chapter 29
TRANSCRIPT
-
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 ex-pressing an exponential identity and the following two ex-pressions 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 loga-rithm
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 I4 _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) 2o 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: logx = - 1 ^ ^
log -0
.-. the given expression
I
I = log a a = 1
ab - log,, b = log,, a&- log,, h = log,, -
logA a b
-
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 follow-ing 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
-
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.