FRANK Solutions Class 9 Maths Chapter 10 Logarithms · 2020. 12. 1. · FRANK Solutions Class 9...

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

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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




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




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


(iii) logx 243 = 5

x5 = 243

x5 = 35

x = 3


(iv) log8 x = (2 / 3)

x = 82 / 3

Taking cube on both sides, we get,

x3 = 82

x3 = 64

x3 = 43

x = 4


(v) log7 x = 3

x = 73

x = 343


(vi) log4 x = - 4

x = 4- 4

x = (1 / 256)

Therefore, the value of x is (1 / 256)

(vii) log2 0.5 = x




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


(ix) log10 0.0001 = x

0.0001 = 10x

10x = 10- 4

x = - 4


(x) log4 0.0625 = x

0.0625 = 4x

4x = 4- 2

x = - 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




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


(iii) log5 3125

Let log5 3125 = x

5x = 3125

5x = 55

We get,

x = 5


(iv) log2 128

Let log2 128 = x

2x = 128

2x = 27

We get,

x = 7


(v) log1 / 5 125

Let log1 / 5 125 = x

(1 / 5)x = 125

5- x = 53




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


(vii) log5 125

Let log5 125 = x

125 = 5x

5x = 53

We get,

x = 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




Logarithms


(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


(xii) log5 (1 / 25)

Let log5 (1 / 25) = x

5x = (1 / 25)

5x = 5-2

We get,

x = -2


(xiii) log2 8

Let log2 8 = x

2x = 8

2x = 23

We get,

x = 3


(xiv) loga a3

Let loga a3 = x

ax = a3




Logarithms

We get,

x = 3


(xv) log0.1 10

Let log0.1 10 = x

(0.1)x = 10

(10-1)x = 101

-x = 1

We get,

x = -1


(xvi)

Let = x

= 3√3

3x / 2 = 31 + 1 / 2

3x / 2 = 33 / 2

(x / 2) = (3 / 2)

We get,

x = 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




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




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)




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




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:




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)




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}




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


log V = log {(4 / 3)πr3}




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


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


= 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


= log (24 × 32) – log (23 × 32) + log (2 × 3 × 52) – log (2 × 52)




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


= 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


= 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




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


= 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


= 3 log (5 / 23) + 2 log {23 / (3 × 5)} – (1 / 2) log (52 / 34) + 3 log 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




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


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




Logarithms

(ii) log (x2 + 36) – 2 log x = 1


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


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


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




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


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,




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



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FRANK Solutions Class 9 Maths Chapter 10 Logarithms · 2020. 12. 1. · FRANK Solutions Class 9...

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