# Concise Selina Solutions for Class 9 Maths Chapter 8-Logarithms

The Selina solutions for the questions given in Chapter 8, Logarithms, of the Class 9 Selina textbooks are available here. In this chapter students learn about the topic of Logarithms in detail. Students can easily score full marks in the exams by solving all the questions present in the Selina textbook.

The Class 9 Selina solutions maths are very easy to understand. These solutions cover all the exercise questions included in the book and are according to the syllabus prescribed by the ICSE or CISCE. Here, the PDF of the Class 9 Maths Chapter 8 Selina solutions is available which can be downloaded as well as viewed online. Students can also avail these Selina solutions and download it for free to practice them offline as well.

## Download PDF of Selina Solutions for Class 9 Maths Chapter 8:-

Exercise 8(A)

1. Express each of the following in logarithmic form:

(i) 53Â = 125

(ii) 3-2Â =Â 1/9

(iii) 10-3Â = 0.001

(iv)Â (81)3/4 = 27

Solution:

We know that,

ab = c â‡’ loga c = b

(i) 53 = 125

log5 125 = 3Â

Â

(ii) 3-2 = 1/9

log3 1/9 = -2Â

Â

(iii) 10-3 = 0.001

log10 0.001 = -3Â

(iv) (81)3/4 = 27

log81 27 = Â¾Â

2. Express each of the following in exponential form:

(i)Â log8 0.125 = -1

(ii) log10 0.01 = -2

(iii)Â loga AÂ = x

(iv) log10 1 = 0

Solution:

We know that,

loga c = b â‡’ ab = c

(i)Â log8 0.125 = -1

8-1 = 0.125

(ii) log10 0.01 = -2

8-1 = 0.125

(iii)Â loga AÂ = x

ax = A

(iv) log10 1 = 0

100 = 1

3. Solve for x:Â log10Â x = -2.

Solution:

We have,

log10Â x = -2

10-2 = x [As loga c = b â‡’ ab = c]

x = 10-2

x = 1/102

x = 1/100

Hence, x = 0.01

4. Find the logarithm of:

(i) 100 to the base 10

(ii) 0.1 to the base 10

(iii) 0.001 to the base 10

(iv) 32 to the base 4

(v) 0.125 to the base 2

(vi)Â 1/16Â to the base 4

(vii) 27 to the base 9

(viii)Â 1/81Â to the base 27

Solution;

(i) Let log10 100 = x

So, 10x = 100

10x = 102

Then,

x = 2 [If am = an; then m = n]

Hence, log10 100 = 2

(ii) Let log10 0.1 = x

So, 10x = 0.1

10x = 1/10

10x = 10-1

Then,

x = -1 [If am = an; then m = n]

Hence, log10 0.1 = -1

(iii) Let log10 0.001 = x

So, 10x = 0.001

10x = 1/1000

10x = 1/103

10x = 10-3

Then,

x = -3 [If am = an; then m = n]

Hence, log10 0.001 = -3

(iv) Let log4 32 = x

So, 4x = 32

4x = 2 x 2 x 2 x 2 x 2

(22)x = 25

22x = 25

Then,

2x = 5 [If am = an; then m = n]

x = 5/2

Hence, log4 32 = 5/2

(v) Let log2 0.125 = x

So, 2x = 0.125

2x = 125/1000

2x = 1/8

2x = (Â½)3

2x = 2-3

Then,

x = -3 [If am = an; then m = n]

Hence, log2 0.125 = -3

(vi) Let log4 1/16 = x

So, 4x = 1/16

4x = (Â¼)-2

4x = 4-2

Then,

x = -2 [If am = an; then m = n]

Hence, log4 1/16 = -2

(vii) Let log9 27 = x

So, 9x = 27

9x = 3 x 3 x 3

(32)x = 33

32x = 33

Then,

2x = 3 [If am = an; then m = n]

x = 3/2

Hence, log9 27 = 3/2

(viii) Let log27 1/81 = x

So, 27x = 1/81

27x = 1/92

(33)x = 1/(32)2

33x = 1/34

33x = 3-4

Then,

3x = -4 [If am = an; then m = n]

x = -4/3

Hence, log27 1/81 = -4/3

5. State, true or false:

(i) If log10Â x = a, then 10xÂ = a

(ii) IfÂ xyÂ = z, then y =Â logz x

(iii)Â log2Â 8 = 3 and log8 2 =Â 1/3

Solution:

(i) We have,

log10Â x = a

So, 10a = x

Thus, the statement 10x = a is false

(ii) We have,

xyÂ = z

So, logx z = y

Thus, the statement y =Â logz x is false

(iii) We have,

log2Â 8 = 3

So, 23 = 8 â€¦ (1)

Now consider the equation,

log8 2 = 1/3

81/3 = 2

(23)1/3 = 2 â€¦ (2)

Both equations (1) and (2) are correct

Thus, the given statements, log2Â 8 = 3 and log8 2 =Â 1/3 are true

6. Find x, if:

(i) log3Â x = 0

(ii)Â logxÂ 2 = -1

(iii) log9243 = x

(iv) log5Â (x – 7) = 1

(v) log432 = x – 4

(vi) log7Â (2x2Â – 1) = 2

Solution:

(i) We have, log3Â x = 0

So, 30 = x

1 = x

Hence, x = 1

(ii)Â we have, logxÂ 2 = -1

So, x-1 = 2

1/x = 2

Hence, x = Â½

(iii) We have, log9 243 = x

9x = 243

(32)x = 35

32x = 35

On comparing the exponents, we get

2x = 5

x = 5/2 = 2Â½

(iv) We have, log5Â (x – 7) = 1

So, 51 = x â€“ 7

5 = x â€“ 7

x = 5 + 7

Hence, x = 12

(v) We have, log4 32 = x â€“ 4

So, 4(x – 4) = 32

(22)(x – 4) = 25

2(2x – 8) = 25

On comparing the exponents, we get

2x â€“ 8 = 5

2x = 5 + 8

Hence,

x = 13/2 = 6Â½

(vi) We have, log7Â (2x2Â – 1) = 2

So, (2x2 – 1) = 72

2x2 – 1 = 49

2x2 = 49 + 1

2x2 = 50

x2 = 25

Taking square root on both side, we get

x = Â±5

Hence, x = 5 (Neglecting the negative value)

Â

7. Evaluate:

(i) log10Â 0.01

(ii) log2Â (1 Ã· 8)

(iii) log5Â 1

(iv) log5Â 125

(v) log16Â 8

(vi) log0.5Â 16

Solution:

(i) Let log10Â 0.01 = x

Then, 10x = 0.01

10x = 1/100 = 1/102

So, 10x = 10-2

On comparing the exponents, we get

x = -2

Hence, log10Â 0.01 = -2

(ii) Let log2Â (1 Ã· 8) = x

Then, 2x = 1/8

2x = 1/23

So, 2x = 2-3

On comparing the exponents, we get

x = -3

Hence, log10Â (1 Ã· 8) = -3

(iii) Let log5Â 1 = x

Then, 5x = 1

5x = 50

On comparing the exponents, we get

x = 0

Hence, log5Â 1 = 0

(iv) Let log5Â 125 = x

Then, 5x = 125

5x = (5 x 5 x 5) = 53

So, 5x = 53

On comparing the exponents, we get

x = 3

Hence, log5Â 125 = 3

(v) Let log16Â 8 = x

Then, 16x = 8

(24)x = (2 x 2 x 2) = 23

So, 24x = 23

On comparing the exponents, we get

4x = 3

x = 3/4

Hence, log16Â 8 = 3/4

(vi) Let log0.5Â 16 = x

Then, 0.5x = 16

(5/10)x = (2 x 2 x 2 x 2)

(1/2)x = 24

So, 2-x = 24

On comparing the exponents, we get

-x = 4

â‡’ x = -4

Hence, log0.5Â 16 = -4

Â

8. If logaÂ m = n, express an – 1Â in terms in terms of a and m.

Solution:

We have, logaÂ m = n

So,

an = m

Dividing by a on both sides, we get

an/a = m/a

an-1 = m/aÂ

9. Given log2 x = m and log5 y = n

(i) Express 2m-3 in terms of x

(ii) Express 53n+2 in terms of yÂ

Solution:

Given, log2 x = m and log5 y = n

So,

2m = x and 5n = y

(i) Taking, 2m = x

2m/23 = x/23

2m-3 = x/8

(ii) Taking, 5n = y

Cubing on both sides, we have

(5n)3 = y3

53n = y3

Multiplying by 52 on both sides, we have

53n x 52 = y3 x 52

53n+2 = 25y3Â

10. If log2 x = a and log3 y = a, write 72a in terms of x and y.Â

Solution:

Given, log2 x = a and log3 y = a

So,

2a = x and 3a = y

Now, the prime factorization of 72 is

72 = 2 x 2 x 2 x 3 x 3 = 23 x 32

Hence,

(72)a = (23 x 32)a

= 23a x 32a

= (2a)3 x (3a)2

= x3y2 [As 2a = x and 3a = y]

11. Solve for x: log (x – 1) + log (x + 1) = log2 1Â

Solution:

We have,

log (x – 1) + log (x + 1) = log2 1

log (x – 1) + log (x + 1) = 0

log [(x – 1) (x + 1)] = 0

Then,

(x – 1) (x + 1) = 1 [As log 1 = 0]

x2 â€“ 1 = 1

x2 = 1 + 1

x2 = 2

x = Â±âˆš2

The value -âˆš2 is not a possible, since log of a negative number is not defined.

Hence, x = âˆš2Â

12. If log (x2Â – 21) = 2, show that x =Â Â± 11.

Solution:

Given, log (x2Â – 21) = 2

So,

x2Â – 21 = 102

x2Â – 21 = 100

x2 = 121

Taking square root on both sides, we get

x = Â±11

Exercise 8(B)

1. Express in terms of log 2 and log 3:

(i) log 36Â

(ii) log 144Â

(iii) log 4.5

(iv) logÂ 26/51Â – logÂ 91/119Â

(v) logÂ 75/16 – 2logÂ 5/9Â + logÂ 32/243

Solution:

(i) log 36 = log (2 x 2 x 3 x 3)

= log (22 x 32)

= log 22 + log 32 [Using loga mn = loga m + loga n]

= 2log 2 + 2log 3 [Using loga mn = nloga m]

(ii) log 144 = log (2 x 2 x 2 x 2 x 3 x 3)

= log (24 x 32)

= log 24 + log 32 [Using loga mn = loga m + loga n]

= 4log 2 + 2log 3 [Using loga mn = nloga m]

(iii) log 4.5 = log 45/10

= log (5 x 3 x 3)/ (5 x 2)

= log 32/2

= log 32 – log 2 [Using loga m/n = loga m – loga n]

= 2log 3 – log 2 [Using loga mn = nloga m]

(iv) logÂ 26/51Â – logÂ 91/119Â = logÂ (26/51)/ (91/119)Â  [Using loga m – loga n = loga m/n]

= log [(26/51) x (119/91)]

= log (2 x 13 x 7 x 117)/ (3 x 17 x 7 x 13)

= log 2/3

= log 2 â€“ log 3 [Using loga m/n = loga m – loga n]

(v) logÂ 75/16 – 2logÂ 5/9Â + logÂ 32/243

= logÂ 75/16 – logÂ (5/9)2Â + logÂ 32/243 [Using nloga m = loga mn]

= logÂ 75/16 – logÂ 25/81Â + logÂ 32/243

= logÂ [(75/16)/ (25/81)]Â + logÂ 32/243 [Using loga m – loga n = loga m/n]

= log (75 x 81)/ (16 x 25) + log 32/243

= log (3 x 81)/16 + log 32/243

= log 243/16 + log 32/243

= log (243/16) x (32/243) [Using loga m + loga n = loga mn]

= log 32/16

= log 2

2. Express each of the following in a form free from logarithm:

(i) 2 log x – log y = 1

(ii) 2 log x + 3 log y = log a

(iii) a log x – b log y = 2 log 3

Solution:

(i) We have, 2 log x – log y = 1

Then,

log x2 – log y = 1 [Using nloga m = loga mn]

log x2/y = 1 [Using loga m – loga n = loga m/n]

Now, on removing log we have

x2/y = 101

â‡’ x2 = 10y

(ii) We have, 2 log x + 3 log y = log a

Then,

log x2 + log y3 = log a [Using nloga m = loga mn]

log x2y3 = log a [Using loga m + loga n = loga mn]

Now, on removing log we have

x2y3 = a

(iii) a log x – b log y = 2 log 3

Then,

log xa – log yb = log 32 [Using nloga m = loga mn]

log xa/yb = log 32 [Using loga m – loga n = loga m/n]

Now, on removing log we have

xa/yb = 32

â‡’ x2 = 9yb

3. Evaluate each of the following without using tables:

(i) log 5 + log 8 – 2log 2

(ii) log10 8 + log10 25 + 2log10 3 – log10 18

(iii) log 4 +Â 1/3 log 125 â€“Â 1/5 log 32

Solution:

(i)Â We have, log 5 + log 8 – 2log 2

= log 5 + log 8 – log 22 [Using nloga m = loga mn]

= log 5 + log 8 – log 4

= log (5 x 8) – log 4 [Using loga m + loga n = loga mn]

= log 40 – log 4

= log 40/4 [Using loga m – loga n = loga m/n]

= log 10

= 1Â

(ii) We have, log10 8 + log10 25 + 2log10 3 – log10 18

= log10 8 + log10 25 + log10 32 – log10 18 [Using nloga m = loga mn]

= log10 8 + log10 25 + log10 9 – log10 18

= log10 (8 x 25 x 9) – log10 18 [Using loga l + loga m + loga n = loga lmn]

= log10 1800 â€“ log10 18

= log10 1800/18 [Using loga m – loga n = loga m/n]

= log10 100

= log10 102

= 2log10 10 [Using loga mn = nloga m]

= 2 x 1

= 2

Â

(iii)Â We have, log 4 +Â 1/3log 125 â€“Â 1/5log 32

= log 4 +Â log (125)1/3 â€“Â log (32)1/5 [Using nloga m = loga mn]

= log 4 + log (53)1/3 â€“ log (25)1/5

= log 4 + log 5 â€“ log 2

= log (4 x 5) â€“ log 2 [Using loga m + loga n = loga mn]

= log 20 â€“ log 2

= log 20/2 [Using loga m – loga n = loga m/n]

= log 10

= 1

4. Prove that:

2log 15/18 â€“ log 25/162 + log 4/9 = log 2

Solution:

Taking L.H.S.,

= 2log 15/18 â€“ log 25/162 + log 4/9

= log (15/18)2 â€“ log 25/162 + log 4/9 [Using nloga m = loga mn]

= log 225/324 â€“ log 25/162 + log 4/9

= log [(225/324)/(25/162)] + log 4/9 [Using loga m – loga n = loga m/n]

= log (225 x 162)/(324 x 25) + log 4/9

= log (9 x 1)/(2 x 1) + log 4/9

= log 9/2 + log 4/9 [Using loga m + loga n = loga mn]

= log (9/2 x 4/9)

= log 2

= R.H.S.

Â

5. Find x, if:

x –Â logÂ 48 + 3 log 2 =Â 1/3 log 125 – log 3.

Solution:

We have,

x –Â logÂ 48 + 3 log 2 =Â 1/3 log 125 – log 3

Solving for x, we have

x = logÂ 48 â€“ 3 log 2 +Â 1/3 log 125 – log 3

= log 48 â€“ log 23 + log 1251/3 â€“ log 3 [Using nloga m = loga mn]

= log 48 â€“ log 8 + log (53)1/3 â€“ log 3

= (log 48 â€“ log 8) + (log 5 â€“ log 3)

= log 48/8 + log 5/3 [Using loga m – loga n = loga m/n]

= log (48/8 x 5/3) [Using loga m + loga n = loga mn]

= log (2 x 5)

= log 10

= 1

Hence, x = 1

Â

6. Express log10 2 + 1 in the form of log10 x.

Solution:

Given, log10 2 + 1

= log10 2 + log10 10 [As, log10 10 = 1]

= log10 (2 x 10) [Using loga m + loga n = loga mn]

= log10 20

7. Solve for x:

(i) log10Â (x – 10) = 1

(ii) log (x2Â – 21) = 2

(iii) log (x – 2) + log (x + 2) = log 5

(iv) log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3

Solution:

(i)Â We have, log10Â (x – 10) = 1

Then,

x â€“ 10 = 101

x = 10 + 10

Hence, x = 20

(ii) We have, log (x2Â – 21) = 2

Then,

x2 â€“ 21 = 102

x2 â€“ 21 = 100

x2 = 100 + 21

x2 = 121

Taking square root on both sides,

Hence, x = Â±11

(iii) We have, log (x – 2) + log (x + 2) = log 5

Then,

log (x – 2)(x + 2) = log 5 [Using loga m + loga n = loga mn]

log (x2 – 22) = log 5 [As (x – a)(x + a) = x2 â€“ a2]

log (x2 – 4) = log 5

Removing log on both sides, we get

x2 â€“ 4 = 5

x2 = 5 + 4

x2 = 9

Taking square root on both sides,

x = Â±3

(iv) We have, log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3

Then,

log (x + 5) + log (x – 5) = log 24 + log 32 [Using nloga m = loga mn]

log (x + 5)(x – 5) = log 16 + log 9 [Using loga m + loga n = loga mn]

log (x2 – 52) = log (16 x 9) [As (x – a)(x + a) = x2 â€“ a2]

log (x2 â€“ 25) = log 144

Removing log on both sides, we have

x2 â€“ 25 = 144

x2 = 144 + 25

x2 = 169

Taking square root on both sides, we get

x = Â±13

8. Solve for x:

(i)Â log 81/log 27 = x

(ii)Â log 128/log 32 = x

(iii)Â log 64/log 8 = log x

(iv)Â log 225/log 15 = log x

Solution:

(i)Â We have, log 81/log 27 = x

x = log 81/log 27

= log (3 x 3 x 3 x 3)/ log (3 x 3 x 3)

= log 34/log 33

= (4log 3)/(3log 3) [Using loga mn = nloga m]

= 4/3

Hence, x = 4/3

(ii)Â We have, log 128/log 32 = x

x = log 128/log 32

= log (2 x 2 x 2 x 2 x 2 x 2 x 2)/ log (2 x 2 x 2 x 2 x 2)

= log 27/log 25

= (7log 2)/(5log 2) [Using loga mn = nloga m]

= 7/5

Hence, x = 7/5

(iii)Â log 64/log 8 = log x

log x = log 64/log 8

= log (2 x 2 x 2 x 2 x 2 x 2)/ log (2 x 2 x 2)

= log 26/log 23

= (6log 2)/(3log 2) [Using loga mn = nloga m]

= 6/3

= 2

So, log x = 2

Hence, x = 102 = 100

(iv)Â We have, log 225/log 15 = log x

log x = log 225/log 15

= log (15 x 15)/ log 15

= log 152/log 15

= (2log 15)/log 15 [Using loga mn = nloga m]

= 2

So, log x = 2

Hence, x = 102 = 100

9. Given log x = m + n and log y = m – n, express the value of log 10x/y2Â in terms of m and n.

Solution:

Given, log x = m + n and log y = m â€“ n

Now consider log 10x/y2,

log 10x/y2 = log 10x â€“ log y2 [Using loga m/n = loga m – loga n]

= log 10x â€“ 2log y

= log 10 + log x â€“ 2 log y

= 1 + (m + n) â€“ 2 (m – n)

= 1 + m + n â€“ 2m + 2n

= 1 + 3n – m

Â

10. State, true or false:

(i) log 1Â Ã— log 1000 = 0

(ii) log x/log y = log x â€“ log y

(iii) IfÂ log 25/log 5 = log x, then x = 2

(iv) log xÂ Ã— log y = log x + log y

Solution:

(i) We have, log 1Â Ã— log 1000 = 0

Now,

log 1 = 0 and

log 1000 = log 103 = 3log 10 = 3 [Using loga mn = nloga m]

So,

log 1 Ã— log 1000 = 0 x 3 = 0

Thus, the statement log 1Â Ã— log 1000 = 0 is true

(ii) We have, log x/log y = log x â€“ log y

We know that,

log x/y = log x â€“ log y

So,

log x/log y â‰  log x â€“ log y

Thus, the statement log x/log y = log x â€“ log y is false

Â

(iii) We have, log 25/log 5 = log x

log (5 x 5)/log 5 = log x

log 52/ log 5 = log x

2log 5/log 5 = log x [Using loga mn = nloga m]

2 = log x

So, x = 102

x = 100

Thus, the statement x = 2 is false

Â

(iv) We know, log x + log y = log xy

So,

log x + log y â‰  log x Ã— log y

Thus, the statement log x + log y = log x Ã— log y is false

Â

11. If log10 2 = a and log10 3 = b; express each of the following in terms of ‘a’ and ‘b’:

(i) log 12

(ii) log 2.25

(iii) logÂ

Â

(iv) log 5.4

(v) log 60

(iv) logÂ

Â

Solution:

Given that log10 2 = a and log10 3 = b â€¦ (1)

(i) log 12 = log (2 x 2 x 3)

= log (2 x 2) + log 3 [Using loga mn = loga m + loga n]

= log 22 + log 3

= 2log 2 + log 3 [Using loga mn = nloga m]

= 2a + b [From 1]

(ii) log 2.25 = log 225/100

= log (25 x 9)/(25 x 4)

= log 9/4

= log (3/2)2

= 2log 3/2 [Using loga mn = nloga m]

= 2(log 3 â€“ log 2) [Using loga m/n = loga m – loga n]

= 2(b – a) [From 1]

= 2b â€“ 2a

(iii) logÂ  = log 9/4

= log (3/2)2

= 2log 3/2 [Using loga mn = nloga m]

= 2(log 3 â€“ log 2) [Using loga m/n = loga m – loga n]

= 2(b – a) [From 1]

= 2b â€“ 2a

(iv) log 5.4 = log 54/10

= log (2 x 3 x 3 x 3)/10

= log (2 x 33) â€“ log 10 [Using loga m/n = loga m – loga n]

= log 2 + log 33 â€“ 1 [Using loga mn = loga m + loga n and log 10 = 1]

= log 2 + 3log 3 â€“ 1 [Using loga mn = nloga m]

= a + 3b â€“ 1 [From 1]

Â

(v) log 60 = log (10 x 3 x 2)

= log 10 + log 3 + log 2 [Using loga lmn = loga l + loga m + log10 n]

= 1 + b + a [From 1]

(vi) log = log 25/8

= log 52/23

= log 52 â€“ log 23 [Using loga m/n = loga m – loga n]

= 2log 5 â€“ 3log 2 [Using loga mn = nloga m]

= 2log 10/2 â€“ 3log 2

= 2(log 10 â€“ log 2) â€“ 3log 2 [Using loga m/n = loga m – loga n]

= 2log 10 â€“ 2log 2 â€“ 3log 2

= 2(1) â€“ 2a â€“ 3a [From 1]

= 2 â€“ 5a

Â

12. If log 2 = 0.3010 and log 3 = 0.4771; find the value of:

(i) log 12

(ii) log 1.2

(iii) log 3.6

(iv) log 15

(v) log 25

(vi)Â 2/3 log 8

Solution:

Given, log 2 = 0.3010 and log 3 = 0.4771

(i) log 12 = log (4 x 3)

= log 4 + log 3 [Using loga mn = loga m + loga n]

= log 22 + log 3

= 2log 2 + log 3 [Using loga mn = nloga m]

= 2 x 0.3010 + 0.4771

= 1.0791

(ii) log 1.2 = log 12/10

= log 12 â€“ log 10 [Using loga m/n = loga m – loga n]

= log (4 x 3) â€“ log 10

= log 4 + log 3 â€“ log 10 [Using loga mn = loga m + loga n]

= log 22 + log 3 â€“ log 10

= 2log 2 + log 3 â€“ log 10 [Using loga mn = nloga m]

= 2 x 0.3010 + 0.4771 â€“ 1 [As log 10 = 1]

= 0.6020 + 0.4771 â€“ 1

= 1.0791 â€“ 1

= 0.0791

(iii) log 3.6 = log 36/10

= log 36 â€“ log 10 [Using loga m/n = loga m – loga n]

= log (2 x 2 x 3 x 3) â€“ 1 [As log 10 = 1]

= log (22 x 32) â€“ 1

= log 22 + log 32 â€“ 1 [Using loga mn = loga m + loga n]

= 2log 2 + 2log 3 â€“ 1 [Using loga mn = nloga m]

= 2 x 0.3010 + 2 x 0.4771 â€“ 1

= 0.6020 + 0.9542 â€“ 1

= 1.5562 â€“ 1

= 0.5562

(iv) log 15 = log (15/10 x 10)

= log 15/10 + log 10 [Using loga mn = loga m + loga n]

= log 3/2 + 1 [As log 10 = 1]

= log 3 â€“ log 2 + 1 [Using loga m/n = loga m – loga n]

= 0.4771 – 0.3010 + 1

= 1.1761

(v) log 25 = log (25/4 x 4)

= log 100/4

= log 100 â€“ log 4 [Using loga m/n = loga m – loga n]

= log 102 â€“ log 22

= 2log 10 â€“ 2log 2 [Using loga mn = nloga m]

= (2 x 1) â€“ (2 x 0.3010)

= 2 â€“ 0.6020

= 1.398

13. Given 2 log10Â x + 1 = log10Â 250, find:

(i) x

(ii) log10Â 2x

Solution:

(i) Given equation, 2log10Â x + 1 = log10Â 250

log10Â x2 + log10 10 = log10Â 250 [Using nloga m = loga mn and log10 10 = 1]

log10 10x2 = log10 250 [Using loga m + loga n = loga mn]

Removing log on both sides, we have

10x2 = 250

x2 = 25

x = Â±5

As x cannot be a negative value, x = -5 is not possible

Hence, x = 5

(ii) Now, from (i) we have x = 5

So,

log10 2x = log10 2(5)

= log10 10

= 1

Â

14. Given 3log x + Â½ log y = 2, express y in term of x.

Solution:

We have, 3log x + Â½ log y = 2

log x3 + log y1/2 = 2 [Using loga m + loga n = loga mn]

log x3y1/2 = 2

Removing logarithm, we get

x3y1/2 = 102

y1/2 = 100/x3

On squaring on both sides, we get

y = 10000/x6

y = 10000x-6

Â

15. If x = (100)a, y = (10000)b and z = (10)c, find log 10âˆšy/x2z3 in terms of a, b and c.

Solution:

We have,

x = (100)a, y = (10000)b and z = (10)c

So,

log x = a log 100, log y = b log 10000 and log z = c log 10

â‡’ log x = a log 102, log y = b log 104 and log z = c log 10

â‡’ log x = 2a log 10, log y = 4b log 10 and log z = c log 10

â‡’ log x = 2a, log y = 4b and log z = c â€¦ (i)

Now,

log 10âˆšy/x2z3 = log 10âˆšy â€“ log x2z3 [Using loga m/n = loga m – loga n]

= (log 10 + log âˆšy) â€“ (log x2 + log z3) [Using loga mn = loga m + loga n]

= 1 + log y1/2 â€“ log x2 â€“ log z3

= 1 + Â½ log y â€“ 2 log x â€“ 3 log z [Using loga mn = nloga m]

= 1 + Â½(4b) â€“ 2(2a) â€“ 3c â€¦ [Using (i)]

= 1 + 2b â€“ 4a â€“ 3c

Â

16. If 3(log 5 â€“ log 3) â€“ (log 5 â€“ 2 log 6) = 2 â€“ log x, find x.

Solution:

We have,

3(log 5 â€“ log 3) â€“ (log 5 â€“ 2 log 6) = 2 â€“ log x

3log 5 â€“ 3log 3 â€“ log 5 + 2log 6 = 2 â€“ log x

3log 5 â€“ 3log 3 â€“ log 5 + 2log (3 x 2) = 2 â€“ log x

2log 5 â€“ 3log 3 + 2(log 3 + log 2) = 2 â€“ log x [Using loga mn = loga m + loga n]

2log 5 â€“ log 3 + 2log 3 + 2log 2 = 2 â€“ log x

2log 5 – log 3 + 2log 2 = 2 â€“ log x

2log 5 – log 3 + 2log 2 + log x = 2

log 52 – log 3 + log 22 + log x = 2 [Using nloga m = loga mn]

log 25 – log 3 + log 4 + log x = 2

log (25 Ã— 4 Ã— x)/3 = 2 [Using loga m + loga n = loga mn & loga m – loga n = loga m/n]

log 100x/3 = 2

On removing logarithm,

100x/3 = 102

100x/3 = 100

Dividing by 100 on both sides, we have

x/3 = 1

Hence, x = 3

Exercise 8(C)

1. If log10Â 8 = 0.90; find the value of:

(i) log10Â 4

(ii) logÂ âˆš32

(iii) log 0.125

Solution:

Given, log10Â 8 = 0.90

log10 (2 x 2 x 2) = 0.90

log10 23 = 0.90

3 log10 2 = 0.90

log10 2 = 0.90/3

log10 2 = 0.30 â€¦ (1)

Â

(i) log 4 = log10 (2 x 2)

= log10 22

= 2 log10 2

= 2 x 0.60 â€¦ [From (1)]

= 1.20

(ii) log âˆš32 = log10 321/2

= Â½ log10 (2 x 2 x 2 x 2 x 2)

= Â½ log10 25

= Â½ x 5 log10 2

= Â½ x 5 x 0.30 â€¦ [From (1)]

= 0.75

(iii) log 0.125 = log 125/1000

= log10 1/8

= log10 1/23

= log10 2-3

= -3 log10 2

= -3 x 0.30 â€¦ [From (1)]

= -0.90

2. If log 27 = 1.431, find the value of:

(i) log 9 (ii) log 300

Solution:

Given, log 27 = 1.431

So, log 33 = 1.431

3log 3 = 1.431

log 3 = 1.431/3

= 0.477 â€¦ (1)

(i) log 9 = log 32

= 2log 3

= 2 x 0.477 â€¦ [From (1)]

= 0.954

(ii) log 300 = log (3 x 100)

= log 3 + log 100

= log 3 + log 102

= log 3 + 2log 10

= log 3 + 2 [As log 10 = 1]

= 0.477 + 2

= 2.477

Â

3. If log10Â a = b, find 103b – 2Â in terms of a.

Solution:

Given, log10 a = b

Now,

Let 103b – 2Â = x

Applying log on both sides,

log 103b â€“ 2 = log x

(3b â€“ 2)log 10 = log x

3b â€“ 2 = log x

3log10 a â€“ 2 = log x

3log10 a â€“ 2log 10 = log x

log10 a3 â€“ log 102 = log x

log10 a3 â€“ log 100 = log x

log10 a3/100 = log x

On removing logarithm, we get

a3/100 = x

Hence, 103b – 2 = a3/100

4. If log5Â x = y, find 52y+ 3Â in terms of x.

Solution:

Given, log5Â x = y

So, 5y = x

Squaring on both sides, we get

(5y)2 = x2

52y = x2

52y Ã— 53 = x2 Ã— 53

Hence,

52y+3 = 125x2

5. Given: log3Â m = x and log3Â n = y.

(i) Express 32x – 3Â in terms of m.

(ii) Write down 31 – 2y +Â 3xÂ in terms of m and n.

(iii) If 2 log3Â A = 5x – 3y; find A in terms of m and n.

Â

Solution:

Given, log3Â m = x and log3Â n = y

So, 3x = m and 3y = nÂ â€¦ (1)

(i) Taking the given expression, 32x – 3

32x – 3Â = 32x . 3-3

= 32x . 1/33

= (3x)2/33

= m2/33 â€¦ [Using (1)]

= m2/27

Hence, 32x – 3Â = m2/27

(ii) Taking the given expression, 31 – 2y +Â 3xÂ

31 – 2y +Â 3xÂ = 31 . 3-2y . 33xÂ

= 3 . (3y)-2 . (3x)3Â

= 3 . n-2 . m3 â€¦ [Using (1)]

= 3m3/n2

Hence, 31 – 2y +Â 3xÂ = 3m3/n2

(iii) Taking the given equation,

2 log3Â A = 5x – 3y

log3 A2 = 5x – 3y

log3 A2 = 5log3Â m – 3log3Â n â€¦ [Using (1)]

log3 A2 = log3Â m5 – log3Â n3

log3 A2 = log3Â m5/n3

Removing logarithm on both sides, we get

A2 = m5/n3

Hence, by taking square root on both sides

A = âˆš(m5/n3) = m5/2/n3/2

Â

6. Simplify:

(i) log (a)3Â – log a

(ii) log (a)3Â Ã·Â log a

Solution:

(i) We have, log (a)3Â – log a

= 3log a – log a

= 2log a

(ii) We have, log (a)3Â Ã·Â log a

= 3log a/ log a

= 3

7. If log (a + b) = log a + log b, find a in terms of b.

Solution:

We have, log (a + b) = log a + log b

Then,

log (a + b) = log ab

So, on removing logarithm, we have

a + b = ab

a â€“ ab = -b

a(1 – b) = -b

a = -b/(1 – b)

Hence,

a = b/(b – 1)

8. Prove that:

(i) (log a)2Â – (log b)2Â = log (a/b). log (ab)

(ii) If a log b + b log a – 1 = 0, thenÂ ba.Â abÂ = 10

Solution:

(i) Taking L.H.S. we have,

= (log a)2Â – (log b)2

= (log aÂ + log b)Â (log aÂ – log b) [As x2 â€“ y2 = (x + y)(x – y)]

= (log ab). (log a/b)

= R.H.S.

– Hence proved

(ii) We have, a log b + b log a – 1 = 0

So,

log ba + log ab – 1 = 0

log ba + log ab = 1

log baab = 1

On removing logarithm, we get

baab = 10

– Hence proved

Â

Â

9. (i) If log (a + 1) = log (4a – 3) – log 3; find a.

(ii) If 2 log y – log x – 3 = 0, express x in terms of y.

(iii) Prove that: log10Â 125 = 3(1 – log102).

Solution:

(i) Given, log (a + 1) = log (4a – 3) – log 3

So,

log (a + 1) = log (4a – 3)/3

On removing logarithm on both sides, we have

a + 1 = (4a – 3)/3

3(a + 1) = 4a â€“ 3

3a + 3 = 4a â€“ 3

Hence, a = 6

(ii) Given, 2log y – log x – 3 = 0

So,

log y2 â€“ log x = 3

log y2/x = 3

On removing logarithm, we have

y2/x = 103 = 1000

Hence, x = y2/1000

(iii) Considering the L.H.S., we have

log10Â 125 = log10Â (5 x 5 x 5)

= log10Â 53

= 3log10Â 5

= 3log10Â 10/2

= 3(log10Â 10 – log10Â 2)

= 3(1 – log10Â 2) [Since, log10 10 = 1]

= R.H.S.Â

– Hence proved

Â

10. Given log x = 2m â€“ n, log y = n â€“ 2m and log z = 3m â€“ 2n. Find in terms of m and n, the value of log x2y3/z4.

Solution:

We have, log x = 2m â€“ n, log y = n â€“ 2m and log z = 3m â€“ 2n

Now, considering

log x2y3/z4 = log x2y3 â€“ log z4

= (log x2 + log y3) â€“ log z4

= 2log x + 3log y â€“ 4log z

= 2(2m – n) + 3(n â€“ 2m) â€“ 4(3m â€“ 2n)

= 4m â€“ 2n + 3n â€“ 6m â€“ 12m + 8n

= -14m + 9n

11. Given logx 25 â€“ logx 5 = 2 â€“ logx 1/125; find x.

Solution:

We have, logx 25 â€“ logx 5 = 2 â€“ logx 1/125

logx (5 x 5) â€“ logx 5 = 2 â€“ logx 1/(5 x 5 x 5)

logx 52 â€“ logx 5 = 2 â€“ logx 1/53

2logx 5 â€“ logx 5 = 2 â€“ logx 1/53

logx 5 = 2 â€“ 3logx 1/5

logx 5 = 2 + 3logx (1/5)-1

logx 5 = 2 + 3logx 5

2 = logx 5 – 3logx 5

2 = -2logx 5

-1 = logx 5

Removing logarithm, we get

x-1 = 5

Hence, x = 1/5

Â

Exercise 8(D)

1. IfÂ 3/2 log a +Â 2/3 log b – 1 = 0, find the value of a9.b4

Solution:

Given equation,

3/2 log a +Â 2/3 log b – 1 = 0

log a3/2 +Â log b2/3 – 1 = 0

log a3/2 Ã— b2/3 – 1 = 0

log a3/2.b2/3 = 1

Removing logarithm, we have

a3/2.b2/3 = 10

On manipulating,

(a3/2.b2/3)6 = 106

Hence,

a9.b4 = 106

2. If x = 1 + log 2 – log 5, y = 2log3 and z = log a – log 5; find the value ofÂ aÂ if x + y = 2z.

Solution:

Given, x = 1 + log 2 – log 5, y = 2log3 and z = log a – log 5

Now, considering the given equation x + y = 2z

(1 + log 2 â€“ log 5) + (2log 3) = 2(log a – log 5)

1 + log 2 â€“ log 5 + 2log 3 = 2log a â€“ 2log 5

1 + log 2 â€“ log 5 + 2log 3 + 2log 5 = 2log a

log 10 + log 2 + log 32 + log 5 = log a2

log 10 + log 2 + log 9 + log 5 = log a2

log (10 x 2 x 9 x 5) = log a2

log 900 = log a2

On removing logarithm on both sides, we have

900 = a2

Taking square root, we get

a = Â±30

Since, a cannot be a negative value,

Hence, a = 30

3. If x = log 0.6; y = log 1.25 and z = log 3 – 2log 2, find the values of:

(i) x + y – zÂ  Â  Â  Â  (ii) 5x + y – z

Solution:

Given,

x = log 0.6, y = log 1.25 and z = log 3 – 2log 2

So, z = log 3 – log 22

= log 3 â€“ log 4

= log Â¾

= log 0.75 â€¦ (1)

(i) Considering,

x + y – zÂ = log 0.6 + log 1.25 â€“ log 0.75 â€¦ [From (1)]

= log (0.6 x 1.25)/0.75

= log 0.75/0.75

= log 1

= 0 â€¦ (2)

Â

(ii) Now, considering

5x+y-z = 50 â€¦ [From (2)]

= 1

Â

4. If a2Â = log x, b3Â = log y and 3a2Â – 2b3Â = 6 log z, express y in terms of x and z.

Solution:

We have, a2Â = log x and b3Â = log y

Now, considering the equation

3a2 â€“ 2b3 = 6log z

3log x â€“ 2log y = 6log z

log x3 â€“ log y2 = log z6

log x3/y2 = log z6

On removing logarithm on both sides, we get

x3/y2 = z6

So,

y2 = x3/z6

Taking square root on both sides, we get

y = âˆš(x3/z6)

Hence, y = x3/2/z3

Â

5. If log (a – b)/2 = Â½ (log a + log b), show that: a2Â + b2Â = 6ab.

Solution:

We have, log (a – b)/2 = Â½ (log a + log b)

log (a – b)/2 = Â½ log a + Â½ log b

= log a1/2 + log b1/2

= log âˆša + log âˆšb

= log âˆš(ab)

Now, removing logarithm on both sides, we get

(a – b)/2 = âˆš(ab)

Squaring on both sides, we get

[(a – b)/2]2 = [âˆš(ab)]2

(a – b)2/4 = ab

(a â€“ b)2 = 4ab

a2 + b2 â€“ 2ab = 4ab

a2 + b2 = 4ab + 2ab

a2 + b2 = 6ab

– Hence proved

6. If a2Â + b2Â = 23ab, show that: logÂ (a + b)/5 = Â½ (log a + log b).

Solution:

Given, a2Â + b2Â = 23ab

a2Â + b2Â + 2ab = 23ab + 2ab

(a + b)2 = 25ab

(a + b)2/25 = ab

[(a + b)/5]2Â = ab

Taking logarithm on both sides, we have

log [(a + b)/5]2Â = log ab

2log (a + b)/5 = log ab

2log (a + b)/5 = log a + log b

Thus,

log (a + b)/5 = Â½ (log a + log b)

7. If m = log 20 and n = log 25, find the value of x, so that: 2log (x – 4) = 2m – n.

Solution:

Given, m = log 20 and n = log 25

Now, considering the given expression

2log (x – 4) = 2m – n

2log (x – 4) = 2log 20 â€“ log 25

log (x – 4)2 = log 202 â€“ log 25

log (x – 4)2 = log 400 â€“ log 25

log (x – 4)2 = log 400/25

Removing logarithm on both sides,

(x – 4)2 = 400/25

x2 â€“ 8x + 16 = 16

x2 â€“ 8x = 0

x(x – 8) = 0

So,

x = 0 or x = 8

If x = 0, then log (x – 4) doesnâ€™t exist

Hence, x = 8

8. Solve for x and y; if x > 0 and y > 0; logÂ xyÂ = logÂ x/y + 2log 2 = 2.

Solution:

We have,

logÂ xyÂ = logÂ x/y + 2log 2 = 2

Considering the equation,

log xy = 2

log xy = 2log 10

log xy = log 102

log xy = log 100

On removing logarithm,

xy = 100 â€¦ (1)

Now, consider the equation

logÂ x/y + 2log 2 = 2

log x/y + log 22 = 2

log x/y + log 4 = 2

log 4x/y = 2

Removing logarithm, we get

4x/y = 102

4x/y = 100

x/y = 25

(xy)/y2 = 25

100/y2 = 25 â€¦ [From (1)]

y2 = 100/25

y2 = 4

y = 2 [Since, y > 0]

From log xy = 2

Substituting the value of y, we get

log 2x = 2

On removing logarithm,

2x = 102

2x = 100

x = 100/2

x = 50

Thus, the values x and y are 50 and 2 respectively

9. Find x, if:

(i)Â logxÂ 625 = -4Â

(ii)Â logxÂ (5x – 6) = 2

Solution:

(i) We have, logx 625 = -4

On removing logarithm,

x-4 = 625

(1/x)4 = 54

Taking the fourth root on both sides,

1/x = 5

Hence, x = 1/5

Â

(ii) We have, logxÂ (5x – 6) = 2

On removing logarithm,

x2 = 5x â€“ 6

x2 â€“ 5x + 6 = 0

x2 â€“ 3x â€“ 2x + 6 = 0

x(x – 3) â€“ 2(x – 2) = 0

(x – 2)(x – 3) = 0

Hence,

x = 2 or 3

10. If p = log 20 and q = log 25, find the value of x, if 2log (x + 1) = 2p â€“ q.Â

Solution:

Given, p = log 20 and q = log 25

Considering the equation,

2log (x + 1) = 2p â€“ q

log (x + 1)2 = 2p â€“ q

log (x + 1)2 = 2log 20 â€“ log 25

log (x + 1)2 = log 202 â€“ log 25

log (x + 1)2 = log 400 â€“ log 25

log (x + 1)2 = log 400/25

Removing logarithm on both sides, we have

(x + 1)2 = 400/25 = 16

(x + 1)2 = (4)2

Taking square root on both sides, we have

x + 1 = 4

x = 4 -1

Hence, x = 3Â Â

Â

11. If log2 (x + y) = log3 (x – y) = log 25/log 0.2, find the value of x and y.

Solution:

Considering the relation, log2 (x + y) = log 25/log 0.2

log2 (x + y) = log0.2 25

= log2/10 52

= 2log1/5 5

= 2log5-1 5

= -2log5 5

= -2 x 1

= -2

So, we have

log2 (x + y) = -2

Removing logarithm, we get

x + y = 2-2

x + y = 1/22

x + y = Â¼ â€¦ (i)

Now, considering the relation log3 (x – y) = log 25/log 0.2

log3 (x – y) = = log0.2 25

= log2/10 52

= 2log1/5 5

= 2log5-1 5

= -2log5 5

= -2 x 1

= -2

So, we have

log3 (x – y) = -2

Removing logarithm, we get

x â€“ y = 3-2

x â€“ y = 1/32

x â€“ y = 1/9 â€¦ (ii)

On adding (i) and (ii), we get

x + y = Â¼

x â€“ y = 1/9

—————-

2x = Â¼ + 1/9

2x = (9 + 4)/36

2x = 13/36

x = 13/(36 x 2)

= 13/72

Now, substituting the value of x in (i), we get

13/72 + y = Â¼

y = Â¼ – 13/72

= (18 – 13)/72

= 5/72

Hence, the values of x and are is 13/72 and 5/72 respectively

Â

12. Given: log x/log y = 3/2 and log xy = 5; find the values of x and y.Â

Solution:

Given, log x/log y = 3/2 â€¦ (i) and log xy = 5 â€¦ (ii)

So,

log xy = log x + log y = 5

And, we have log y = (2log x)/3 â€¦ [From (i)]

Now,

log x + (2log x)/3 = 5

3log x + 2log x = 5 x 3

5log x = 15

log x = 15/5

log x = 3

Removing logarithm, we get

x = 103 = 1000

Substituting value of x in (ii), we get

log xy = 5

Removing logarithm, we get

xy = 105

(103). y = 105

y = 105/103

y = 102

y = 100

Â

13. Given log10 x = 2a and log10 yÂ =Â b/2

(i) Write 10aÂ in terms of x

(ii) Write 102b + 1Â in terms of yÂ

(iii) If log10 p = 3a â€“ 2b, express p in terms of x and y.

Solution:

Given, log10 x = 2a and log10 yÂ =Â b/2

(i) Taking log10 x = 2a

Removing logarithm on both sides,

x = 102a

Taking square root on both sides, we get

x1/2 = 102a/2

Hence, 10a = x1/2

(ii) Taking log10 yÂ =Â b/2

Removing logarithm on both sides,

y = 10b/2

On manipulating,

y4 = 10b/2 x 4

y4 = 102b

10y4 = 102b x 10

Hence, 102b + 1Â = 10y4

(iii) We have, 10a = x1/2

and y = 10b/2

Considering the equation, log10 p = 3a â€“ 2b

log10 p = 3a â€“ 2b

Removing logarithm, we get

p = 103a â€“ 2b

p = 103a/102b

p = (10a)3/(10b/2)4

p = (x1/2)3/(y)4

Hence, p = x3/2/y4

14. Solve:

log5 (x + 1) – 1 = 1 + log5 (x – 1).

Solution:

Considering the given equation,

log5 (x + 1) – 1 = 1 + log5 (x – 1)

log5 (x + 1) – log5 (x – 1) = 1 + 1

log5 (x + 1)/(x – 1) = 2

Removing logarithm, we have

(x + 1)/(x – 1) = 52

(x + 1)/(x – 1) = 25

(x + 1) = 25(x – 1)

x + 1 = 25x â€“ 25

25x â€“ x = 25 + 1

24x = 26

x = 26/24

Hence, x = 13/12

Â

15. Solve for x, if:

logx 49 â€“ logx 7 + logx 1/343 + 2 = 0Â

Solution:

We have,

logx 49 â€“ logx 7 + logx 1/343 + 2 = 0Â

logx 49/(7 x 343) + 2 = 0

logx 1/49 = -2

logx 1/72 = -2

logx 7-2 = -2

-2logx 7 = -2

So,

logx 7 = 1

Removing logarithm, we get

x = 7

16. If a2 = log x, b3 = log y and a2/2 â€“ b3/3 = log c, find c in terms of x and y. Â

Solution:

Given,

a2 = log x, b3 = log y

Considering the given equation,

a2/2 â€“ b3/3 = log c

(log x)/2 â€“ (log y)/3 = log c

Â½ log x â€“ 1/3 log y = log c

log x1/2 â€“ log y1/3 = log c

log x1/2/y1/3 = log c

On removing logarithm, we get

x1/2/y1/3 = c

Hence, c = x1/2/y1/3 is the required relationÂ

17. Given: x = log10 12, y = log4 2 x log10 9 and z = log10 0.4, find

(i) x â€“ y â€“ z

(ii) 13x – y – zÂ

Solution:

(i) Considering, x â€“ y â€“ z

= log10 12 â€“ (log4 2 x log10 9) – log10 0.4

= log10 12 â€“ (log4 2 x log10 9) – log10 0.4

= log10 (4 x 3) â€“ (log10 2/ log10 4 x log10 9) – log10 0.4

= log10 4 + log10 3 â€“ (log10 2 x log10 32)/ log10 22 â€“ log10 4/10

= log10 4 + log10 3 â€“ (log10 2 x 2log10 3)/ 2log10 2 â€“ (log10 4 â€“ log10 10)

= log10 4 + log10 3 â€“ log10 3 â€“ log10 4 + log10 10

= log10 4 + log10 3 â€“ log10 3 â€“ log10 4 + 1

= 1

(ii) Now,

13x – y – zÂ = 131 = 13Â

18. Solve for x, logx 15âˆš5 = 2 â€“ logx 3âˆš5Â

Solution:

Considering the given equation,

logx 15âˆš5 = 2 â€“ logx 3âˆš5Â

logx 15âˆš5 + logx 3âˆš5Â = 2

logx (15âˆš5 x 3âˆš5)Â = 2

logx (45 x 5)Â = 2

logx 225Â = 2

Removing logarithm, we get

x2 = 225

Taking square root on both sides,

x = 15Â

19. Evaluate:

(i) logb a x logc b x loga c

(ii) log3 8 Ã· log9 16

(iii) log5 8/(log25 16 x log100 10)Â

Solution:

Using logb a = 1/loga b and logx a/logx b = logb a, we have

(i) logb a x logc b x loga c

= 1

(ii) log3 8 Ã· log9 16

= log3 8/ log9 16

= 3 x Â½

= 3/2

(iii) log5 8/(log25 16 x log100 10)

= 3 x Â½ x 2

= 3 Â

20. Show that:

loga m Ã· logab m = 1 + loga bÂ

Solution:

Considering the L.H.S.,

loga m Ã· logab m = loga m/logab m

= logm ab/logm a [As logb a = 1/loga b]

= loga ab [As logx a/logx b = logb a]

= loga a + loga b

= 1 + loga bÂ

21. If logâˆš27 x = 2 2/3, find x.

Solution:

We have,

logâˆš27 x = 2 2/3

logâˆš27 x = 8/3

Removing logarithm, we get

x = âˆš278/3

= 271/2 x 8/3

= 274/3

= 33 x 4/3

= 34

Hence, x = 81

22. Evaluate:

1/(loga bc + 1) + 1/(logb ca + 1) + 1/(logc ab + 1)

Solution:

We have,

## Selina Solutions for Class 9 Maths Chapter 8- Logarithms

The Chapter 8, Logarithms, contains 4 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

8.1 Introduction

8.2 Interchanging

8.3 Laws of logarithms with use

8.4 Expansion of expressions with the help of laws of logarithms