5
LO GARI TH MS
CHAPTER
Important Facts and Formulae
Ex.3:
If log
Sol. :
log
8
I. Logarithm : If a is a positive real number, other than 1 and am = x, then we write : m = loga x and we say that the value of log x to the base a is m. Example : (i) 103 = 1000 log10 1000 = 3 (ii) 34 = 81 log3 81 = 4 1 1 (i) 2–3 = log2 = – 3 8 8 (i) (.1)2 = 0.1 log(.1) 0.1 = 2
Ex.1:
Evalute :
Sol. :
1 (iii) log100(0.01) (ii) log 7 343 (i) Let log3 27 = n Then, 3n = 27 = 33 or n = 3 log3 27 = 3
Ex.4:
or n = –1 Ex.2: Sol.:
Evaluate (i) log53 × log2725 (ii) log927 – log279
log 3 log 25 (i) log53 × log2725 = log 5 log 27
(ii) Let log927 = n Then, 9n = 27 32n = 33 2n = 3
3 2 Again, let log279 = m n
1 1 3 7 – 3 or n = –3 343 7
Then, 27m = 9 33m = 32 3m = 2 m
1 log 7 3 343 (0.01) = n
Then, (100)n = 0.01 =
2 5 32
log 3 log(5 2 ) log 3 2 log 5 2 = log 5 log(33 ) log 5 3 log 3 3
1 n (ii) Let log 7 343
(iii) Let log100
10 x ( 8 )10 / 3 3 3 10 3
(i) log327
Then, 7n =
x
3 / 2 10 / 3 2 2 x = (2 )
Sol. :
EXAMPLES
8
1 x 3 , find the value of x. 3
3 2 5 log927 – log279 = (n–m) = 2 3 6
1 = (100)–1 100
log100 (0.01) = –1
Evalute : (i) log7 1 = 0 (ii) log34 34 (iii) 36log64 (i) We know that loga 1 = 0, so log7 1= 0 (ii) We know that loga a = 1, so log34 34 =0 (iii) We know that a log a x x . Now,
Ex.5:
75 5 32 Simplify : log 2 log log 16 9 243
Sol. :
log
2
75 5 32 75 32 5 2 log log log log log 16 9 243 16 243 9
= log
36 log 6 4 (6 2 ) log 6 4 6 2 ( log 6 4 ) 6 log6 ( 4
2
)
6 log 6 16 16
2 3
75 25 32 log log 16 81 243
75 32 81 log 2 = log 16 243 25
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Ex.6: Sol. :
Find the value of x which satisfies the relation log10 3 + log10 (4x + 1) = log10 (x + 1) + 1 log10 3 + log10 (4x + 1) = log10 (x + 1) + 1 log10 3 + log10 (4x + 1) = log10 (x + 1) + log10 10 log10 {3 (4x + 1) = log10 [10 (x + 1)] 3 (4x + 1) = 10 (x + 1) 12x + 3 = 10x + 10 2x + 7 x
Ex.7:
7 2
Simplify :
1 1 1 log xy ( xyz) log yz ( xyz) log zx ( xyz) Sol. :
Given expression = log xyz (xy) + log
xyz
(yz) + log xyz (zx) 1 log a x log x a
= logxyz (xy × yz × zx) = logxyz (xyz)2
Ex.9:
Sol. :
If log 2 = 0.3010, and log 3 = 0.4771, find the values of: (i) log 25 (ii) log 4.5 100 = log 100 – log 4 (i) log 25 = log 4 = 2 – 2 log 2 = (2 – 2 × 0.3010) = 1.398 9 (ii) log 4.5 = log 2 = log 9 – log 2 = 2 log 3 – log 2 = (2 × 0.4771 – 0.3010) = 0.6532
Ex.10: If log 2 = 0.30103, find the number of digits in 256 Sol. : log (256) = 56 log 2 = (56 × 0.30103) = 16.85768 Its characteristic is 16. Hence, the number of digits in 256 is 17
= 2 logxyz (xyz) = 2 × 1 = 2 Ex.8:
If log10 2 = 0.30103, find the value of log10 50.
Sol. :
100 = log10100– log10 2 log10 50 = log10 2 = 2 – 0.30103 = 1.69897
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EXERCISE Q.1
Q.2
Q.3
The value of log2 16 is 1 (A) (B) 4 (C) 8 8 The value of log343 7 is 1 1 (A) (B) –3 (C) 3 3
Q.12 (D) 16
(D) 3
1 is: The value of log 5 125
(A) 3
(B) –3
(C)
1 3
Q.13 (D)
The value of log (A)
Q.5
(B) 5
32 is: (C) 10
1 4
(B)
1 4
(C) – 4
(D)
1 10
1 3
(B)
1 3
(C)
3 2
3 2
(D)
The logarithm of 0.0625 to the base 2 is: (A) –4 (B) –2 (C) 0.25 (D) 0.5
Q.8
If log3 x = – 2, then x is equal to:
Q.9
(A) –9
(B) –6
If log 8 x
2 , then the value of x is: 3
(A)
Q.10
3 4
4 3
(C) 3
(D)
3 4
(B)
3 4
(C) 3
(C)
1 1000
(D)
1 10000
If log x 4
1 , then x is equal to: 4 (B) 64 (C) 128 (D) 256
If log x (0.1)
1 , then the value of x is: 3 (B) 100 1 (D) 1000
Q.15
If log 32 x 0.8 , then x is equal to: (A) 25.6 (B) 16 (C) 10 (D) 12.8
Q.16
If log x y 100 and log2 x = 10, then the value of y is: (A) 210 (B) 2100 1000 (C) 2 (D) 210000
Q.17
The value of log (1/ 3) 81 is equal to: (A) –27 (B) –4 (C) 4 (D) 27
Q.18
The value of log 2 (A) 3 (C) 6
1 9
Q.19
(D)
(A)
(1728) is: (B) 5 (D) 9
8
(B)
1 4
(C)
1 2
(D)
1 8
Q.20
Which of the following statements is not correct? (A) log10 10 = 1 (B) log (2 + 3) = log (2 × 3) (C) log10 1 = 0 (D) log (1 + 2 + 3) = log 1 + log 2 + log 3
Q.21
The value of log2 (log5 625) is : (A) 2 (B) 5 (C) 10 (D) 15
256 81
If logx 4 = 4.0 then the value of x is: (A) 1 (B) 4 (C) 16 (D) 32
3
log 8 is equal to: log 8 1
(D) 4
1 9 If log x , then x is equal to: 16 2
(A) Q.11
(B)
1 100
(D) 4
Q.7
(C) –8
(B)
(C) 1000
The value of log(.01) (1000) is : (A)
1 10
(A) 10
The value of log10 (.0001) is (A)
Q.6
5 2
2
1 , then x is equal to: 4
(A)
(A) 16
1 3
Q.14 Q.4
If log10000 x
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Q.22 Q.23 Q.24
If log2 [log3 (log2 x)] = 1, then x is equal to: (A) 0 (B) 12 (C) 128 (D) 512 The value of log2 log2 log3 log3 (A) 0 (B) 1 (C) 2 If
ax
=
by,
log a x (B) log b y
Q.27
Q.28
Q.35
Q.36
1 The value of log10 125 2 log10 4 log10 32 3 is : 4 (A) 0 (B) (C) 1 (D) 2 5 1 2 log10 5 + log108 log10 4 = ? 2 (A) 2 (B) 4 (C) 2 + 2 log10 2 (D) 4 – 4 log10 2
If loga (ab) = x, then logb (ab) is:
x x (B) (C) x 1 1 x
2 1 (C) 2 x y z 3 3
Q.30
Q.31
1 1 , then the value of x is: 6 3 (B) 16 (C) 18 (D) 24
(A)
1 3
(B) .064
(C) –3
(D) 3
(D) 4
(B)
3 2
(C) 2
(D) 5
If log12 27 = a, then log6 16 is:
3a (A) 4(3 a )
3 a (B) 4(3 a )
4(3 a ) (C) (3 a )
4(3 a ) (D) (3 a )
If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1, then x is equal to: (A) 1 (B) 3 (C) 5 (D) 10
Q.38
If log5 (x2 + x) – log5 (x + 1) = 2, then the value of x is: (A) 5 (B) 10 (C) 25 (D) 32
Q.39
1 1 1 The value of log 60 log 60 log 60 is: 3 4 5 (A) 0 (B) 1 (C) 5 (D) 60
Q.40
The value of (log3 4) (log4 5) (log5 6) (log6 7) (log7 8) (log8 9) is: (A) 2 (B) 7 (C) 8 (D) 33
Q.41
The value of 16 log 4 5 is: (A)
Q.42
5 64
Q.43
Q.44
(B) 5
(C) 16
(D) 25
If log x + log y = log (x + y), then (A) x = y (B) xy = 1 (C) y
If log 8 x log 8
If log10 125 + log10 8 = x, then x is equal to:
(D) 7
(log5 5) (log4 9) (log3 2) is equal to:
If log4 x + log2 x = 6, then x is equal to: (A) 2 (B) 4 (C) 8 (D) 16
(A) 12 Q.32
2 1 (B) 2x y z 3 3 2 1 (D) 2x y z 3 3
(C) 4
(log5 3) × (log3 625) equals: (A) 1 (B) 2 (C) 3
of log 4.3 63 is: 2 1 (A) 2 x y z 3 3
19 6
Q.37
x (D) x 1
If log 2 = x, log 3 = y and log 7 = z, then the value
(B)
(D) none of these
log 360 is equal to: (A) 2 log 2 + 3 log 3 (B) 3 log 2 + 2 log 3 (C) 3 log 2 + 2 log 3 – log 5 (D) 3 log 2 + 2 log 3 + log 5
1 (A) x Q.29
Q.34
7 2
(A) 1
log a y (C) log b x
Q.26
is: (D) 3
The value of (log9 27 + log8 32) is: (A)
273
then:
a x (A) log b y
Q.25
Q.33
x 1 x
(D) y
x x 1
a b log log(a b) , then: b a (A) a + b = 1 (B) a – b = 1 (C) a = b (D) a2 – b2 = 1
If log
a2 b2 c 2 log log log is equal to: ac ab bc (A) 0 (B) 1 (C) 2 (D) abc
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Q.45
Q.46
(logb a × logc b× loga c) is equal to: (A) 0 (B) 1 (C) abc (D) a + b + c
1 1 1 (log bc ) 1 (log ca ) 1 (log ab ) 1 a b c is equal to: (A) 1
Q.47
Q.48
(B)
3 2
(C) 2
1 If log10 7 = a, then log10 is equal to: 70 (A) – (1 + a) (B) (1 + a)–1
(C)
(D) 3
(D)
1 10a
Q.49
If a = bx, b = cy and c = az, then the value of xyz is equal to: (A) –1 (B) 0 (C) 1 (D) abc
Q.50
If log 27 = 1.431, then the value of log 9 is: (A) 0.934 (B) 0.945 (C) 0.954 (D) 0.958
The value of
1 1 1 is log ( p / q ) x log (q / r ) x log ( r / p ) x (A) 0 (B) 1 (C) 2 (D) 3
a 10
ANSWER KEY Q.No 1 Ans. B Q.No 21 Ans. A Q.No 41 Ans. D
2 A 22 D 42 D
3 B 23 A 43 A
4 C 24 C 44 A
5 C 25 D 45 B
6 D 26 C 46 A
7 A 27 A 47 A
8 D 28 D 48 A
9 D 29 B 49 C
10 D 30 D 50 C
11 D 31 A
12 A 32 D
13 D 33 B
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14 C 34 D
15 B 35 A
16 C 36 D
17 B 37 B
18 C 38 C
19 C 39 B
20 B 40 A
Hints & Solution Sol.1
Sol.2
Sol.7
Let log2 16 = n. Then, 2n = 16 = 24 n log2 16 = 4
Then, 2n = 0.0625 = 1 16 2n = 2–4 n = –4 log2 0.0625 = –4
2 n
Let log343 7 = n. Then, (343)n = 7 (73)n = 7 3n = 1 n
1 3
Sol.8
1 log 343 7 3
Sol.3
Let log2 0.0625 = n
log3 x = –2 x 3 2
1 n then, Let log 5 125
Sol.9
Sol.4
Let log
2 3 3
Sol.10
n 5 2 n = 10
Sol.5
1
16 9 2
256 16 x 81 9
32 10 2
Let log10 (0.0001) = n Then, 10n = .0001
Sol.11 logx4 = 0.4
1 1 10 4 10000 10 10n = 10–4 n = –4 log10 (.0001) = – 4
x2/5 = 4
log x 4
x = 4
n
1 3 10 100
= (22)
x 2 x = 32 Sol.12
3 2
5/2
4 2 10 5
5 2 2
Let log(.01) = (1000) = n Then (.01)n =(1000)
–2n = 3 n
9 16
9 16
x
x
n
Sol.6
1 9 log x 2 16 1 / 2 x
( 2 ) n 32 (2)n/2 = 25
log
x= 2 x = 22 = 4
32 n Then,
2
1 9
x =82/3 = (23)2/3
1 5 125 5n = 5–3 n = –3 1 log 5 3 125
3
2
2 3
log 8 x
n
1
25
1 4 x = (10000)–1/4 x = (104)–1/4 log10000 x
x = 10–1 =
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1 10
625 10000
Sol.13
Sol.14
Sol.22 log2 [log3(log2 x)] = 1 = log22 log3 (log2 x) = 2 log2 x = 32 = 9 x = 29 = 512
1 4 x 1/4 = 4 x = 44 = 256 log x 4
Sol.23 log2 log2 log3 (log3 273) = log2 log2 log3 [log3(33)3] = log2 log2 log3 [log3(3)9] = log2 log2 log3 (9 log3 3) = log2 log2 log3 9 [log3 3 = 1] = log2 log2 [log3 (3)2] = log2 log2 (2 log3 3) = log2 log2 2 = log2 1 = 0
1 3 = 0.1
log x (0.1)
x
–1/3
1 x
1/ 3
0 .1
Sol.24 ax = by log ax = log by
1 10 x1/ 3 0.1 x = (10)
Sol.25 360 = (2 × 2 × 2) × (3 × 3) × 5 So, log 360 = log (23 × 32 × 5) = log 23 + log 32 + log 5 = 3 log 2 + 2 log 3 + log 5
Sol.15 log32 x = 0.8 x = (32)0.8 x = (25) 4/5 = 24 = 16 Sol.16 log2 x = 10 x logx y = 100 y = x100 = (210)100y = 21000
Sol.26
Sol.17 Let log(–1/3)81= x. Then, x
1 1 4 4 81 3 (3) 3 3 x = –4 i.e., log (–1/3) 81= –4 Sol.18 Let log 2
3
= log10 5 log10 16 log10 32 5 32 log10 10 1 = log10 16
Then, ( 2 3 ) 1728 (12 ) [( 2 3 ) ] 3
(2 3 ) = (2 3 ) 6
Sol.19
3
Sol.27
2 log
2 3
x
x = 6, i.e. log 2
1 log10 125 2 log10 4 log10 32 3
= log10 (125)1/ 3 log10 (4) 2 log10 32
4
(1728) x x
log a y log b x
x log a = y log b
10
5 log
10
8
1 log 2
10
4
log10 (52 ) + log10 8 – log10(41/2) 25 8 log10 25 log10 8 log10 2 log10 2 = log10 100 2
(1728) 6
1 log8 log 8 log(8)1/ 2 2 1 log8 log8 log8 2
Sol.20 (a) Since loga a = 1, So log10 10 = 1 (b) log (2 + 3) = 5 and log (2 × 3) = log 6 = log2 + log3 log (2 + 3) log (2 × 3) (c) Since loga1= 0, so log10 1= 0 (d) log (1 + 2 + 3) = log 6 = log (1 × 2 × 3) = log 1 + log 2 + log 3 So, (b) is incorrect Sol.21 Let log5 625 = x . Then 5x = 625 = 54 or x = 4 Let log2 (log5 625) = y. Then, log24 = y or 2y = 4 or y =12 log2 (log5 625) = 2
Sol.28 loga (ab) = x
log ab x log a
log a log b x log a
log b x log a
log b x 1 log a
1
log a 1 log b x 1
log b log a x log b x 1
log(ab) x log b x 1
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1
log a 1 1 log b x 1
log b (ab)
x x 1
Sol.29 log (4.3 63) log 4 log(3 63) log 4 log(63)1/ 3 = log(2 ) log(7 3 ) 2
1 2 1 2 = 2 log 2 log 7 log 3 2 x z y 3 3 3 3
Sol.30
=
log 32 log 2
2
Sol.36 log12 27 = a
log x log x 6 2 log 2 log 2 3 log x = 12 log 2 log x = 4 log 2 log x = log (24) = log 16 x = 16
1 1 log8 x log8 6 3 1 log log x 6 1 log 8 log 8 3 1 1 log 8 6 3
log x log
1 log(81/ 3 ) log 2 6
log x log 2 log
log 33 log(3 2 ) 2
x = 12 Sol.32 log10 125 + log10 8 = x log10 (125 × 8) = x x = log10 (1000)log10(10)3 = 3 log10 10 = 3 Sol.33 Let log 9 27 x Then,
9x 27 32 x 33 2x 3 x
3 2
a
log 3 2 log 2 1 3 log 3 a
2 log 2 1 1 3 a 3 log 3 a 3 3a
log 2 3 a log 3 2a
1 6 2 log 12 6 1
log 27 a log12
log 6 16
log x log
log 9 log 2 [log5 5 = 1] log 4 log 3
log 2 2 log 3 log 2 1 log 3 2 log 2 log 3
log 4 x log 2 x 6 log x log x 6 log 4 log 2
Sol.31
Sol.35 Given expression =
2 1/ 3
3 log 3 a log 3 2 log 2
log 3 2 log 2 1 3 log 3 3 log 3 a
2a log 3 log 2 3a
log16 log 2 4 log 6 log(2 3)
4 log 2 log 2 log 3
4 log 2 2a log 21 3 a
4 4(3 a ) 3 a (3 a ) 3a
Sol.37 log10 5 + log10 (5x + 1) = log10 (x + 5) + 1 log10 5 + log10 (5x + 1)= log10 (x + 5) + log 10 10 log10 [5(5x + 1)] = log10 [10 (x + 5)] 5x + 1 (x + 5) 5x + 1 = 2x + 10 3x = 9 x
Let log8 32 = y. Then, yy = 25 3y = 5 y
3 5 19 2 3 6
log9 27 + log8 32 =
log 3 log 625 Sol.34 Given expression = log 5 log 3
log 625 log(5 4 ) 4 log 5 4 = log 5 log 5 log 5
5 3
x2 x 2 Sol.38 log5(x2 + x) – log5 (x + 1) = 2 log 5 x 1 x ( x 1) log 5 2 x 1
log5 x 2 x 5 2 25 Sol.39 Given expression = log60 3 + log60 4 +log60 5 = log60 (3 × 4 × 5) = log60 60 = 1
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Sol.40 Given expression =
Sol.46 Given expression
log 4 log 5 log 6 log 7 log 8 log 9 log 3 log 4 log 5 log 6 log 7 log 8 =
=
1 1 1 log a bc log a a log b ca log b b log c ab log c c
log 9 log 32 2 log 3 2 log 3 log 3 log 3
=
1 1 1 log a (abc) log b (abc) log c (abc)
= log abc a log abc b log abc c
log abc (abc) 1
Sol.41 We know that : a log a x x . 16log 4 5 (4 2 ) log 4 5 4 2 log 4 5 4 log 4 (5 ) 4 log 4 25 25 2
Sol.47 Given expression
p r q = log x q log x r log x p
Sol.42 log x + log y = log (x + y) log(x + y log (xy) x + y = xy y (x –1) = x y Sol.43
log
p q r = log x q r p log x 1 0
x x 1
a b log log (a b) b a
Sol.48
a b log (a b) log log 1 b a
So, a + b = 1
a 2 b2 c2 Sol.44 Given expression = log log1 0 bc ac ab log a log b log c Sol.45 Given expression = log b log c log a 1
1 log10 log10 1 log10 70 – log10 (7 10) 70 = (log10 7 log10 10) (a 1)
Sol.49 a = bx, b =cy, c = az x = logb a, y = logc b, z = logac xyz = (logb a)× (logc b )× (logac)
log a log b log c 1 xyz log b log c log a
Sol.50 log 27 = 1.431 log(33) = 1.431 3 log 3 = 1.431 log 3 = 0.477 log 9 = log(32) = 2 log3 = (2 × 0.477) = 0.954
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