frank solutions class 9 maths chapter 10 logarithms · 2020. 12. 1. · frank solutions class 9...
Post on 07-Aug-2021
10 Views
Preview:
TRANSCRIPT
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
1. Express each of the following in the logarithmic form:
(i) 33 = 27
(ii) 54 = 625
(iii) 90 = 1
(iv) (1 / 8) = 2-3
(v) 112 = 121
(vi) 3-2 = (1 / 9)
(vii) 10-4 = 0.0001
(viii) 70 = 1
(ix) (1 / 3)4 = (1 / 81)
(x) 9- 4 = (1 / 6561)
Solution:
The logarithmic forms of the given expressions are as follows:
(i) 33 = 27
log3 27 = 3
(ii) 54 = 625
log5 625 = 4
(iii) 90 = 1
log9 1 = 0
(iv) (1 / 8) = 2- 3
log2 (1 / 8) = - 3
(v) 112 = 121
log11 121 = 2
(vi) 3-2 = (1 / 9)
log3 (1 / 9) = - 2
(vii) 10-4 = 0.0001
log10 0.0001 = - 4
(viii) 70 = 1
log7 1 = 0
(ix) (1 / 3)4 = (1 / 81)
log1 / 3 (1 / 81) = 4
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(x) 9-4 = (1 / 6561)
log9 (1 / 6561) = - 4
2. Express each of the following in the exponential form:
(i) log2 128 = 7
(ii) log3 81 = 4
(iii) log10 0.001 = - 3
(iv) log2 (1 / 32) = - 5
(v) logb a = c
(vi) log2 (1 / 2) = - 1
(vii) log5 a = 3
(viii)
(ix)
(x)
(xi)
(xii) – 2 = log2 (0.25)
Solution:
(i) log2 128 = 7
128 = 27
Hence, the exponential form of log2 128 = 7 is 27
(ii) log3 81 = 4
81 = 34
Hence, the exponential form of log3 81 = 4 is 34
(iii) log10 0.001 = - 3
0.001 = 10-3
Hence, the exponential form of log10 0.001 = - 3 is 10-3
(iv) log2 (1 / 32) = - 5
(1 / 32) = 2- 5
Hence, the exponential form of log2 (1 / 32) = - 5 is 2-5
(v) logb a = c
a = bc
Hence, the exponential form of logb a = c is bc
(vi) log2 (1 / 2) = - 1
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(1 / 2) = 2- 1
Hence, the exponential form of log2 (1 / 2) = - 1 is 2-1
(vii) log5 a = 3
a = 53
Hence, the exponential form of log5 a = 3 is 53
(viii)
27 =
Hence, the exponential form of is
(ix)
Hence, the exponential form of is 251 / 4
(x)
p = aq
Hence, the exponential form of is aq
(xi)
Hence, the exponential form of is
(xii) -2 =
We get,
2-2 = 0.25
3. Find x in each of the following when:
(i) logx 49 = 2
(ii) logx 125 = 3
(iii) logx 243 = 5
(iv) log8 x = (2 / 3)
(v) log7 x = 3
(vi) log4 x = - 4
(vii) log2 0.5 = x
(viii) log3 243 = x
(ix) log10 0.0001 = x
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(x) log4 0.0625 = x
Solution:
(i) logx 49 = 2
x2 = 49
x = 7
Therefore, the value of x is 7
(ii) logx 125 = 3
x3 = 125
x3 = 53
x = 5
Therefore, the value of x is 5
(iii) logx 243 = 5
x5 = 243
x5 = 35
x = 3
Therefore, the value of x is 3
(iv) log8 x = (2 / 3)
x = 82 / 3
Taking cube on both sides, we get,
x3 = 82
x3 = 64
x3 = 43
x = 4
Therefore, the value of x is 4
(v) log7 x = 3
x = 73
x = 343
Therefore, the value of x is 343
(vi) log4 x = - 4
x = 4- 4
x = (1 / 256)
Therefore, the value of x is (1 / 256)
(vii) log2 0.5 = x
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
2x = 0.5
2x = (1 / 2)
2x = 2-1
x = - 1
Therefore, the value of x is -1
(viii) log3 243 = x
243 = 3x
35 = 3x
x = 5
Therefore, the value of x is 5
(ix) log10 0.0001 = x
0.0001 = 10x
10x = 10- 4
x = - 4
Therefore, the value of x is -4
(x) log4 0.0625 = x
0.0625 = 4x
4x = 4- 2
x = - 2
Therefore, the value of x is -2
4. Find the values of:
(i) log10 1000
(ii) log3 81
(iii) log5 3125
(iv) log2 128
(v) log1 / 5 125
(vi) log10 0.0001
(vii) log5 125
(viii) log8 2
(ix) log1 / 2 16
(x) log0.01 10
(xi) log3 81
(xii) log5 (1 / 25)
(xiii) log2 8
(xiv) loga a3
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(xv) log0.1 10
(xvi)
Solution:
(i) log10 1000
Let log10 1000 = x
10x = 1000
10x = 103
We get,
x = 3
Hence, the value of x is 3
(ii) log3 81
Let log3 81 = x
3x = 81
3x = 34
We get,
x = 4
Hence, the value of x is 4
(iii) log5 3125
Let log5 3125 = x
5x = 3125
5x = 55
We get,
x = 5
Hence, the value of x is 5
(iv) log2 128
Let log2 128 = x
2x = 128
2x = 27
We get,
x = 7
Hence, the value of x is 7
(v) log1 / 5 125
Let log1 / 5 125 = x
(1 / 5)x = 125
5- x = 53
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
- x = 3
We get,
x = - 3
Hence, the value of x is -3
(vi) log10 0.0001
Let log10 0.0001 = x
0.0001 = 10x
10x = 10- 4
We get,
x = - 4
Hence, the value of x is -4
(vii) log5 125
Let log5 125 = x
125 = 5x
5x = 53
We get,
x = 3
Hence, the value of x is 3
(viii) log8 2
Let log8 2 = x
2 = 8x
This can be written as,
(23)x = 2
23x = 21
3x = 1
We get,
x = (1 / 3)
Hence, the value of x is (1 / 3)
(ix) log1 / 2 16
Let log1 / 2 16 = x
16 = (1 / 2)x
2- x = 24
- x = 4
We get,
x = -4
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
Hence, the value of x is -4
(x) log0.01 10
Let log0.01 10 = x
(0.01)x = 10
(10-2)x = 101
10-2x = 101
-2x = 1
We get,
x = (- 1 / 2)
Hence, the value of x is (-1 / 2)
(xi) log3 81
Let log3 81 = x
3x = 81
3x = 34
We get,
x = 4
Hence, the value of x is 4
(xii) log5 (1 / 25)
Let log5 (1 / 25) = x
5x = (1 / 25)
5x = 5-2
We get,
x = -2
Hence, the value of x is -2
(xiii) log2 8
Let log2 8 = x
2x = 8
2x = 23
We get,
x = 3
Hence, the value of x is 3
(xiv) loga a3
Let loga a3 = x
ax = a3
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
We get,
x = 3
Hence, the value of x is 3
(xv) log0.1 10
Let log0.1 10 = x
(0.1)x = 10
(10-1)x = 101
-x = 1
We get,
x = -1
Hence, the value of x is -1
(xvi)
Let = x
= 3√3
3x / 2 = 31 + 1 / 2
3x / 2 = 33 / 2
(x / 2) = (3 / 2)
We get,
x = 3
Hence, the value of x is 3
5. If log10 x = a, express the following in terms of x:
(i) 102a
(ii) 10a + 3
(iii) 10- a
(iv) 102a – 3
Solution:
(i) 102a
log10 x = a
x = 10a
Hence,
102a = (10a)2
102a = x2
(ii) 10a + 3
log10 x = a
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
x = 10a
Hence,
10a + 3 = 10a. 103
10a + 3 = x.1000
10a + 3 = 1000x
(iii) 10-a
log10 x = a
x = 10a
Hence,
10-a = x-1
10-a = (1 / x)
(iv) 102a – 3
log10 x = a
x = 10a
Hence,
102a – 3 = 102a.10-3
102a – 3 = (10a)2 10- 3
102a – 3 = (x2 / 1000)
6. If log10 m = n, express the following in terms of m:
(i) 10n – 1
(ii) 102n – 1
(iii) 10- 3n
Solution:
(i) 10n - 1
log10 m = n
m = 10n
Therefore,
10n – 1 = 10n.10- 1
10n – 1 = (m / 10) [m = 10n]
(ii) 102n + 1
log10 m = n
m = 10n
Therefore,
102n + 1 = 102n. 101
102n + 1 = (10n)2 . 10
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
102n + 1 = (m)2 . 10 [m = 10n]
102n + 1 = 10m2
(iii) 10-3n
log10 m = n
m = 10n
Therefore,
10-3n = (10n)- 3
10-3n = (m)-3 [m = 10n]
10-3n = (1 / m3)
7. If log10 x = p, express the following in terms of x:
(i) 10p
(ii) 10p + 1
(iii) 102p – 3
(iv) 102 – p
Solution:
(i) 10p
log10 x = p
We get,
x = 10p
(ii) 10p + 1
log10 x = p
x = 10p
Therefore,
10p + 1 = 10p.101
10p + 1 = (x). 10 [x = 10p]
We get,
10p + 1 = 10x
(iii) 102p – 3
log10 x = p
x = 10p
Therefore,
102p – 3 = 102p. 10-3
102p – 3 = (10p)2. 10-3
102p – 3 = (x)2.10-3 [x = 10p]
102p – 3 = (x2 / 1000)
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(iv) 102 – p
log10 x = p
x = 10p
Therefore,
102 – p = 102.10-p
102 – p = 100.x-1
102 – p = (100 / x)
8. If log10 x = a, log10 y = b and log10 z = 2a – 3b, express z in terms of x and y.
Solution:
log10 x = a
x = 10a
log10 y = b
y = 10b
log10 z = 2a – 3b
z = 102a – 3b
Therefore,
z = 102a – 3b
z = (10a)2.(10b)-3
z = (x)2.(y)-3
z = (x2 / y3)
9. Express the following in terms of log 2 and log 3:
(i) log 36
(ii) log 54
(iii) log 144
(iv) log 216
(v) log 648
(vi) log 128
Solution:
(i) log 36
log 36 = log (2 × 2 × 3 × 3)
log 36 = log (22 × 32)
log 36 = log 22 + log 32
We get,
log 36 = 2 log 2 + 2 log 3
(ii) log 54
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
log 54 = log (2 × 3 × 3 × 3)
log 54 = log (2 × 33)
log 54 = log 2 + log 33
We get,
log 54 = log 2 + 3 log 3
(iii) log 144
log 144 = log (24 × 32)
log 144 = log 24 + log 32
We get,
log 144 = 4 log 2 + 2 log 3
(iv) log 216
log 216 = log (23 × 33)
log 216 = log 23 + log 33
We get,
log 216 = 3 log 2 + 3 log 3
(v) log 648
log 648 = log (23 × 34)
log 648 = log 23 + log 34
We get,
log 648 = 3 log 2 + 4 log 3
(vi) log 128
log 128 = log (3 × 22)8
log 128 = 8 log (3 × 22)
log 128 = 8 {log 3 + log 22}
We get,
log 128 = 8 {log 3 + 2 log 2}
10. Express the following in terms of log 5 and / or log 2:
(i) log 20
(ii) log 80
(iii) log 125
(iv) log 160
(v) log 500
(vi) log 250
Solution:
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(i) log 20
log 20 = log (22 × 5)
log 20 = log 22 + log 5
We get,
log 20 = 2 log 2 + log 5
(ii) log 80
log 80 = log (24 × 5)
log 80 = log 24 + log 5
We get,
log 80 = 4 log 2 + log 5
(iii) log 125
log 125 = log 53
We get,
log 125 = 3 log 5
(iv) log 160
log 160 = log (25 × 5)
log 160 = log 25 + log 5
We get,
log 160 = 5 log 2 + log 5
(v) log 500
log 500 = log (22 × 53)
log 500 = log 22 + log 53
We get,
log 500 = 2 log 2 + 3 log 5
(vi) log 250 = log (53 × 2)
log 250 = log 53 + log 2
We get,
log 250 = 3 log 5 + log 2
11. Express the following in terms of log 2 and log 3:
(i)
(ii)
(iii)
(iv) log (26 / 51) – log (91 / 119)
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(v) log (225 / 16) – 2 log (5 / 9) + log (2 / 3)5
Solution:
(i)
= log (144)1 / 3
= (1 / 3) log 144
= (1 / 3) log (24 × 32)
= (1 / 3) log 24 + (1 / 3) log 32
We get,
= (4 / 3) log 2 + (2 / 3) log 3
(ii)
= log (216)1 / 5
= (1 / 5) log 216
= (1 / 5) log (23 × 33)
= (1 / 5) log 23 + (1 / 5) log 33
We get,
= (3 / 5) log 2 + (3 / 5) log 3
(iii)
= log (648)1 / 4
= (1 / 4) log 648
= (1 / 4) log (23 × 34)
= (1 / 4) log 23 + (1 / 4) log 34
= (3 / 4) log 2 + (4 / 4) log 3
We get,
= (3 / 4) log 2 + 1 log 3
= (3 / 4) log 2 + log 3
(iv) log (26 / 51) – log (91 / 119)
= log {(2 × 13) / (3 × 17)} – log {(7 x 13) / (7 x 17)}
= log {(2 × 13) / (3 × 17)} – log (13 / 17)
= (log 13 + log 2 – log 3 – log 17) – (log 13 – log 17)
= log 13 + log 2 – log 3 – log 17 – log 13 + log 17
We get,
= log 2 – log 3
(v) log (225 / 16) – 2 log (5 / 9) + log (2 / 3)5
= log (225 / 16) – 2 log (5 / 9) + 5 log (2 / 3)
= log 225 – log 16 – 2 {log 5 – log 9} + 5 {log 2 – log 3}
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
= log (52 × 32) – log 24 – 2 {log 5 – log 32} + 5 {log 2 – log 3}
= log 52 + log 32 – 4 log 2 – 2 {log 5 – 2 log 3} + 5 {log 2 – log 3}
= 2 log 5 + 2 log 3 – 4 log 2 – 2 log 5 + 4 log 3 + 5 log 2 – 5 log 3
We get,
= log 2 + log 3
12. Write the logarithmic equation for:
(i) F = {G (m1m2) / d2}
(ii) E = (1 / 2) mv2
(iii)
(iv) V = (4 / 3) πr3
(v)
Solution:
(i) F = {G (m1m2) / d2}
Taking log on both the sides, we get,
log F = log [{G (m1m2)} / d2]
log F = log (Gm1m2) – log d2
We get,
log F = log G + log m1 + log m2 – 2 log d
(ii) E = (1 / 2) mv2
Taking log on both the sides, we get,
log E = log {(1 / 2) mv2}
log E = log (1 / 2) + log m + log v2
We get,
log E = log 1 – log 2 + log m + 2 log v
(iii)
n = (M.g / m.l)1 / 2
On taking log on both the sides, we get,
log n = log (M.g / m.l)1 / 2
log n = (1 / 2) log (M.g / m. l)
log n = (1 / 2) {log (M.g) – log (m.l)}
We get,
log n = (1 / 2) {log M + log g – log m – log l}
(iv) V = (4 / 3) πr3
On taking log on both the sides, we get,
log V = log {(4 / 3)πr3}
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
log V = log 4 + log π + log r3 – log 3
log V = log 22 + log π + 3 log r – log 3
log V = 2 log 2 – log 3 + log π + 3 log r
(v)
V = 1 / Dl (T / πr)1 / 2
On taking log on both the sides, we get,
log V = log {1 / Dl (T / πr)1 / 2}
log V = log (1 / Dl) + log (T / πr)1 / 2
log V = (log 1 – log D – log l) + (1 / 2) log (T / πr)
log V = (0 – log D – log l) + (1 / 2) {(log T – log π – log r)}
We get,
log V = (1 / 2) (log T – log π – log r) – log D – log l
13. Express the following as a single logarithm:
(i) log 18 + log 25 – log 30
(ii) log 144 – log 72 + log 150 – log 50
(iii) 2 log 3 – (1 / 2) log 16 + log 12
(iv) 2 + (1 / 2) log 9 – 2 log 5
(v) 2 log (9 / 5) – 3 log (3 / 5) + log (16 / 20)
(vi) 2 log (15 / 18) – log (25 / 162) + log (4 / 9)
(vii) 2 log (16 / 25) – 3 log (8 / 5) + log 90
(viii) (1 / 2) log 25 – 2 log 3 + log 36
(ix) log (81 / 8) – 2 log (3 / 5) + 3 log (2 / 5) + log (25 / 9)
(x) 3 log (5 / 8) + 2 log (8 / 15) – (1 / 2) log (25 / 81) + 3
Solution:
(i) log 18 + log 25 – log 30
This can be written as,
= log (2 × 32) + log 52 – log (2 × 3 × 5)
= log 2 + log 32 + 2 log 5 – {log 2 + log 3 + log 5}
= log 2 + 2 log 3 + 2 log 5 – log 2 – log 3 – log 5
= log 3+ log 5
= log (3 × 5)
We get,
= log 15
(ii) log 144 – log 72 + log 150 – log 50
This can be written as,
= log (24 × 32) – log (23 × 32) + log (2 × 3 × 52) – log (2 × 52)
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
= log 24 + log 32 – {log 23 + log 32) + log 2 + log 3 + log 52 – {log 2 + log 52}
= 4 log 2 + 2 log 3 – 3 log 2 – 2 log 3 + log 2 + log 3 + 2 log 5 – log 2 – 2 log 5
We get,
= log 2 + log 3
= log (2 × 3)
We get,
= log 6
(iii) 2 log 3 – (1 / 2) log 16 + log 12
= 2 log 3 – (1 / 2) log 24 + log (22 × 3)
= 2 log 3 – (1 / 2) × 4 log 2 + log 22 + log 3
We get,
= 2 log 3 – 2 log 2 + 2 log 2 + log 3
= 3 log 3
= log 33
We get,
= log 27
(iv) 2 + (1 / 2) log 9 – 2 log 5
This can be written as,
= 2 + (1 / 2) log 32 – 2 log 5
= 2 log 10 + (1 / 2) × 2 log 3 – 2 log 5
= log 102 + log 3 – log 52
= log 100 + log 3 – log 25
= log {(100 × 3) / 25}
We get,
= log 12
(v) 2 log (9 / 5) – 3 log (3 / 5) + log (16 / 20)
= 2 log 9 – 2 log 5 – 3 log 3 + 3 log 5 + log 16 – log 20
This can be written as,
= 2 log (32) – 2 log 5 – 3 log 3 + 3 log 5 + log (42) – log (5 × 4)
= 4 log 3 – 2 log 5 – 3 log 3 + 3 log 5 + 2 log 4 – log 5 - log 4
= (4 – 3) log 3 + (-2 -1 + 3) log 5 + log 4
= log 3 + log 4
= log (3 × 4)
We get,
= log 12
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(vi) 2 log (15 / 18) – log (25 / 162) + log (4 / 9)
= 2 log {5 / (2 × 3)} – log {52 / (2 × 34)} + log (22 / 32)
= 2 log 5 – 2 log 2 – 2 log 3 – {log 52 – log 2 – log 34} + log 22 – log 32
= 2 log 5 – 2 log 2 – 2 log 3 – 2 log 5 + log 2 + 4 log 3 + 2 log 2 – 2 log 3
We get,
= log 2
(vii) 2 log (16 / 25) – 3 log (8 / 5) + log 90
This can be written as,
= 2 log (24 / 52) – 3 log (23 / 5) + log (2 × 5 × 32)
= 2 log 24 – 2 log 52 – 3 {log 23 – log 5) + log 2 + log 5 + log 32
= 4 × 2 log 2 – 2 × 2 log 5 – 3 × 3 log 2 + 3 log 5 + log 2 + log 5 + 2 log 3
= 8 log 2 – 4 log 5 – 9 log 2 + 3 log 5 + log 2 + log 5 + 2 log 3
= 2 log 3
= log 32
We get,
= log 9
(viii) (1 / 2) log 25 – 2 log 3 + log 36
= (1 / 2) log 52 – 2 log 3 + log (22 × 32)
= (1 / 2) × 2 log 5 – 2 log 3 + log 22 + log 32
= log 5 - log 32 + 2 log 2 + log 32
= log 5 + 2 log 2
= log 5 + log 22
= log 5 + log 4
= log (5 × 4)
We get,
= log 20
(ix) log (81 / 8) – 2 log (3 / 5) + 3 log (2 / 5) + log (25 / 9)
= log (34 / 23) – 2 log (3 / 5) + 3 log (2 / 5) + log (52 / 32)
= log 34 – log 23 – 2 log 3 + 2 log 5 + 3 log 2 – 3 log 5 + log 52 – log 32
= 4 log 3 – 3 log 2 – 2 log 3 + 2 log 5 + 3 log 2 – 3 log 5 + 2 log 5 – 2 log 3
We get,
= log 5
(x) 3 log (5 / 8) + 2 log (8 / 15) – (1 / 2) log (25 / 81) + 3
This can be written as,
= 3 log (5 / 23) + 2 log {23 / (3 × 5)} – (1 / 2) log (52 / 34) + 3 log 10
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
= 3 log 5 – 3 log 23 + 2 log 23 – 2 log 3 – 2 log 5 – (1 / 2) log 52 + (1 / 2) log 34 + 3 log (2
× 5)
= 3 log 5 – 3 × 3 log 2 + 2 × 3 log 2 – 2 log 3 – 2 log 5 – (1 / 2) × 2 log 5 + (1 / 2) × 4 log
3 + 3 log 2 + 3 log 5
= 3 log 5 - 9 log 2 + 6 log 2 – 2 log 3 – 2 log 5 – log 5 + 2 log 3 + 3 log 2 + 3 log 5
= 3 log 5
= log 53
We get,
= log 125
14. Simplify the following:
(i) 2 log 5 + log 8 – (1 / 2) log 4
(ii) 2 log 7 + 3 log 5 – log (49 / 8)
(iii) 3 log (32/27) + 5 log (125/ 24) – 3 log (625/ 243) + log (2 / 75)
(iv) 12 log (3/2) + 7 log (125 / 27) – 5 log (25 / 36) – 7 log 25 + log (16 / 3)
Solution:
(i) 2 log 5 + log 8 – (1 / 2) log 4
= 2 log 5 + log 23 – (1 / 2) log 22
= 2 log 5 + 3 log 2 – (1 / 2) × 2 log 2
= 2 log 5 + 3 log 2 – log 2
= 2 log 5 + 2 log 2
= 2 (log 5 + log 2)
= 2 log (5 × 2)
We get,
= 2 log 10
= 2 × 1
= 2
(ii) 2 log 7 + 3 log 5 – log (49 / 8)
= 2 log 7 + 3 log 5 – log 49 + log 8
= 2 log 7 + 3 log 5 – log 72 + log 23
= 2 log 7 + 3 log 5 – 2 log 7 + 3 log 2
= 3 log 5 + 3 log 2
= 3 (log 5 + log 2)
= 3 log (5 × 2)
= 3 log 10
We get,
= 3 × 1
= 3
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(iii) 3 log (32/27) + 5 log (125/ 24) – 3 log (625/ 243) + log (2 / 75)
= 3 log (25 / 33) + 5 log {53 / (23 x 3)} – 3 log (54 / 35) + log {2 / (3 x 52)}
= 3 log 25 – 3 log 33 + 5 log 53 – 5 log 23 – 5 log 3 – 3 log 54 + 3 log 35 + log 2 – log 3 -
log 52
= 15 log 2 – 9 log 3 + 15 log 5 – 15 log 2 – 5 log 3 – 12 log 5 + 15 log 3 + log 2 – log 3 -
2 log 5
= log 2 + log 5
(iv) 12 log (3/2) + 7 log (125 / 27) – 5 log (25 / 36) – 7 log 25 + log (16 / 3)
= 12 log (3/2) + 7 log (53 / 33) – 5 log {52 / (22 x 32)} – 7 log 52 + log (24 / 3)
= 12 log 3 – 12 log 2 + 7 log 53 – 7 log 33 – 5 log 52 + 5 log 22 + 5 log 32 – 7 log 52 + log
24 – log 3
= 12 log 3 – 12 log 2 + 21 log 5 – 21 log 3 – 10 log 5 + 10 log 2 + 10 log 3 – 14 log 5 + 4
log 2 – log 3
We get,
= 2log 2 – 3 log 5
15. Solve the following:
(i) log (3 – x) – log (x – 3) = 1
(ii) log (x2 + 36) – 2 log x = 1
(iii) log 7 + log (3x – 2) = log (x + 3) + 1
(iv) log (x + 1) + log (x – 1) = log 11 + 2 log 3
(v) log4 x + log4 (x – 6) = 2
(vi) log8 (x2 – 1) – log8 (3x + 9) = 0
(vii) log (x + 1) + log (x – 1) = log 48
(viii) log2 x + log4 x + log16 x = (21 / 4)
Solution:
(i) log (3 – x) – log (x – 3) = 1
This can be written as,
log {(3 – x) / (x – 3)} = 1
log {(3 – x) / (x – 3)} = log 10
(3 – x) / (x – 3) = 10
On calculating further, we get,
(3 – x) = 10 (x – 3)
(3 – x) = 10 x – 30
11x = 33
We get,
x = 3
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(ii) log (x2 + 36) – 2 log x = 1
This can be written as,
log (x2 + 36) – log x2 = 1
log {(x2 + 36) / x2} = 1
log {(x2 + 36) / x2} = log 10
{(x2 + 36) / x2} = 10
On further calculation, we get,
x2 + 36 = 10x2
9x2 = 36
x2 = 4
We get,
x = 2
(iii) log 7 + log (3x – 2) = log (x + 3) + 1
log 7 + log (3x – 2) – log (x + 3) = 1
This can be written as,
log {7.(3x – 2) / (x + 3)} = log 10
{7. (3x – 2) / (x + 3)} = 10
On further calculation, we get,
21x – 14 = 10 (x + 3)
21x – 10x = 30 + 14
11x = 44
x = (44 / 11)
We get,
x = 4
(iv) log (x + 1) + log (x – 1) = log 11 + 2 log 3
This can be written as,
log {(x + 1) (x – 1)} = log 11 + log 32
log (x2 – 1) = log (11. 9)
log (x2 – 1) = log 99
x2 – 1 = 99
x2 = 99 + 1
x2 = 100
Hence,
x = 10 or -10
Here, negative value is rejected
Therefore,
x = 10
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
(v) log4 x + log4 (x – 6) = 2
log4 {x (x – 6)} = 2 log4 4
log4 {x2 – 6x} = log4 42
x2 – 6x = 16
x2 – 6x – 16 = 0
x2 – 8x + 2x – 16 = 0
x (x – 8) + 2 (x – 8) = 0
(x – 8) (x + 2) = 0
We get,
x = 8 or -2
Negative value is rejected
Hence,
x = 8
(vi) log8 (x2 – 1) – log8 (3x + 9) = 0
log8 {(x2 – 1) / (3x + 9)} = log8 1
(x2 – 1) / (3x + 9) = 1
x2 – 1 = 3x + 9
On calculating further, we get,
x2 – 3x – 10 = 0
x2 – 5x + 2x – 10 = 0
x (x – 5) + 2 (x – 5) = 0
(x – 5) (x + 2) = 0
x = 5 or x = -2
negative value is rejected,
Hence,
x = 5
(vii) log (x + 1) + log (x – 1) = log 48
This can be written as,
log {(x + 1) (x – 1)} = log 48
log (x2 – 1) = log 48
x2 – 1 = 48
x2 = 48 + 1
x2 = 49
x = 7 or -7
neglecting the negative value
Therefore,
FRANK Solutions Class 9 Maths Chapter 10
Logarithms
x = 7
(viii) log2 x + log4 x + log16 x = (21 / 4)
(1 / logx 2) + (1 / logx 22) + (1 / logx 24) = (21 / 4)
(1 / logx 2) + (1 / 2 logx 2) + (1 / 4 logx 2) = (21 / 4)
Taking (1 / logx 2) as common, we get,
(1 / logx 2) {1 + (1 / 2) + (1 / 4)} = (21 / 4)
We get,
(1 / logx 2) (7 / 4) = (21 / 4)
logx 2= (7 / 4) × (4 / 21)
logx 2 = (1 / 3)
So,
x1 / 3 = 2
We get,
x = 23
x = 8
top related