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**Exercise 8(A)**

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

**(i) 5 ^{3}Â = 125**

**(ii) 3 ^{-2}Â =Â 1/9**

**(iii) 10 ^{-3}Â = 0.001**

**(iv)Â (81) ^{3/4} = 27**

**Solution:**

We know that,

a^{b} = c â‡’ log_{a }c = b

(i) 5^{3} = 125

log_{5} 125 = 3Â

Â

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

log_{3} 1/9 = -2Â

Â

(iii) 10^{-3} = 0.001

log_{10} 0.001 = -3Â

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

log_{81} 27 = Â¾Â

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

**(i)Â log _{8 }0.125 = -1**

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

**(iii)Â log _{a }AÂ = x**

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

**Solution:**

We know that,

log_{a} c = b â‡’ a^{b} = c

(i)Â log_{8 }0.125 = -1

8^{-1} = 0.125

(ii) log_{10 }0.01 = -2

8^{-1} = 0.125

(iii)Â log_{a }AÂ = x

a^{x} = A

(iv) log_{10 }1 = 0

10^{0} = 1

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

**Solution:**

We have,

log_{10}Â x = -2

10^{-2} = x [As log_{a} c = b â‡’ 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)Â 1/16Â to the base 4**

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

**(viii)Â 1/81Â 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} = (Â½)^{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} = (Â¼)^{-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}Â x = a, then 10^{x}Â = a**

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

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

**Solution:**

(i) We have,

log_{10}Â x = a

So, 10^{a} = x

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

(ii) We have,

x^{y}Â = z

So, log_{x} z = y

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

(iii) We have,

log_{2}Â 8 = 3

So, 2^{3} = 8 â€¦ (1)

Now consider the equation,

log_{8} 2 = 1/3

8^{1/3} = 2

(2^{3})^{1/3} = 2 â€¦ (2)

Both equations (1) and (2) are correct

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

**6. Find x, if:**

**(i) log _{3}Â x = 0**

**(ii)Â log _{x}Â 2 = -1**

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

**(iv) log _{5}Â (x – 7) = 1**

**(v) log _{4}32 = x – 4**

**(vi) log _{7}Â (2x^{2}Â – 1) = 2**

**Solution:**

(i) We have, log_{3}Â x = 0

So, 3^{0} = x

1 = x

Hence, x = 1

(ii)Â we have, log_{x}Â 2 = -1

So, x^{-1} = 2

1/x = 2

Hence, x = Â½

(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 = 2Â½

(iv) We have, log_{5}Â (x – 7) = 1

So, 5^{1} = x â€“ 7

5 = x â€“ 7

x = 5 + 7

Hence, x = 12

(v) We have, log_{4 }32 = x â€“ 4

So, 4^{(x – 4)} = 32

(2^{2})^{(x – 4)} = 2^{5}

2^{(2x – 8)} = 2^{5}

On comparing the exponents, we get

2x â€“ 8 = 5

2x = 5 + 8

Hence,

x = 13/2 = 6Â½

(vi) We have, log_{7}Â (2x^{2}Â – 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 = Â±5

Hence, x = 5 (Neglecting the negative value)

Â

**7. Evaluate:**

**(i) log _{10Â }0.01**

**(ii) log _{2}Â (1 Ã· 8)**

**(iii) log _{5}Â 1**

**(iv) log _{5}Â 125**

**(v) log _{16}Â 8**

**(vi) log _{0.5}Â 16**

**Solution:**

(i) Let log_{10Â }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_{10Â }0.01 = -2

(ii) Let log_{2}Â (1 Ã· 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_{10Â }(1 Ã· 8) = -3

(iii) Let log_{5}Â 1 = x

Then, 5^{x} = 1

5^{x} = 5^{0}

On comparing the exponents, we get

x = 0

Hence, log_{5Â }1 = 0

(iv) Let log_{5}Â 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_{5Â }125 = 3

(v) Let log_{16}Â 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_{16Â }8 = 3/4

(vi) Let log_{0.5}Â 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

â‡’ x = -4

Hence, log_{0.5Â }16 = -4

Â

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

**Solution:**

We have, log_{a}Â m = n

So,

a^{n} = m

Dividing by a on both sides, we get

a^{n}/a = m/a

a^{n-1} = m/aÂ

**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 yÂ **

**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}Â

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

**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} 1Â **

**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} â€“ 1 = 1

x^{2} = 1 + 1

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

**Solution:**

Given, log (x^{2}Â – 21) = 2

So,

x^{2}Â – 21 = 10^{2}

x^{2}Â – 21 = 100

x^{2} = 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 (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)** **logÂ 26/51Â – logÂ 91/119Â = logÂ (26/51)/ (91/119)Â [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 â€“ log 3 [Using log_{a} m/n = log_{a} m – log_{a} n]

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

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

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

= logÂ [(75/16)/ (25/81)]Â + logÂ 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}

â‡’ 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}

â‡’ 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 +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 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

= 1Â

(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 â€“ 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

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 nlog_{a} m = log_{a} m^{n}]

= log 4 + log (5^{3})^{1/3} â€“ log (2^{5})^{1/5}

= log 4 + log 5 â€“ log 2

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

= log 20 â€“ 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 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 nlog_{a} m = log_{a} m^{n}]

= log 225/324 â€“ 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.

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 2^{3} + log 125^{1/3} â€“ log 3 [Using nlog_{a} m = log_{a} m^{n}]

= log 48 â€“ log 8 + log (5^{3})^{1/3} â€“ log 3

= (log 48 â€“ log 8) + (log 5 â€“ 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

Â

**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}Â (x – 10) = 1**

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

Then,

x â€“ 10 = 10^{1}

x = 10 + 10

Hence, x = 20

(ii) We have, log (x^{2}Â – 21) = 2

Then,

x^{2} â€“ 21 = 10^{2}

x^{2} â€“ 21 = 100

x^{2} = 100 + 21

x^{2} = 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 log_{a} m + log_{a} n = log_{a} mn]

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

log (x^{2} – 4) = log 5

Removing log on both sides, we get

x^{2} â€“ 4 = 5

x^{2} = 5 + 4

x^{2} = 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 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} â€“ a^{2}]

log (x^{2} â€“ 25) = log 144

Removing log on both sides, we have

x^{2} â€“ 25 = 144

x^{2} = 144 + 25

x^{2} = 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 3^{4}/log 3^{3 }

= (4log 3)/(3log 3) [Using log_{a} m^{n} = nlog_{a} 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 2^{7}/log 2^{5 }

= (7log 2)/(5log 2) [Using log_{a} m^{n} = nlog_{a} 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 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)Â 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}Â in terms of m and n.**

**Solution:**

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

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

log 10x/y^{2}** **= log 10x â€“ log y^{2 } [Using log_{a} m/n = log_{a} m – log_{a} 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 10^{3} = 3log 10 = 3 [Using log_{a} m^{n} = nlog_{a} 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 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

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 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) logÂ **

**Â **

**(iv) log 5.4**

**(v) log 60**

**(iv) logÂ **

**Â **

**Solution:**

Given that log_{10} 2 = a and log_{10} 3 = b â€¦ (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 â€“ log 2) [Using log_{a} m/n = log_{a} m – log_{a} n]

= 2(b – a) [From 1]

= 2b â€“ 2a

(iii) logÂ = log 9/4

= log (3/2)^{2}

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

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

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

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

= a + 3b â€“ 1 [From 1]

Â

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

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

= 2log 10/2 â€“ 3log 2

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

= log (4 x 3) â€“ log 10

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

= log 2^{2} + log 3 â€“ log 10

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

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

= log (2^{2} x 3^{2}) â€“ 1

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

= 2log 2 + 2log 3 â€“ 1 [Using log_{a} m^{n} = nlog_{a} 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 log_{a} mn = log_{a} m + log_{a} n]

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

= log 3 â€“ 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 â€“ log 4 [Using log_{a} m/n = log_{a} m – log_{a} n]

= log 10^{2} â€“ log 2^{2}

= 2log 10 â€“ 2log 2 [Using log_{a} m^{n} = nlog_{a} m]

= (2 x 1) â€“ (2 x 0.3010)

= 2 â€“ 0.6020

= 1.398

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

**(i) x**

**(ii) log _{10}Â 2x**

**Solution:**

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

log_{10}Â x^{2} + log_{10} 10 = log_{10}Â 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 = 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,

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

= log_{10} 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 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}

Â

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

â‡’ log x = a log 10^{2}, log y = b log 10^{4} 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/x^{2}z^{3} = log 10âˆšy â€“ log x^{2}z^{3 } [Using log_{a} m/n = log_{a} m – log_{a} n]

= (log 10 + log âˆšy) â€“ (log x^{2} + log z^{3}) [Using log_{a} mn = log_{a} m + log_{a} n]

= 1 + log y^{1/2} â€“ log x^{2} â€“ log z^{3}

= 1 + Â½ log y â€“ 2 log x â€“ 3 log z [Using log_{a} m^{n} = nlog_{a} 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 log_{a} mn = log_{a} m + log_{a} 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 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 Ã— 4 Ã— 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}Â 8 = 0.90; find the value of:**

**(i) log _{10}Â 4**

**(ii) logÂ âˆš32**

**(iii) log 0.125**

**Solution:**

Given, log_{10}Â 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 â€¦ (1)

Â

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

= log_{10} 2^{2}

= 2 log_{10} 2

= 2 x 0.60 â€¦ [From (1)]

= 1.20

(ii) log âˆš32 = log_{10} 32^{1/2}

= Â½ log_{10}^{ }(2 x 2 x 2 x 2 x 2)

= Â½ log_{10}^{ }2^{5}

= Â½ x 5 log_{10} 2

= Â½ x 5 x 0.30 â€¦ [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 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 3^{3} = 1.431

3log 3 = 1.431

log 3 = 1.431/3

= 0.477 â€¦ (1)

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

= 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 10^{2}

= log 3 + 2log 10

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

= 0.477 + 2

= 2.477

Â

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

**Solution:**

Given, log_{10} a = b

Now,

Let 10^{3b – 2}Â = x

Applying log on both sides,

log 10^{3b â€“ 2} = log x

(3b â€“ 2)log 10 = log x

3b â€“ 2 = log x

3log_{10} a â€“ 2 = log x

3log_{10} a â€“ 2log 10 = log x

log_{10} a^{3} â€“ log 10^{2} = log x

log_{10} a^{3} â€“ 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}Â x = y, find 5^{2y+ 3}Â in terms of x.**

**Solution:**

Given, log_{5}Â x = y

So, 5^{y} = x

Squaring on both sides, we get

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

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

5^{2y} Ã— 5^{3 }= x^{2} Ã— 5^{3}

Hence,

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

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

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

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

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

**Â **

**Solution:**

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

So, 3^{x} = m and 3^{y} = nÂ â€¦ (1)

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

3^{2x – 3}Â = 3^{2x} . 3^{-3}

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

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

= m^{2}/3^{3} â€¦ [Using (1)]

= m^{2}/27

Hence, 3^{2x – 3}Â = m^{2}/27

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

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

= 3 . (3^{y})^{-2 }. (3^{x})^{3}Â

= 3 . n^{-2} . m^{3} â€¦ [Using (1)]

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

Hence, 3^{1 – 2y +Â 3x}Â = 3m^{3}/n^{2}

(iii) Taking the given equation,

2 log_{3}Â A = 5x – 3y

log_{3} A^{2}_{ }= 5x – 3y

log_{3} A^{2}_{ }= 5log_{3}Â m – 3log_{3}Â n â€¦ [Using (1)]

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

log_{3} A^{2}_{ }= log_{3}Â 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 = âˆš(m^{5}/n^{3}) = m^{5/2}/n^{3/2}

Â

**6. Simplify:**

**(i) log (a) ^{3}Â – log a**

**(ii) log (a) ^{3}Â Ã·Â log a**

**Solution:**

(i) We have, log (a)^{3}Â – log a

= 3log a – log a

= 2log a

(ii) We have, log (a)^{3}Â 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) ^{2}Â – (log b)^{2}Â = log (a/b). log (ab)**

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

**Solution:**

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

= (log a)^{2}Â – (log b)^{2}

= (log aÂ + log b)Â (log aÂ – log b) [As x^{2} â€“ 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

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: log _{10}Â 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 â€“ 3

3a + 3 = 4a â€“ 3

Hence, a = 6

(ii) Given, 2log y – log x – 3 = 0

So,

log y^{2} â€“ 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}Â 125 = log_{10}Â (5 x 5 x 5)

= log_{10}Â 5^{3}

= 3log_{10}Â 5

= 3log_{10}Â 10/2

= 3(log_{10}Â 10 – log_{10}Â 2)

= 3(1 – log_{10}Â 2) [Since, log_{10} 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 x ^{2}y^{3}/z^{4}.**

**Solution:**

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

Now, considering

log x^{2}y^{3}/z^{4} = log x^{2}y^{3} â€“ log z^{4}

= (log x^{2 }+ log y^{3}) â€“ log z^{4}

= 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 log _{x} 25 â€“ log_{x} 5 = 2 â€“ log_{x} 1/125; find x.**

**Solution:**

We have, log_{x} 25 â€“ log_{x} 5 = 2 â€“ log_{x} 1/125

log_{x} (5 x 5) â€“ log_{x} 5 = 2 â€“ log_{x} 1/(5 x 5 x 5)

log_{x} 5^{2} â€“ log_{x} 5 = 2 â€“ log_{x} 1/5^{3}

2log_{x} 5 â€“ log_{x} 5 = 2 â€“ log_{x} 1/5^{3}

log_{x} 5 = 2 â€“ 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

Â

**Exercise 8(D)**

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

**Solution:**

Given equation,

3/2 log a +Â 2/3 log b – 1 = 0

log a^{3/2} +Â log b^{2/3} – 1 = 0

log a^{3/2} Ã— 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 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 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 = 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) 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 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

5^{x+y-z} = 5^{0} â€¦ [From (2)]

= 1

Â

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

**Solution:**

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

Now, considering the equation

3a^{2} â€“ 2b^{3} = 6log z

3log x â€“ 2log y = 6log z

log x^{3} â€“ 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 = âˆš(x^{3}/z^{6})

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

Â

**5. If log (a – b)/2 = Â½ (log a + log b), show that: a ^{2Â }+ b^{2}Â = 6ab.**

**Solution:**

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

log (a – b)/2 = Â½ log a + Â½ log b

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

= 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

a^{2} + b^{2} â€“ 2ab = 4ab

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

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

– Hence proved

**6. If a ^{2}Â + b^{2}Â = 23ab, show that: logÂ (a + b)/5 = Â½ (log a + log b).**

**Solution:**

Given, a^{2}Â + b^{2}Â = 23ab

Adding 2ab on both sides,

a^{2}Â + b^{2}Â + 2ab = 23ab + 2ab

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

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

[(a + b)/5]^{2}Â = ab

Taking logarithm on both sides, we have

log [(a + b)/5]^{2}Â = 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 20^{2} 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

x^{2} â€“ 8x + 16 = 16

x^{2} 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 10^{2}

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 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 â€¦ [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)Â log _{x}Â 625 = -4Â **

**(ii)Â log _{x}Â (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

Â

(ii) We have, log_{x}Â (5x – 6) = 2

On removing logarithm,

x^{2} = 5x â€“ 6

x^{2} â€“ 5x + 6 = 0

x^{2} 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 20^{2} 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 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 = Â¼ â€¦ (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 â€“ y = 3^{-2}

x â€“ y = 1/3^{2}

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 = 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

_{Â }

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

**(i) Write 10 ^{a}Â in terms of x**

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

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

**Solution:**

Given, log_{10 }x = 2a and log_{10 }yÂ =Â 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 }yÂ =Â 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}Â = 10y^{4}

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

and y = 10^{b/2}

Considering the equation, log_{10} p = 3a â€“ 2b

log_{10} p = 3a â€“ 2b

Removing logarithm, we get

p = 10^{3a â€“ 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 â€“ 25

25x â€“ x = 25 + 1

24x = 26

x = 26/24

Hence, x = 13/12

Â

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

**log _{x} 49 â€“ log_{x} 7 + log_{x} 1/343 + 2 = 0_{Â }**

**Solution:**

We have,

log_{x} 49 â€“ log_{x} 7 + log_{x} 1/343 + 2 = 0_{Â }

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 â€“ b^{3}/3 = log c, find c in terms of x and y. Â **

**Solution:**

Given,

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

Considering the given equation,

a^{2}/2 â€“ b^{3}/3 = log c

(log x)/2 â€“ (log y)/3 = log c

Â½ log x â€“ 1/3 log y = log c

log x^{1/2} â€“ 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 relationÂ

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

**(i) x â€“ y â€“ z**

**(ii) 13 ^{x – y – z}Â **

**Solution:**

(i) Considering, x â€“ y â€“ z

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

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

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

= log_{10} 4 + log_{10} 3 â€“ (log_{10} 2 x log_{10} 3^{2})/ log_{10} 2^{2} â€“ log_{10} 4/10

= log_{10} 4 + log_{10} 3 â€“ (log_{10} 2 x 2log_{10} 3)/ 2log_{10} 2 â€“ (log_{10} 4 â€“ log_{10} 10)

= log_{10} 4 + log_{10} 3 â€“ log_{10} 3 â€“ log_{10} 4 + log_{10} 10

= log_{10} 4 + log_{10} 3 â€“ log_{10} 3 â€“ log_{10} 4 + 1

= 1

(ii) Now,

13^{x – y – z}Â = 13^{1} = 13Â

**18. Solve for x, log _{x} 15âˆš5 = 2 â€“ log_{x} 3âˆš5Â **

**Solution:**

Considering the given equation,

log_{x} 15âˆš5 = 2 â€“ log_{x} 3âˆš5Â

log_{x} 15âˆš5 + log_{x} 3âˆš5Â = 2

log_{x} (15âˆš5 x 3âˆš5)Â = 2

log_{x} (45 x 5)Â = 2

log_{x} 225Â = 2

Removing logarithm, we get

x^{2} = 225

Taking square root on both sides,

x = 15Â

**19. Evaluate:**

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

**(ii) log _{3} 8 Ã· log_{9} 16**

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

**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 Ã· log_{9} 16

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

= 3 x Â½

= 3/2

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

= 3 x Â½ x 2

= 3 Â

**20. Show that:**

**log _{a} m Ã· log_{ab} m = 1 + log_{a} b**Â

**Solution:**

Considering the L.H.S.,

log_{a} m Ã· 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} bÂ

**21. If log _{âˆš27} x = 2 2/3, find x.**

**Solution:**

We have,

log_{âˆš27} x = 2 2/3

log_{âˆš27} x = 8/3

Removing logarithm, we get

x = âˆš27^{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.