x 2 + x - \ = o

7 a: Hint: - l''g| | ] = l o ^ f x 7 l o g l = 0

8. a; Hint: Given expression = '°S

Rule 6

r* ->, 2 2 A abc 2i.2 2

ytl b C j = l o g l = 0

log = log(x) - log( y)

Illustrative Example Ex.: I f log| 0(/w)=6 + log|0(/7),findthevalueofm.

Soln: We have, log, 0 m = b+ log 1 0 n

=> log, 0 w - l o g , 0 n = b

log ///

= b

" = 1 0 *

.-. « = /7 1 0"

. , t ni 1 log/?/ Note: log — | * • ».J I02/7

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700 P R A C T I C E B O O K ON Q U I C K E R MATHS

Exercise I . I f log 1 0 2 = 0.301, then the value of log l 0(50) is

a) 0.699 b) 1.301 c) 1.699 d) 2.301 [SI Delhi Police Exam, 1997)

2. I f log, 0 2 = 0.3010 and log, 0 7 = 0.8451, then the value

of log| 0 2.8 is . a) 0.4471 b) 1.4471 c) 2.4471 d) 14.471

[SSC Graduate Level PT Exam, 1999]

3. I f log2 = 0.3010 , then log5 equals to a) 0.3010 b) 0.6990 c) 0.7525 d) Given log 2, it is not possible to calculate log 5

|SSC Graduate Level PT Exam, 2000|

75 5 32 4. The simplified form of l o g — - 2 l o g - + l o g — — j s

16 9 343

a) log 2 b) 2 log 2 c) log3 d) log5

|SSC Graduate Level PT Exam, 2000|

9 27 3 5. The value of log - - log — + log - j s .

o 32 4 a)0

Answers b ) l

1. c; Hint: log I 0 50 = log

c)2

50x2

d)3

10 = log!00-log2

= l o g 1 0 2 - l o g 2 = 2-0.301 = 1.699

28 2. a; Hint: log t 0 2.8 = log 1 0 — = log28-log 10

log(7 x 4 ) - log 10 = log 7 + 2 log 2 - log 10

= 0.8451 + 2x0.3010-1

= 0.8451 + 0.6020-1 =0.4471

3. b; Hint: log5 = l o g y = log l0 - log2

= 1-0.3010 = 0.6990

, 7 5 * i 5 . 32 4. a;Hint: log — - 2 1 o g - + l o g ­

ic 9 343 . 25x3 , 25 , 16x2

= log l o g — + log 6 4x4 s 8 1 81x3

= log(25 x 3 ) - log(4 x 4 ) - log(25) + log81 + log(l6x2)-log(81x3)

= log25 + log3 - log 16 - log25 + log81 + log 16 H log2- log81- log3

= log2 5. a; Hint: Given expression

'9 27 3 l , ( 9 3 32 = log —x —x —

1 8 4 27 log - + — x —

8 32 4

= logl = 0

Rule 7

l 0 g f l X =

Proof: Let

log/,* log* a

log/,«

or, l o g A x = v log A a

or, \oghx = \ogh(ay)

or, l o g „ W = l o g a ( a y ) = y = log/, a

Illustrative Examples Ex. I: I f log| 0 m = Z>log]0 n, find the value of m.

logio m Soln: We have,

logio " = b

=> log,, m = b

:. m = n Ex. 2: I f log2 = 0.3010 and log3 = 0.4771, then what value of

x satisfies

the equation y + i =135 (approximately)?

Soln : We have, 3 x+3 135

or, 3xx33 =135

, 3 - 1 ^ = 5 27

=> log 3 5 = x

_a ' ° g i o 5 _ r

log 3 ^ t n e a D O v e I ° r r n u l a }

_ log 1 0(10 + 2)

logio 3

l o g ) 0 - l o g 1 0 2

log,o 3

= X

= X

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Logarithm 701

1-0.3010 => = x

0.4771

.". x « 1 . 5 (approximately)

Ex. 3: log,, a \ogc b log,, c = ? Soln: logA ax log,.bx\oga c

\ogca log c b

log c /3xlog a c Since, \ogx y = log. y log. x

Since, \ogL. a = = log c .axlog a c = l

log a e

Exercise 1. Given that log 1 0 2 = 0.3010 , then log, 10 is equal to

| Assistant's Grade Exam, 1997]

a) 0.3010 b) 0.6990 c) 1000 301 d)

699 301

2. I f log l 2 27 * a then log 6 16 is

[Assistant's Grade Exam, 1997|

4(3 -a) 4(3 +a) 3 + a b)

3 + a - 3-a C ; 4(3-4 d ) 4 ( 3 + a)

3. I f log x 4 = 0.4, then the value of x is .

[ Assistant's Grade Exam, 1998] a)4 b) 16 c ) l d)32

4. I f log t y = 100 and log 2 x = 10 , then the value of y is

3 - a

[SSC Graduate Level PT Exam, 1999|

a) 2»o b) 2 1 0 0 0 c) 2 1 0 0 d) 2 1 0 0 0 0

5. I f ax = b> b}" = c> c z = a - then the value of xyz is

a)0 b ) l c)2 d)4

6. The value of log, 3 x log 3 2 x log, 4 x log 4 3 is

a)l b)2 c)3 d)4

log., x 7. The value of — 5 2 log u b is .

log f l /, x a)0

Answers

l.c; H i n t : ' ° g 2 1 0

b) 1 c) a d) ab

loglO _ 1 _ 1.0000 _ J000 log2 " log2 ~ 0.3010 ~ 301

log 27 .a; Hint: iog 1 227 = a or, , o o ] 2 = a

or, a logl2 = log3

or, a log(3 x 4) = 3 log 3

or, a[log 3 + log 4] = 3 log 3

or, alog4 + alog3 = 31og3

or, a log2 2 = (3-a)log3

or, 2alog2 = (3-a)log3

log2 3 - a or, log 3 •(»)

Now, log 61 6 • l o g l 6 _ log2 4 4 log 2 log 6 log(2x3) log2 + log3

I 0 ! 2 . / 3 - a ) log3 _ ( 2a] _ 4(3-a)

+ 1

3. d; Hint: l o S . v 4 :

log 2 log 3

log 4 2

3-a 2a

+ 1 3 + a

\ogx 5

2 log 2 _ 2 o r ' logx~ ~ 5 o r ' l o 8 x = 5 l o § 2 = l o 8 2 5 r >og32

.-. x = 32

4. b; Hint: log,, y = 100, log 2 x = 10

logy i n n log* l f l

or, 1 = 1 0 0 and - = 1 0

' logx

logy

log 2

= 100x10 = 1000 ° n log2

or, l o g 2 y = 1000 or, y = 2 1 0 0 0

5. b; Hint: a* =b o r > '°ga b = x

or, t>y = c or, log* c = y

or, c - a ff or, log,, a = 2

.-. x x y x 2 SB log,, b x log/, cx log,, a = 1

[See Illustrative Ex. 3]

log3 log2 log4 log3 6. a; Hint: — ^ x — ? — x — ^ x — ^ - = 1

log 2 log 3 log 3 log 4 7. b; Hint: log„x = - 1 ^ ^

log<,/> -0

.-. the given expression

I

I = log a a = 1

ab - log,, b = log,, a&- log,, h = log,, -

log„A a b

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702

Rule 8 y

I f log (x +y) = log (x) + log 0 ) , then x = _ j

Proof: log (x + y) = log (xy) or, x + y = xy

v-1

Illustrative Example Ex.: I f log (x + 2) = log (.v) + log (2), then find the value of

x Soln: Detail Method:

We have, log (x + 2) = log (x) + log (2) = log (2x) or, x + 2 = 2x :. x = 2 Quicker Method: Applying the above formula, we have

2 - 1

Exercise 1. I f log(x + 5) = log(5)+log(x), then find the value ofx.

a) 5 b)25 c ) - $ |

2. If log(x + 3)= log(3)+ log(x), then find the value ofx.

a) | b ) y c)3 d)4

3. I f log(x + 4)= log(4)+log(x) and

Iog(x + 6)= log(y)+log(6), then which of the follow­ing is correct? a )x=y b ) x < y c ) x > y d) Can't say

Answers l . d 2.a

4 4 5 5 3.c; Hint:x = — = y , y = — = -

.'. x>y

Rule 9

to# fx-y) = tog x - logy, then x = .

Illustrative Example Ex: I f log (x - 2) = logx - log (2), then find the value ofx. Soln: Detail Method:

We have, log (x - 2) - log (x) - log (2)

x or, x - 2 = —

P R A C T I C E B O O K ON Q U I C K E R MATHS

or, 2x -4 = x .-. x = 4 Quicker Method: Applying the above formula, we have

(2) 2

x = = 4 . 2-1

Exercise I f log(x--3) = log(x)- log(3), then, find the value ofx.

9 3 ) 2

b)9 9

c)4 d) - -

If log(x--4) = log(x)-- log(4). then find the value ofx.

16 a ) y

b)5 c)4 d ) - l

If log(x--5) = log(x)-- log(5) and

log(x - 6) = log(x)- log(6), then which of the following

is correct. a) x > y b) x < y c) x = y d) Can't say

Answers l .a 2. a

3 b ; Hint:-v = — = 6 - a n d > = — = 7 -4 4 5 5

Rule 10 To find the number of digits in a h .

No. of digits = [integral part of (b log 1 0 <3)]+1

Illustrative Example Ex.: Find the no. of digits in 2 4 7 • (Given that

log 1 0 2 =0.3010) Soln: Applying the above rule, we have

the required answer = (Integral part of 47 l o g 1 0 2 ) + 1

= (47 x 0.3010)+1 =[14.1470+1] = 14+ 1 = 15.

Exercise 1. Find the no. of digits in s 5 7 (given that

log l 0 2 = 0.3010) a)52 b)50 c)51 d)53

2. Find the number o f digits in g 1 0 . (Given that

log l 0 2 = 0.3010) [ RRB, Calcutta Supervisor (P Way) Exam, 2000]

a) 19 b)20 c)17 d) 10

3. I f log 2 = 0.3010 , then the number of digits in 2 6 4 is

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Logarithm 703

[CBI and CPO Exam, 1997] a) 18 b) 19 c)20 d)21

Answers

La; Hint: 8 5 7 = ( 2 3 ) 5 7 = 2 , 7 ,

.-. required answer = [l711og,0 2 + l ]

= [l71x0.3010]+l = [51.4710]+l = 51 + l = 52

2. d; H i n t : 8 1 0 = ( 2 3 ) ' 0 = 2 3 0

.-. required answer = [301og1 02 + l]

= [30x0.3010]+l =(9.03)+l = 9 + l = 10

3. c; Hint: Required answer • [641og I 0 2]+l

= [64x0.3010]+l = [l9.264]+l = 19 + 1 =20 .

Rule 11

Illustrative Example Ex.: Find the value of 3 l o g5 7

a) 7>°g5 3 b) 5 l o g J 7 c) 7 l c , g3 5 d) None of these

Soln: Applying the above rule, we have

3 log,7 _ y l o g s 3

Hence, (a) is the correct answer.

Exercise

1. I f A = log 2 7 625 + 7 l 0 8 " 1 3 a n d B = log 9 125+13 l o g " 7 ,

then which of the following is true a) A > B b) A < B c)A = B d) Can't say

Hint:b; A = log 2 7 625 + 7 l o g " 1 3 = log 3 , 5 4 + 7 l o g " 1 3

= | l o g 3 5 + 7 l o g ^ 3

B = log9125 + 13 l o g " 7 = l o g 3 2 5 3 +13 l o g " 7

= | t e f o 5+13'°«" 7

Let log 3 5 = x and by the above rule

7 I 0 8 1 1 I 3 _ ]3log,,7

Therefore,A= J A : + 13 , o b " 7 a n d B = | * + 13 , o g " 7

Clearly, A < B. Hence (b) is the correct answer.

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