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.

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**Exercise 8(A)**

**1. Express each of the following in logarithmic form:**

**(i) 5 ^{3} = 125**

**(ii) 3 ^{-2} = 1/9**

**(iii) 10 ^{-3} = 0.001**

**(iv) (81) ^{3/4} = 27**

**Solution:**

We know that,

a^{b} = c ⇒ log_{a }c = b

(i) 5^{3} = 125

log_{5} 125 = 3

(ii) 3^{-2} = 1/9

log_{3} 1/9 = -2

(iii) 10^{-3} = 0.001

log_{10} 0.001 = -3

(iv) (81)^{3/4} = 27

log_{81} 27 = ¾

**2. Express each of the following in exponential form:**

**(i) log _{8 }0.125 = -1**

**(ii) log _{10 }0.01 = -2**

**(iii) log _{a }A = x**

**(iv) log _{10 }1 = 0**

**Solution:**

We know that,

log_{a} c = b ⇒ a^{b} = c

(i) log_{8 }0.125 = -1

8^{-1} = 0.125

(ii) log_{10 }0.01 = -2

8^{-1} = 0.125

(iii) log_{a }A = x

a^{x} = A

(iv) log_{10 }1 = 0

10^{0} = 1

**3. Solve for x: log _{10} x = -2.**

**Solution:**

We have,

log_{10} x = -2

10^{-2} = x [As log_{a} c = b ⇒ a^{b} = c]

x = 10^{-2}

x = 1/10^{2}

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) 1/16 to the base 4**

**(vii) 27 to the base 9**

**(viii) 1/81 to the base 27**

**Solution;**

(i) Let log_{10} 100 = x

So, 10^{x} = 100

10^{x} = 10^{2}

Then,

x = 2 [If a^{m} = a^{n}; then m = n]

Hence, log_{10} 100 = 2

(ii) Let log_{10} 0.1 = x

So, 10^{x} = 0.1

10^{x} = 1/10

10^{x} = 10^{-1}

Then,

x = -1 [If a^{m} = a^{n}; then m = n]

Hence, log_{10} 0.1 = -1

(iii) Let log_{10} 0.001 = x

So, 10^{x} = 0.001

10^{x} = 1/1000

10^{x} = 1/10^{3}

10^{x} = 10^{-3}

Then,

x = -3 [If a^{m} = a^{n}; then m = n]

Hence, log_{10} 0.001 = -3

(iv) Let log_{4} 32 = x

So, 4^{x} = 32

4^{x} = 2 x 2 x 2 x 2 x 2

(2^{2})^{x} = 2^{5}

2^{2x} = 2^{5}

Then,

2x = 5 [If a^{m} = a^{n}; then m = n]

x = 5/2

Hence, log_{4} 32 = 5/2

(v) Let log_{2} 0.125 = x

So, 2^{x} = 0.125

2^{x} = 125/1000

2^{x} = 1/8

2^{x} = (½)^{3}

2^{x} = 2^{-3}

Then,

x = -3 [If a^{m} = a^{n}; then m = n]

Hence, log_{2} 0.125 = -3

(vi) Let log_{4} 1/16 = x

So, 4^{x} = 1/16

4^{x} = (¼)^{-2}

4^{x} = 4^{-2}

Then,

x = -2 [If a^{m} = a^{n}; then m = n]

Hence, log_{4} 1/16 = -2

(vii) Let log_{9} 27 = x

So, 9^{x} = 27

9^{x} = 3 x 3 x 3

(3^{2})^{x} = 3^{3}

3^{2x} = 3^{3}

Then,

2x = 3 [If a^{m} = a^{n}; then m = n]

x = 3/2

Hence, log_{9} 27 = 3/2

(viii) Let log_{27} 1/81 = x

So, 27^{x} = 1/81

27^{x} = 1/9^{2}

(3^{3})^{x} = 1/(3^{2})^{2}

3^{3x} = 1/3^{4}

3^{3x} = 3^{-4}

Then,

3x = -4 [If a^{m} = a^{n}; then m = n]

x = -4/3

Hence, log_{27} 1/81 = -4/3

**5. State, true or false:**

**(i) If log _{10} x = a, then 10^{x} = a**

**(ii) If x ^{y} = z, then y = log_{z }x**

**(iii) log _{2} 8 = 3 and log_{8} 2 = 1/3**

**Solution:**

(i) We have,

log_{10} x = a

So, 10^{a} = x

Thus, the statement 10^{x} = a is false

(ii) We have,

x^{y} = z

So, log_{x} z = y

Thus, the statement y = log_{z }x is false

(iii) We have,

log_{2} 8 = 3

So, 2^{3} = 8 … (1)

Now consider the equation,

log_{8} 2 = 1/3

8^{1/3} = 2

(2^{3})^{1/3} = 2 … (2)

Both equations (1) and (2) are correct

Thus, the given statements, log_{2} 8 = 3 and log_{8} 2 = 1/3 are true

**6. Find x, if:**

**(i) log _{3} x = 0**

**(ii) log _{x} 2 = -1**

**(iii) log _{9}243 = x**

**(iv) log _{5} (x – 7) = 1**

**(v) log _{4}32 = x – 4**

**(vi) log _{7} (2x^{2} – 1) = 2**

**Solution:**

(i) We have, log_{3} x = 0

So, 3^{0} = x

1 = x

Hence, x = 1

(ii) we have, log_{x} 2 = -1

So, x^{-1} = 2

1/x = 2

Hence, x = ½

(iii) We have, log_{9 }243 = x

9^{x} = 243

(3^{2})^{x} = 3^{5}

3^{2x} = 3^{5}

On comparing the exponents, we get

2x = 5

x = 5/2 = 2½

(iv) We have, log_{5} (x – 7) = 1

So, 5^{1} = x – 7

5 = x – 7

x = 5 + 7

Hence, x = 12

(v) We have, log_{4 }32 = x – 4

So, 4^{(x – 4)} = 32

(2^{2})^{(x – 4)} = 2^{5}

2^{(2x – 8)} = 2^{5}

On comparing the exponents, we get

2x – 8 = 5

2x = 5 + 8

Hence,

x = 13/2 = 6½

(vi) We have, log_{7} (2x^{2} – 1) = 2

So, (2x^{2} – 1) = 7^{2}

2x^{2} – 1 = 49

2x^{2} = 49 + 1

2x^{2} = 50

x^{2} = 25

Taking square root on both side, we get

x = ±5

Hence, x = 5 (Neglecting the negative value)

**7. Evaluate:**

**(i) log _{10 }0.01**

**(ii) log _{2} (1 ÷ 8)**

**(iii) log _{5} 1**

**(iv) log _{5} 125**

**(v) log _{16} 8**

**(vi) log _{0.5} 16**

**Solution:**

(i) Let log_{10 }0.01 = x

Then, 10^{x} = 0.01

10^{x} = 1/100 = 1/10^{2}

So, 10^{x} = 10^{-2}

On comparing the exponents, we get

x = -2

Hence, log_{10 }0.01 = -2

(ii) Let log_{2} (1 ÷ 8) = x

Then, 2^{x} = 1/8

2^{x} = 1/2^{3}

So, 2^{x} = 2^{-3}

On comparing the exponents, we get

x = -3

Hence, log_{10 }(1 ÷ 8) = -3

(iii) Let log_{5} 1 = x

Then, 5^{x} = 1

5^{x} = 5^{0}

On comparing the exponents, we get

x = 0

Hence, log_{5 }1 = 0

(iv) Let log_{5} 125 = x

Then, 5^{x} = 125

5^{x} = (5 x 5 x 5) = 5^{3}

So, 5^{x} = 5^{3}

On comparing the exponents, we get

x = 3

Hence, log_{5 }125 = 3

(v) Let log_{16} 8 = x

Then, 16^{x} = 8

(2^{4})^{x} = (2 x 2 x 2) = 2^{3}

So, 2^{4x} = 2^{3}

On comparing the exponents, we get

4x = 3

x = 3/4

Hence, log_{16 }8 = 3/4

(vi) Let log_{0.5} 16 = x

Then, 0.5^{x} = 16

(5/10)^{x} = (2 x 2 x 2 x 2)

(1/2)^{x} = 2^{4}

So, 2^{-x} = 2^{4}

On comparing the exponents, we get

-x = 4

⇒ x = -4

Hence, log_{0.5 }16 = -4

**8. If log _{a} m = n, express a^{n – 1 }in terms in terms of a and m.**

**Solution:**

We have, log_{a} m = n

So,

a^{n} = m

Dividing by a on both sides, we get

a^{n}/a = m/a

a^{n-1} = m/a

**9. Given log _{2} x = m and log_{5} y = n**

**(i) Express 2 ^{m-3} in terms of x**

**(ii) Express 5 ^{3n+2} in terms of y **

**Solution:**

Given, log_{2} x = m and log_{5} y = n

So,

2^{m} = x and 5^{n} = y

(i) Taking, 2^{m} = x

2^{m}/2^{3} = x/2^{3}

2^{m-3} = x/8

(ii) Taking, 5^{n} = y

Cubing on both sides, we have

(5^{n})^{3} = y^{3}

5^{3n} = y^{3}

Multiplying by 5^{2} on both sides, we have

5^{3n} x 5^{2} = y^{3} x 5^{2}

5^{3n+2} = 25y^{3}

**10. If log _{2} x = a and log_{3} y = a, write 72^{a} in terms of x and y. **

**Solution:**

Given, log_{2} x = a and log_{3} y = a

So,

2^{a} = x and 3^{a} = y

Now, the prime factorization of 72 is

72 = 2 x 2 x 2 x 3 x 3 = 2^{3} x 3^{2}

Hence,

(72)^{a} = (2^{3} x 3^{2})^{a}

= 2^{3a} x 3^{2a}

= (2^{a})^{3} x (3^{a})^{2}

= x^{3}y^{2} [As 2^{a} = x and 3^{a} = y]

**11. Solve for x: log (x – 1) + log (x + 1) = log _{2} 1 **

**Solution:**

We have,

log (x – 1) + log (x + 1) = log_{2} 1

log (x – 1) + log (x + 1) = 0

log [(x – 1) (x + 1)] = 0

Then,

(x – 1) (x + 1) = 1 [As log 1 = 0]

x^{2} – 1 = 1

x^{2} = 1 + 1

x^{2} = 2

x = ±√2

The value -√2 is not a possible, since log of a negative number is not defined.

Hence, x = √2

**12. If log (x ^{2} – 21) = 2, show that x = ± 11.**

**Solution:**

Given, log (x^{2} – 21) = 2

So,

x^{2} – 21 = 10^{2}

x^{2} – 21 = 100

x^{2} = 121

Taking square root on both sides, we get

x = ±11

**Exercise 8(B)**

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

**(i) log 36 **

**(ii) log 144 **

**(iii) log 4.5**

**(iv) log 26/51 – log 91/119 **

**(v) log 75/16 – 2log 5/9 + log 32/243**

**Solution:**

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

= log (2^{2} x 3^{2})

= log 2^{2} + log 3^{2} [Using log_{a} mn = log_{a} m + log_{a} n]

= 2log 2 + 2log 3 [Using log_{a} m^{n} = nlog_{a} m]

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

= log (2^{4} x 3^{2})

= log 2^{4} + log 3^{2} [Using log_{a} mn = log_{a} m + log_{a} n]

= 4log 2 + 2log 3 [Using log_{a} m^{n} = nlog_{a} m]

(iii) log 4.5 = log 45/10

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

= log 3^{2}/2

= log 3^{2} – log 2 [Using log_{a} m/n = log_{a} m – log_{a} n]

= 2log 3 – log 2 [Using log_{a} m^{n} = nlog_{a} m]

(iv)** **log 26/51 – log 91/119 = log (26/51)/ (91/119) [Using log_{a} m – log_{a} n = log_{a} 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 – log 3 [Using log_{a} m/n = log_{a} m – log_{a} n]

(v) log 75/16 – 2log 5/9 + log 32/243

= log 75/16 – log (5/9)^{2} + log 32/243 [Using nlog_{a} m = log_{a} m^{n}]

= log 75/16 – log 25/81 + log 32/243

= log [(75/16)/ (25/81)] + log 32/243 [Using log_{a} m – log_{a} n = log_{a} 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 log_{a} m + log_{a} n = log_{a} 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 x^{2} – log y = 1 [Using nlog_{a} m = log_{a} m^{n}]

log x^{2}/y = 1 [Using log_{a} m – log_{a} n = log_{a} m/n]

Now, on removing log we have

x^{2}/y = 10^{1}

⇒ x^{2} = 10y

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

Then,

log x^{2} + log y^{3} = log a [Using nlog_{a} m = log_{a} m^{n}]

log x^{2}y^{3} = log a [Using log_{a} m + log_{a} n = log_{a} mn]

Now, on removing log we have

x^{2}y^{3} = a

(iii) a log x – b log y = 2 log 3

Then,

log x^{a} – log y^{b} = log 3^{2} [Using nlog_{a} m = log_{a} m^{n}]

log x^{a}/y^{b} = log 3^{2} [Using log_{a} m – log_{a} n = log_{a} m/n]

Now, on removing log we have

x^{a}/y^{b} = 3^{2}

⇒ x^{2} = 9y^{b}

**3. Evaluate each of the following without using tables:**

**(i) log 5 + log 8 – 2log 2**

**(ii) log _{10 }8 + log_{10 }25 + 2log_{10 }3 – log_{10 }18**

**(iii) log 4 + 1/3 log 125 – 1/5 log 32**

**Solution:**

(i) We have, log 5 + log 8 – 2log 2

= log 5 + log 8 – log 2^{2} [Using nlog_{a} m = log_{a} m^{n}]

= log 5 + log 8 – log 4

= log (5 x 8) – log 4 [Using log_{a} m + log_{a} n = log_{a} mn]

= log 40 – log 4

= log 40/4 [Using log_{a} m – log_{a} n = log_{a} m/n]

= log 10

= 1

(ii) We have, log_{10 }8 + log_{10 }25 + 2log_{10 }3 – log_{10 }18

= log_{10 }8 + log_{10 }25 + log_{10 }3^{2} – log_{10 }18 [Using nlog_{a} m = log_{a} m^{n}]

= log_{10 }8 + log_{10 }25 + log_{10 }9 – log_{10 }18

= log_{10 }(8 x 25 x 9) – log_{10 }18 [Using log_{a} l + log_{a} m + log_{a} n = log_{a} lmn]

= log_{10} 1800 – log_{10} 18

= log_{10} 1800/18 [Using log_{a} m – log_{a} n = log_{a} m/n]

= log_{10} 100

= log_{10} 10^{2}

= 2log_{10} 10 [Using log_{a} m^{n} = nlog_{a} m]

= 2 x 1

= 2

(iii) We have, log 4 + 1/3log 125 – 1/5log 32

= log 4 + log (125)^{1/3} – log (32)^{1/5} [Using nlog_{a} m = log_{a} m^{n}]

= log 4 + log (5^{3})^{1/3} – log (2^{5})^{1/5}

= log 4 + log 5 – log 2

= log (4 x 5) – log 2 [Using log_{a} m + log_{a} n = log_{a} mn]

= log 20 – log 2

= log 20/2 [Using log_{a} m – log_{a} n = log_{a} m/n]

= log 10

= 1

**4. Prove that:**

**2log 15/18 – log 25/162 + log 4/9 = log 2**

**Solution:**

Taking L.H.S.,

= 2log 15/18 – log 25/162 + log 4/9

= log (15/18)^{2} – log 25/162 + log 4/9 [Using nlog_{a} m = log_{a} m^{n}]

= log 225/324 – log 25/162 + log 4/9

= log [(225/324)/(25/162)] + log 4/9 [Using log_{a} m – log_{a} n = log_{a} 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 log_{a} m + log_{a} n = log_{a} mn]

= log (9/2 x 4/9)

= log 2

= R.H.S.

**5. Find x, if:**

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

**Solution:**

We have,

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

Solving for x, we have

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

= log 48 – log 2^{3} + log 125^{1/3} – log 3 [Using nlog_{a} m = log_{a} m^{n}]

= log 48 – log 8 + log (5^{3})^{1/3} – log 3

= (log 48 – log 8) + (log 5 – log 3)

= log 48/8 + log 5/3 [Using log_{a} m – log_{a} n = log_{a} m/n]

= log (48/8 x 5/3) [Using log_{a} m + log_{a} n = log_{a} mn]

= log (2 x 5)

= log 10

= 1

Hence, x = 1

**6. Express log _{10 }2 + 1 in the form of log_{10 }x.**

**Solution:**

Given, log_{10 }2 + 1

= log_{10 }2 + log_{10} 10 [As, log_{10} 10 = 1]

= log_{10 }(2 x 10) [Using log_{a} m + log_{a} n = log_{a} mn]

= log_{10} 20

**7. Solve for x:**

**(i) log _{10} (x – 10) = 1**

**(ii) log (x ^{2} – 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) We have, log_{10} (x – 10) = 1

Then,

x – 10 = 10^{1}

x = 10 + 10

Hence, x = 20

(ii) We have, log (x^{2} – 21) = 2

Then,

x^{2} – 21 = 10^{2}

x^{2} – 21 = 100

x^{2} = 100 + 21

x^{2} = 121

Taking square root on both sides,

Hence, x = ±11

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

Then,

log (x – 2)(x + 2) = log 5 [Using log_{a} m + log_{a} n = log_{a} mn]

log (x^{2} – 2^{2}) = log 5 [As (x – a)(x + a) = x^{2} – a^{2}]

log (x^{2} – 4) = log 5

Removing log on both sides, we get

x^{2} – 4 = 5

x^{2} = 5 + 4

x^{2} = 9

Taking square root on both sides,

x = ±3

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

Then,

log (x + 5) + log (x – 5) = log 2^{4} + log 3^{2 } [Using nlog_{a} m = log_{a} m^{n}]

log (x + 5)(x – 5) = log 16 + log 9 [Using log_{a} m + log_{a} n = log_{a} mn]

log (x^{2} – 5^{2}) = log (16 x 9) [As (x – a)(x + a) = x^{2} – a^{2}]

log (x^{2} – 25) = log 144

Removing log on both sides, we have

x^{2} – 25 = 144

x^{2} = 144 + 25

x^{2} = 169

Taking square root on both sides, we get

x = ±13

**8. Solve for x:**

**(i) log 81/log 27 = x**

**(ii) log 128/log 32 = x**

**(iii) log 64/log 8 = log x**

**(iv) log 225/log 15 = log x**

**Solution:**

(i) 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 3^{4}/log 3^{3 }

= (4log 3)/(3log 3) [Using log_{a} m^{n} = nlog_{a} m]

= 4/3

Hence, x = 4/3

(ii) 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 2^{7}/log 2^{5 }

= (7log 2)/(5log 2) [Using log_{a} m^{n} = nlog_{a} m]

= 7/5

Hence, x = 7/5

(iii) 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 2^{6}/log 2^{3 }

= (6log 2)/(3log 2) [Using log_{a} m^{n} = nlog_{a} m]

= 6/3

= 2

So, log x = 2

Hence, x = 10^{2} = 100

(iv) We have, log 225/log 15 = log x

log x = log 225/log 15

= log (15 x 15)/ log 15

= log 15^{2}/log 15^{ }

= (2log 15)/log 15 [Using log_{a} m^{n} = nlog_{a} m]

= 2

So, log x = 2

Hence, x = 10^{2} = 100

**9. Given log x = m + n and log y = m – n, express the value of log 10x/y ^{2} in terms of m and n.**

**Solution:**

Given, log x = m + n and log y = m – n

Now consider log 10x/y^{2},

log 10x/y^{2}** **= log 10x – log y^{2 } [Using log_{a} m/n = log_{a} m – log_{a} n]

= log 10x – 2log y

= log 10 + log x – 2 log y

= 1 + (m + n) – 2 (m – n)

= 1 + m + n – 2m + 2n

= 1 + 3n – m

**10. State, true or false:**

**(i) log 1 × log 1000 = 0**

**(ii) log x/log y = log x – log y**

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

**(iv) log x × log y = log x + log y**

**Solution:**

(i) We have, log 1 × log 1000 = 0

Now,

log 1 = 0 and

log 1000 = log 10^{3} = 3log 10 = 3 [Using log_{a} m^{n} = nlog_{a} m]

So,

log 1 × log 1000 = 0 x 3 = 0

Thus, the statement log 1 × log 1000 = 0 is true

(ii) We have, log x/log y = log x – log y

We know that,

log x/y = log x – log y

So,

log x/log y ≠ log x – log y

Thus, the statement log x/log y = log x – log y is false

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

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

log 5^{2}/ log 5 = log x

2log 5/log 5 = log x [Using log_{a} m^{n} = nlog_{a} m]

2 = log x

So, x = 10^{2}

x = 100

Thus, the statement x = 2 is false

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

So,

log x + log y ≠ log x × log y

Thus, the statement log x + log y = log x × log y is false

**11. If log _{10 }2 = a and log_{10 }3 = b; express each of the following in terms of ‘a’ and ‘b’:**

**(i) log 12**

**(ii) log 2.25**

**(iii) log **

** **

**(iv) log 5.4**

**(v) log 60**

**(iv) log **

** **

**Solution:**

Given that log_{10} 2 = a and log_{10} 3 = b … (1)

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

= log (2 x 2) + log 3 [Using log_{a} mn = log_{a} m + log_{a} n]

= log 2^{2} + log 3

= 2log 2 + log 3 [Using log_{a} m^{n} = nlog_{a} 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 log_{a} m^{n} = nlog_{a} m]

= 2(log 3 – log 2) [Using log_{a} m/n = log_{a} m – log_{a} n]

= 2(b – a) [From 1]

= 2b – 2a

(iii) log = log 9/4

= log (3/2)^{2}

= 2log 3/2 [Using log_{a} m^{n} = nlog_{a} m]

= 2(log 3 – log 2) [Using log_{a} m/n = log_{a} m – log_{a} n]

= 2(b – a) [From 1]

= 2b – 2a

(iv) log 5.4 = log 54/10

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

= log (2 x 3^{3}) – log 10 [Using log_{a} m/n = log_{a} m – log_{a} n]

= log 2 + log 3^{3} – 1 [Using log_{a} mn = log_{a} m + log_{a} n and log 10 = 1]

= log 2 + 3log 3 – 1 [Using log_{a} m^{n} = nlog_{a} m]

= a + 3b – 1 [From 1]

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

= log 10 + log 3 + log 2 [Using log_{a} lmn = log_{a} l + log_{a} m + log_{10} n]

= 1 + b + a [From 1]

(vi) log = log 25/8

= log 5^{2}/2^{3}

= log 5^{2} – log 2^{3} [Using log_{a} m/n = log_{a} m – log_{a} n]

= 2log 5 – 3log 2 [Using log_{a} m^{n} = nlog_{a} m]

= 2log 10/2 – 3log 2

= 2(log 10 – log 2) – 3log 2 [Using log_{a} m/n = log_{a} m – log_{a} n]

= 2log 10 – 2log 2 – 3log 2

= 2(1) – 2a – 3a [From 1]

= 2 – 5a

**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) 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 log_{a} mn = log_{a} m + log_{a} n]

= log 2^{2} + log 3

= 2log 2 + log 3 [Using log_{a} m^{n} = nlog_{a} m]

= 2 x 0.3010 + 0.4771

= 1.0791

(ii) log 1.2 = log 12/10

= log 12 – log 10 [Using log_{a} m/n = log_{a} m – log_{a} n]

= log (4 x 3) – log 10

= log 4 + log 3 – log 10 [Using log_{a} mn = log_{a} m + log_{a} n]

= log 2^{2} + log 3 – log 10

= 2log 2 + log 3 – log 10 [Using log_{a} m^{n} = nlog_{a} m]

= 2 x 0.3010 + 0.4771 – 1 [As log 10 = 1]

= 0.6020 + 0.4771 – 1

= 1.0791 – 1

= 0.0791

(iii) log 3.6 = log 36/10

= log 36 – log 10 [Using log_{a} m/n = log_{a} m – log_{a} n]

= log (2 x 2 x 3 x 3) – 1 [As log 10 = 1]

= log (2^{2} x 3^{2}) – 1

= log 2^{2} + log 3^{2} – 1 [Using log_{a} mn = log_{a} m + log_{a} n]

= 2log 2 + 2log 3 – 1 [Using log_{a} m^{n} = nlog_{a} m]

= 2 x 0.3010 + 2 x 0.4771 – 1

= 0.6020 + 0.9542 – 1

= 1.5562 – 1

= 0.5562

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

= log 15/10 + log 10 [Using log_{a} mn = log_{a} m + log_{a} n]

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

= log 3 – log 2 + 1 [Using log_{a} m/n = log_{a} m – log_{a} n]

= 0.4771 – 0.3010 + 1

= 1.1761

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

= log 100/4

= log 100 – log 4 [Using log_{a} m/n = log_{a} m – log_{a} n]

= log 10^{2} – log 2^{2}

= 2log 10 – 2log 2 [Using log_{a} m^{n} = nlog_{a} m]

= (2 x 1) – (2 x 0.3010)

= 2 – 0.6020

= 1.398

**13. Given 2 log _{10} x + 1 = log_{10} 250, find:**

**(i) x**

**(ii) log _{10} 2x**

**Solution:**

(i) Given equation, 2log_{10} x + 1 = log_{10} 250

log_{10} x^{2} + log_{10} 10 = log_{10} 250 [Using nlog_{a} m = log_{a} m^{n } and log_{10} 10 = 1]

log_{10} 10x^{2} = log_{10} 250 [Using log_{a} m + log_{a} n = log_{a} mn]

Removing log on both sides, we have

10x^{2} = 250

x^{2} = 25

x = ±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,

log_{10} 2x = log_{10} 2(5)

= log_{10} 10

= 1

**14. Given 3log x + ½ log y = 2, express y in term of x.**

**Solution:**

We have, 3log x + ½ log y = 2

log x^{3} + log y^{1/2} = 2 [Using log_{a} m + log_{a} n = log_{a} mn]

log x^{3}y^{1/2} = 2

Removing logarithm, we get

x^{3}y^{1/2} = 10^{2}

y^{1/2} = 100/x^{3}

On squaring on both sides, we get

y = 10000/x^{6}

y = 10000x^{-6}

**15. If x = (100) ^{a}, y = (10000)^{b} and z = (10)^{c}, find log 10√y/x^{2}z^{3} 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

⇒ log x = a log 10^{2}, log y = b log 10^{4} and log z = c log 10

⇒ log x = 2a log 10, log y = 4b log 10 and log z = c log 10

⇒ log x = 2a, log y = 4b and log z = c … (i)

Now,

log 10√y/x^{2}z^{3} = log 10√y – log x^{2}z^{3 } [Using log_{a} m/n = log_{a} m – log_{a} n]

= (log 10 + log √y) – (log x^{2} + log z^{3}) [Using log_{a} mn = log_{a} m + log_{a} n]

= 1 + log y^{1/2} – log x^{2} – log z^{3}

= 1 + ½ log y – 2 log x – 3 log z [Using log_{a} m^{n} = nlog_{a} m]

= 1 + ½(4b) – 2(2a) – 3c … [Using (i)]

= 1 + 2b – 4a – 3c

**16. If 3(log 5 – log 3) – (log 5 – 2 log 6) = 2 – log x, find x.**

**Solution:**

We have,

3(log 5 – log 3) – (log 5 – 2 log 6) = 2 – log x

3log 5 – 3log 3 – log 5 + 2log 6 = 2 – log x

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

2log 5 – 3log 3 + 2(log 3 + log 2) = 2 – log x [Using log_{a} mn = log_{a} m + log_{a} n]

2log 5 – log 3 + 2log 3 + 2log 2 = 2 – log x

2log 5 – log 3 + 2log 2 = 2 – log x

2log 5 – log 3 + 2log 2 + log x = 2

log 5^{2} – log 3 + log 2^{2} + log x = 2 [Using nlog_{a} m = log_{a} m^{n}]

log 25 – log 3 + log 4 + log x = 2

log (25 × 4 × x)/3 = 2 [Using log_{a} m + log_{a} n = log_{a} mn & log_{a} m – log_{a} n = log_{a} m/n]

log 100x/3 = 2

On removing logarithm,

100x/3 = 10^{2}

100x/3 = 100

Dividing by 100 on both sides, we have

x/3 = 1

Hence, x = 3

**Exercise 8(C)**

**1. If log _{10} 8 = 0.90; find the value of:**

**(i) log _{10} 4**

**(ii) log √32**

**(iii) log 0.125**

**Solution:**

Given, log_{10} 8 = 0.90

log_{10} (2 x 2 x 2) = 0.90

log_{10} 2^{3} = 0.90

3 log_{10} 2 = 0.90

log_{10} 2 = 0.90/3

log_{10} 2 = 0.30 … (1)

(i) log 4 = log_{10} (2 x 2)

= log_{10} 2^{2}

= 2 log_{10} 2

= 2 x 0.60 … [From (1)]

= 1.20

(ii) log √32 = log_{10} 32^{1/2}

= ½ log_{10}^{ }(2 x 2 x 2 x 2 x 2)

= ½ log_{10}^{ }2^{5}

= ½ x 5 log_{10} 2

= ½ x 5 x 0.30 … [From (1)]

= 0.75

(iii) log 0.125 = log 125/1000

= log_{10} 1/8

= log_{10} 1/2^{3}

= log_{10} 2^{-3}

= -3 log_{10} 2

= -3 x 0.30 … [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 3^{3} = 1.431

3log 3 = 1.431

log 3 = 1.431/3

= 0.477 … (1)

(i) log 9 = log 3^{2}

= 2log 3

= 2 x 0.477 … [From (1)]

= 0.954

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

= log 3 + log 100

= log 3 + log 10^{2}

= log 3 + 2log 10

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

= 0.477 + 2

= 2.477

**3. If log _{10} a = b, find 10^{3b – 2} in terms of a.**

**Solution:**

Given, log_{10} a = b

Now,

Let 10^{3b – 2} = x

Applying log on both sides,

log 10^{3b – 2} = log x

(3b – 2)log 10 = log x

3b – 2 = log x

3log_{10} a – 2 = log x

3log_{10} a – 2log 10 = log x

log_{10} a^{3} – log 10^{2} = log x

log_{10} a^{3} – log 100 = log x

log_{10} a^{3}/100 = log x

On removing logarithm, we get

a^{3}/100 = x

Hence, 10^{3b – 2} = a^{3}/100

**4. If log _{5} x = y, find 5^{2y+ 3} in terms of x.**

**Solution:**

Given, log_{5} x = y

So, 5^{y} = x

Squaring on both sides, we get

(5^{y})^{2} = x^{2}

5^{2y} = x^{2}

5^{2y} × 5^{3 }= x^{2} × 5^{3}

Hence,

5^{2y+3} = 125x^{2}

**5. Given: log _{3} m = x and log_{3} n = y.**

**(i) Express 3 ^{2x – 3} in terms of m.**

**(ii) Write down 3 ^{1 – 2y + 3x} in terms of m and n.**

**(iii) If 2 log _{3} A = 5x – 3y; find A in terms of m and n.**

** **

**Solution:**

Given, log_{3} m = x and log_{3} n = y

So, 3^{x} = m and 3^{y} = n … (1)

(i) Taking the given expression, 3^{2x – 3}

3^{2x – 3} = 3^{2x} . 3^{-3}

= 3^{2x} . 1/3^{3}

= (3^{x})^{2}/3^{3}

= m^{2}/3^{3} … [Using (1)]

= m^{2}/27

Hence, 3^{2x – 3} = m^{2}/27

(ii) Taking the given expression, 3^{1 – 2y + 3x}

3^{1 – 2y + 3x} = 3^{1 }. 3^{-2y }. 3^{3x}

= 3 . (3^{y})^{-2 }. (3^{x})^{3}

= 3 . n^{-2} . m^{3} … [Using (1)]

= 3m^{3}/n^{2}

Hence, 3^{1 – 2y + 3x} = 3m^{3}/n^{2}

(iii) Taking the given equation,

2 log_{3} A = 5x – 3y

log_{3} A^{2}_{ }= 5x – 3y

log_{3} A^{2}_{ }= 5log_{3} m – 3log_{3} n … [Using (1)]

log_{3} A^{2}_{ }= log_{3} m^{5} – log_{3} n^{3}

log_{3} A^{2}_{ }= log_{3} m^{5}/n^{3}

Removing logarithm on both sides, we get

A^{2} = m^{5}/n^{3}

Hence, by taking square root on both sides

A = √(m^{5}/n^{3}) = m^{5/2}/n^{3/2}

**6. Simplify:**

**(i) log (a) ^{3} – log a**

**(ii) log (a) ^{3} ÷ log a**

**Solution:**

(i) We have, log (a)^{3} – log a

= 3log a – log a

= 2log a

(ii) We have, log (a)^{3} ÷ 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 – ab = -b

a(1 – b) = -b

a = -b/(1 – b)

Hence,

a = b/(b – 1)

**8. Prove that:**

**(i) (log a) ^{2} – (log b)^{2} = log (a/b). log (ab)**

**(ii) If a log b + b log a – 1 = 0, then b ^{a}. a^{b} = 10**

**Solution:**

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

= (log a)^{2} – (log b)^{2}

= (log a + log b) (log a – log b) [As x^{2} – y^{2} = (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 b^{a} + log a^{b} – 1 = 0

log b^{a} + log a^{b} = 1

log b^{a}a^{b} = 1

On removing logarithm, we get

b^{a}a^{b} = 10

– Hence proved

**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: log _{10} 125 = 3(1 – log_{10}2).**

**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 – 3

3a + 3 = 4a – 3

Hence, a = 6

(ii) Given, 2log y – log x – 3 = 0

So,

log y^{2} – log x = 3

log y^{2}/x = 3

On removing logarithm, we have

y^{2}/x = 10^{3} = 1000

Hence, x = y^{2}/1000

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

log_{10} 125 = log_{10} (5 x 5 x 5)

= log_{10} 5^{3}

= 3log_{10} 5

= 3log_{10} 10/2

= 3(log_{10} 10 – log_{10} 2)

= 3(1 – log_{10} 2) [Since, log_{10} 10 = 1]

= R.H.S.

– Hence proved

**10. Given log x = 2m – n, log y = n – 2m and log z = 3m – 2n. Find in terms of m and n, the value of log x ^{2}y^{3}/z^{4}.**

**Solution:**

We have, log x = 2m – n, log y = n – 2m and log z = 3m – 2n

Now, considering

log x^{2}y^{3}/z^{4} = log x^{2}y^{3} – log z^{4}

= (log x^{2 }+ log y^{3}) – log z^{4}

= 2log x + 3log y – 4log z

= 2(2m – n) + 3(n – 2m) – 4(3m – 2n)

= 4m – 2n + 3n – 6m – 12m + 8n

= -14m + 9n

**11. Given log _{x} 25 – log_{x} 5 = 2 – log_{x} 1/125; find x.**

**Solution:**

We have, log_{x} 25 – log_{x} 5 = 2 – log_{x} 1/125

log_{x} (5 x 5) – log_{x} 5 = 2 – log_{x} 1/(5 x 5 x 5)

log_{x} 5^{2} – log_{x} 5 = 2 – log_{x} 1/5^{3}

2log_{x} 5 – log_{x} 5 = 2 – log_{x} 1/5^{3}

log_{x} 5 = 2 – 3log_{x} 1/5

log_{x} 5 = 2 + 3log_{x} (1/5)^{-1}

log_{x} 5 = 2 + 3log_{x} 5

2 = log_{x} 5 – 3log_{x} 5

2 = -2log_{x} 5

-1 = log_{x} 5

Removing logarithm, we get

x^{-1} = 5

Hence, x = 1/5

**Exercise 8(D)**

**1. If 3/2 log a + 2/3 log b – 1 = 0, find the value of a ^{9}.b^{4}**

**Solution:**

Given equation,

3/2 log a + 2/3 log b – 1 = 0

log a^{3/2} + log b^{2/3} – 1 = 0

log a^{3/2} × b^{2/3} – 1 = 0

log a^{3/2}.b^{2/3} = 1

Removing logarithm, we have

a^{3/2}.b^{2/3} = 10

On manipulating,

(a^{3/2}.b^{2/3})^{6} = 10^{6}

Hence,

a^{9}.b^{4} = 10^{6}

**2. If x = 1 + log 2 – log 5, y = 2log3 and z = log a – log 5; find the value of a 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 – log 5) + (2log 3) = 2(log a – log 5)

1 + log 2 – log 5 + 2log 3 = 2log a – 2log 5

1 + log 2 – log 5 + 2log 3 + 2log 5 = 2log a

log 10 + log 2 + log 3^{2} + log 5 = log a^{2}

log 10 + log 2 + log 9 + log 5 = log a^{2}

log (10 x 2 x 9 x 5) = log a^{2}

log 900 = log a^{2}

On removing logarithm on both sides, we have

900 = a^{2}

Taking square root, we get

a = ±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 – z (ii) 5 ^{x + y – z}**

**Solution:**

Given,

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

So, z = log 3 – log 2^{2}

= log 3 – log 4

= log ¾

= log 0.75 … (1)

(i) Considering,

x + y – z = log 0.6 + log 1.25 – log 0.75 … [From (1)]

= log (0.6 x 1.25)/0.75

= log 0.75/0.75

= log 1

= 0 … (2)

(ii) Now, considering

5^{x+y-z} = 5^{0} … [From (2)]

= 1

**4. If a ^{2} = log x, b^{3} = log y and 3a^{2} – 2b^{3} = 6 log z, express y in terms of x and z.**

**Solution:**

We have, a^{2} = log x and b^{3} = log y

Now, considering the equation

3a^{2} – 2b^{3} = 6log z

3log x – 2log y = 6log z

log x^{3} – log y^{2} = log z^{6}

log x^{3}/y^{2} = log z^{6}

On removing logarithm on both sides, we get

x^{3}/y^{2} = z^{6}

So,

y^{2} = x^{3}/z^{6}

Taking square root on both sides, we get

y = √(x^{3}/z^{6})

Hence, y = x^{3/2}/z^{3}

**5. If log (a – b)/2 = ½ (log a + log b), show that: a ^{2 }+ b^{2} = 6ab.**

**Solution:**

We have, log (a – b)/2 = ½ (log a + log b)

log (a – b)/2 = ½ log a + ½ log b

= log a^{1/2} + log b^{1/2}

= log √a + log √b

= log √(ab)

Now, removing logarithm on both sides, we get

(a – b)/2 = √(ab)

Squaring on both sides, we get

[(a – b)/2]^{2} = [√(ab)]^{2}

(a – b)^{2}/4 = ab

(a – b)^{2 }= 4ab

a^{2} + b^{2} – 2ab = 4ab

a^{2} + b^{2} = 4ab + 2ab

a^{2} + b^{2} = 6ab

– Hence proved

**6. If a ^{2} + b^{2} = 23ab, show that: log (a + b)/5 = ½ (log a + log b).**

**Solution:**

Given, a^{2} + b^{2} = 23ab

Adding 2ab on both sides,

a^{2} + b^{2} + 2ab = 23ab + 2ab

(a + b)^{2} = 25ab

(a + b)^{2}/25 = ab

[(a + b)/5]^{2} = ab

Taking logarithm on both sides, we have

log [(a + b)/5]^{2} = log ab

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

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

Thus,

log (a + b)/5 = ½ (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 – log 25

log (x – 4)^{2} = log 20^{2} – log 25

log (x – 4)^{2} = log 400 – log 25

log (x – 4)^{2} = log 400/25

Removing logarithm on both sides,

(x – 4)^{2} = 400/25

x^{2} – 8x + 16 = 16

x^{2} – 8x = 0

x(x – 8) = 0

So,

x = 0 or x = 8

If x = 0, then log (x – 4) doesn’t exist

Hence, x = 8

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

**Solution:**

We have,

log xy = log x/y + 2log 2 = 2

Considering the equation,

log xy = 2

log xy = 2log 10

log xy = log 10^{2}

log xy = log 100

On removing logarithm,

xy = 100 … (1)

Now, consider the equation

log x/y + 2log 2 = 2

log x/y + log 2^{2} = 2

log x/y + log 4 = 2

log 4x/y = 2

Removing logarithm, we get

4x/y = 10^{2}

4x/y = 100

x/y = 25

(xy)/y^{2} = 25

100/y^{2} = 25 … [From (1)]

y^{2} = 100/25

y^{2} = 4

y = 2 [Since, y > 0]

From log xy = 2

Substituting the value of y, we get

log 2x = 2

On removing logarithm,

2x = 10^{2}

2x = 100

x = 100/2

x = 50

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

**9. Find x, if:**

**(i) log _{x} 625 = -4 **

**(ii) log _{x} (5x – 6) = 2**

**Solution:**

(i) We have, log_{x} 625 = -4

On removing logarithm,

x^{-4} = 625

(1/x)^{4} = 5^{4}

Taking the fourth root on both sides,

1/x = 5

Hence, x = 1/5

(ii) We have, log_{x} (5x – 6) = 2

On removing logarithm,

x^{2} = 5x – 6

x^{2} – 5x + 6 = 0

x^{2} – 3x – 2x + 6 = 0

x(x – 3) – 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 – q. **

**Solution:**

Given, p = log 20 and q = log 25

Considering the equation,

2log (x + 1) = 2p – q

log (x + 1)^{2} = 2p – q

log (x + 1)^{2} = 2log 20 – log 25

log (x + 1)^{2} = log 20^{2} – log 25

log (x + 1)^{2} = log 400 – 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 = 3

**11. If log _{2} (x + y) = log_{3} (x – y) = log 25/log 0.2, find the value of x and y.**

**Solution:**

Considering the relation, log_{2} (x + y) = log 25/log 0.2

log_{2} (x + y) = log_{0.2} 25

= log_{2/10} 5^{2}

= 2log_{1/5} 5

= 2log_{5}^{-1} 5

= -2log_{5} 5

= -2 x 1

= -2

So, we have

log_{2} (x + y) = -2

Removing logarithm, we get

x + y = 2^{-2}

x + y = 1/2^{2}

x + y = ¼ … (i)

Now, considering the relation log_{3} (x – y) = log 25/log 0.2

log_{3} (x – y) = = log_{0.2} 25

= log_{2/10} 5^{2}

= 2log_{1/5} 5

= 2log_{5}^{-1} 5

= -2log_{5} 5

= -2 x 1

= -2

So, we have

log_{3} (x – y) = -2

Removing logarithm, we get

x – y = 3^{-2}

x – y = 1/3^{2}

x – y = 1/9 … (ii)

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

x + y = ¼

x – y = 1/9

—————-

2x = ¼ + 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 = ¼

y = ¼ – 13/72

= (18 – 13)/72

= 5/72

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

_{ }

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

**Solution:**

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

So,

log xy = log x + log y = 5

And, we have log y = (2log x)/3 … [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 = 10^{3} = 1000

Substituting value of x in (ii), we get

log xy = 5

Removing logarithm, we get

xy = 10^{5}

(10^{3}). y = 10^{5}

y = 10^{5}/10^{3}

y = 10^{2}

y = 100

_{ }

**13. Given log _{10 }x = 2a and log_{10 }y = b/2**

**(i) Write 10 ^{a} in terms of x**

**(ii) Write 10 ^{2b + 1} in terms of y **

**(iii) If log _{10} p = 3a – 2b, express p in terms of x and y.**

**Solution:**

Given, log_{10 }x = 2a and log_{10 }y = b/2

(i) Taking log_{10} x = 2a

Removing logarithm on both sides,

x = 10^{2a}

Taking square root on both sides, we get

x^{1/2} = 10^{2a/2}

Hence, 10^{a} = x^{1/2}

(ii) Taking log_{10 }y = b/2

Removing logarithm on both sides,

y = 10^{b/2}

On manipulating,

y^{4} = 10^{b/2 x 4}

y^{4} = 10^{2b}

10y^{4} = 10^{2b} x 10

Hence, 10^{2b + 1} = 10y^{4}

(iii) We have, 10^{a} = x^{1/2}

and y = 10^{b/2}

Considering the equation, log_{10} p = 3a – 2b

log_{10} p = 3a – 2b

Removing logarithm, we get

p = 10^{3a – 2b}

p = 10^{3a}/10^{2b}

p = (10^{a})^{3}/(10^{b/2})^{4}

p = (x^{1/2})^{3}/(y)^{4}

Hence, p = x^{3/2}/y^{4}

**14. Solve:**

**log _{5 }(x + 1) – 1 = 1 + log_{5 }(x – 1).**

**Solution:**

Considering the given equation,

log_{5 }(x + 1) – 1 = 1 + log_{5 }(x – 1)

log_{5 }(x + 1) – log_{5 }(x – 1) = 1 + 1

log_{5 }(x + 1)/(x – 1) = 2

Removing logarithm, we have

(x + 1)/(x – 1) = 5^{2}

(x + 1)/(x – 1) = 25

(x + 1) = 25(x – 1)

x + 1 = 25x – 25

25x – x = 25 + 1

24x = 26

x = 26/24

Hence, x = 13/12

**15. Solve for x, if:**

**log _{x} 49 – log_{x} 7 + log_{x} 1/343 + 2 = 0_{ }**

**Solution:**

We have,

log_{x} 49 – log_{x} 7 + log_{x} 1/343 + 2 = 0_{ }

log_{x} 49/(7 x 343) + 2 = 0

log_{x} 1/49 = -2

log_{x} 1/7^{2} = -2

log_{x} 7^{-2} = -2

-2log_{x} 7 = -2

So,

log_{x} 7 = 1

Removing logarithm, we get

x = 7

**16. If a ^{2} = log x, b^{3} = log y and a^{2}/2 – b^{3}/3 = log c, find c in terms of x and y. **

**Solution:**

Given,

a^{2} = log x, b^{3} = log y

Considering the given equation,

a^{2}/2 – b^{3}/3 = log c

(log x)/2 – (log y)/3 = log c

½ log x – 1/3 log y = log c

log x^{1/2} – log y^{1/3} = log c

log x^{1/2}/y^{1/3} = log c

On removing logarithm, we get

x^{1/2}/y^{1/3} = c

Hence, c = x^{1/2}/y^{1/3} is the required relation

**17. Given: x = log _{10} 12, y = log_{4} 2 x log_{10} 9 and z = log_{10} 0.4, find **

**(i) x – y – z**

**(ii) 13 ^{x – y – z} **

**Solution:**

(i) Considering, x – y – z

= log_{10} 12 – (log_{4} 2 x log_{10} 9) – log_{10} 0.4

= log_{10} 12 – (log_{4} 2 x log_{10} 9) – log_{10} 0.4

= log_{10} (4 x 3) – (log_{10} 2/ log_{10} 4 x log_{10} 9) – log_{10} 0.4

= log_{10} 4 + log_{10} 3 – (log_{10} 2 x log_{10} 3^{2})/ log_{10} 2^{2} – log_{10} 4/10

= log_{10} 4 + log_{10} 3 – (log_{10} 2 x 2log_{10} 3)/ 2log_{10} 2 – (log_{10} 4 – log_{10} 10)

= log_{10} 4 + log_{10} 3 – log_{10} 3 – log_{10} 4 + log_{10} 10

= log_{10} 4 + log_{10} 3 – log_{10} 3 – log_{10} 4 + 1

= 1

(ii) Now,

13^{x – y – z} = 13^{1} = 13

**18. Solve for x, log _{x} 15√5 = 2 – log_{x} 3√5 **

**Solution:**

Considering the given equation,

log_{x} 15√5 = 2 – log_{x} 3√5

log_{x} 15√5 + log_{x} 3√5 = 2

log_{x} (15√5 x 3√5) = 2

log_{x} (45 x 5) = 2

log_{x} 225 = 2

Removing logarithm, we get

x^{2} = 225

Taking square root on both sides,

x = 15

**19. Evaluate:**

**(i) log _{b} a x log_{c} b x log_{a} c **

**(ii) log _{3} 8 ÷ log_{9} 16**

**(iii) log _{5} 8/(log_{25} 16 x log_{100} 10) **

**Solution:**

Using log_{b} a = 1/log_{a} b and log_{x} a/log_{x} b = log_{b} a, we have

(i) log_{b} a x log_{c} b x log_{a} c

= 1

(ii) log_{3} 8 ÷ log_{9} 16

= log_{3} 8/ log_{9} 16

= 3 x ½

= 3/2

(iii) log_{5} 8/(log_{25} 16 x log_{100} 10)

= 3 x ½ x 2

= 3

**20. Show that:**

**log _{a} m ÷ log_{ab} m = 1 + log_{a} b**

**Solution:**

Considering the L.H.S.,

log_{a} m ÷ log_{ab} m = log_{a} m/log_{ab} m

= log_{m} ab/log_{m} a [As log_{b} a = 1/log_{a} b]

= log_{a} ab [As log_{x} a/log_{x} b = log_{b} a]

= log_{a} a + log_{a} b

= 1 + log_{a} b

**21. If log _{√27} x = 2 2/3, find x.**

**Solution:**

We have,

log_{√27} x = 2 2/3

log_{√27} x = 8/3

Removing logarithm, we get

x = √27^{8/3}

= 27^{1/2 x 8/3}

= 27^{4/3}

= 3^{3 x 4/3}

= 3^{4}

Hence, x = 81

**22. Evaluate:**

**1/(log _{a} bc + 1) + 1/(log_{b} ca + 1) + 1/(log_{c} 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

8.5 More about logarithms

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

The Chapter 8 of class 9 teaches the students the method of calculating using logarithm.Logarithms are used to make long and complicated calculations easy. Logarithm is related to exponents or indices. I.e., while a^{b}=c is called the exponential form, log_{a}c=b is called the logarithmic form. Read and learn the Chapter 8 of Selina textbook to get to know more about Logarithms. Learn the Selina Solutions for Class 9 effectively to attain excellent result in the examination.