## Rajasthan Board RBSE Class 11 Maths Chapter 9 Logarithms Miscellaneous Exercise

Question 1.

log_{√2}x = 4 then value of x will be :

(A) 4^{√2}

(B) \(\frac { 1 }{ 4 }\)

(C) 4

(D) 4 x √2

Solution :

∵ log_{a} n = x

∴ a^{x} = n

Given log_{√2}x= 4

Then (√2)^{4} = x

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

Hence, option (C) is correct.

Question 2.

log_{x} 243 = 2.5, then value of x will be :

(A) 9

(B) 3

(C) 1

(D) 81

Solution:

∵ log_{a}n = x

⇒ a^{x} = n

Similarly log_{x} 243 = 2.5

⇒ 243 = x^{2.5}

⇒ x^{5/2} = 243 = (3)^{5}

Squaring on both sides,

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

⇒ x^{5/2} = (3^{2})^{2}

On comparing, x = 3^{2} = 9

Hence, option (A) is correct.

Question 3.

The value of log (1 + 2 x 3):

(A) 2 log 3

(B) log 1.log 2.log 3

(C) log1 + log2 + log3

(D) log 7

Solution:

log (1 + 2 x 3) = log (1 + 6) = log 7

Hence, option (D) is correct.

Question 4.

The value of log (m + n) is :

(A) log m + log n

(B) log mn

(C) log m x log n

(D) none of these

Solution:

log m + log n = log mn ≠ log (m + n)

log mn = log m + log n ≠ log (m + n)

log m x log n ≠ log (m + n)

Hence, option (D) is correct.

Question 5.

The value of log_{b}a.log_{c}b.log_{a}c is :

(A) 0

(B) log abc

(C) 1

(D) log (a^{a} b^{b} c^{c})

Solution:

log_{b}o.log_{c}6.log_{a}c

Hence, option (C) is correct.

Question 6.

If a > 1 then value of log_{a }0 is :

(A) – ∞

(B) ∞

(C) 0

(D) 1

Solution:

log_{a}0 = – ∞

If a > 1

Hence, option (A) is correct.

Question 7.

If a < 0 then value of log_{a}0 is :

(A) – ∞

(B) ∞

(C) 0

(D) 1

Solution:

log 0 = ∞

If a < 0

Hence, option (B) is correct.

Question 8.

Other form of Iog_{a}b :

(A) a^{b}

(B) b^{a}

(C) \(\frac { 1 }{ { log }_{ b }a } \)

(D) log_{a} b

Solution:

Hence option (C) is correct.

Question 9.

Number log_{2}7 is :

(A) Integer

(B) Rational

(C) Irrational

(D) Prime

Solution:

log_{2} 7 = x

7 = 2^{x}

Taking log on both sides,

log 7 = log 2^{x}

⇒ x log 2 = log 7

= 2.8076

= irrational number

Hence, option (C) is correct.

Question 10.

If a = log_{3 }5 and b = log_{7 }25 then correct option is :

(A) a < b

(B) a > b

(C) a = b

(D) none of these

Solution:

Hence, option (A) is correct.

Question 11.

If log_{2 }x + log_{2 }(x – 1) = 1, then find the value of x.

Solution:

log_{2}x + log_{2}(x – 1) = 1

⇒ log_{2} x(x – 1) = 1

⇒ x(x – 1) = 2^{1}

⇒ x^{2} – x – 2 = 0

⇒ x^{2} – 2x + x – 2 = 0

⇒ (x – 2) (x + 1) = 0

⇒ x = 2, x = – 1

Negative value for algorithm is not possible

∴ x = 2

Question 12.

If log (a- b) = log a – log b, then what will be value of a in terms of b?

Solution:

log (a – b) = log a – log b

⇒ log (a – b) = log(\(\frac { a }{ b }\))

Comparing on both sides,

⇒ a – b = \(\frac { a }{ b }\)

⇒ b(a -b) = a

⇒ ab – b^{2} = a

⇒ ab – a = b^{2}

⇒ a(b – 1) = b^{2}

⇒ a = b^{2}/b – 1

Hence, value of a in terms of b is b^{2}/b – 1.

Question 13.

If , then find relation among a,b and c.

Solution:

According to question,

⇒ log_{x}a + log_{x}c= 2 log_{x}b

⇒ log_{x}(ac) = log_{x}(b)^{2}

On comparing, ac = b^{2 }or b^{2} = ac

Hence, a, b, c are in G.P. and G.M. between a and c is b.

Question 14.

If log 2 = 0.3010, then find the value of log 200.

Solution:

log 2 = 0.3010 ……… (i)

log 200= log (2 x 10^{2})

= log 2 + log 10^{2}

= log 2 + 2 log 10

= 0.3010 + 2 x 1

= 2 + 0.3010

= 2.3010

Hence, log 200 = 2.3010

Question 15.

Find the value of log 0.001.

Solution:

= log (10^{-3}) = -3 log 10

= -3 x 1 = -3 or 3

Question 16.

If log 7 = 0.8451 and log 3 = 0.4771, then find log (21)^{5}.

Solution:

Given, log 7 = 0.8451 and log 3 = 0.4771

log (21)^{5} = log (3 x 7)^{5}

= 5 log (3 x 7)

= 5 log 3 + 5 log 7

= 5 x 0.4771 + 5 x 0.8451

= 2.3855 + 4.2255

= 6.6110

Hence, log (21)^{5} = 6.6110

Question 17.

Find the value of log 6 + 2 log 5 + log 4 – log 3 – log 2.

Solution:

log 6 + 2 log 5 + log 4 – log 3 – log 2

= (log 6 + log 5^{2} + log 4) – (log 3 + log 2)

= log (6 x 5^{2} x 4) – log 3 x 2

= log (5^{2} x 4)

= log (25 x 4) = log 100 = 2

Hence, log 6 + 2 log 5 + log 4 – log 3 – log 2 = 2

Question 18.

, then find the value of x.

Solution:

On comparing,

x = 100

Hence, x=100

Question 19.

Prove that:

log_{10} tan 1°. log_{10} tan 2°……. log_{10} tan 89° = 0

Solution :

L.H.S.

= log_{10} tan 1°. log_{10} tan 2°…. log_{10} tan 45° log_{10} tan 89°

= log_{10} tan 1°. log_{10} tan 2°…. log_{10} (1)…. log_{10} tan 89°

= log_{10}tan 1°- log_{10} tan 2°…. x 0 x…. log_{10} tan 89°

= 0

= R.H.S. Hence Proved.

Question 20.

Prove that:

log_{3}4 . log_{4}5 . log_{5} 6 . log_{6}7 . log_{7}8. log_{8} 9 = 2

Solution:

L.H.S.

= (log_{3}4 – log_{4}5)- (log_{5}6 . log_{6}7) (log_{7}8- log_{8}9)

= (log_{3}5 x log_{5} 7) x log_{7} 9

= log_{3}7 x log_{7}9

= log_{3}9 = log_{3}3^{2}

= 2 x 1= R.H.S. Hence Proved.

Question 21.

If log 52.04 = 1.7163, log 80.65 = 1.9066 and log 9.753 = 0.9891, then find the value of

Solution:

Given,

log 52.04= 1.7163

log 80.65 = 1.9066

log 9.753 = 0.9891

= log (52.04 x 80.65) – log 9.753

= log 52.04 + log 80.65 – log 9.735

= 1.7163 + 1.9066-0.9891

= 3.6229 – 0.9891

= 2.6338

Question 22.

If log 32.9= 1.5172, log 568.1 = 2.7544 and log 13.28 = 1.1232, then find the value of

.

Solution:

Given, log 32.9 = 1.5172, log 568.1 =2.7544 and log 13.28 = 1.1232

= log (13.28)_{3} – log (32.9 x 568.1)

= 3 log 13.28 – log 32.9 – log 568.1

= 3 x 1.1232 – 1.5172-2.7544

= 3.3696-4.2716

= – 0.9020 = – 1 + 0.0980

= 1 + 0.098 = 1.0980

Question 23.

log 2 = 0.3010 and log 3 = 0.4771, then find the value of log (0.06)6

Solution:

Given,

log 2 = 0.3010 and log 3 = 0.4771

= log (6 x 10^{-2})^{6}

= 6 log (2 x 3 x 10_{-2})

= 6 log 2 + 6 log 3 – 12 log 10

= 6 x 0.3010 + 6 x 0.4771 – 12 x 1

= 1.8060 + 2.8626 – 12

= 4.6686 – 12 = -7.3314

= – 8 + 0.6686

= \(\overline { 8 }\) + 0.6686 = \(\overline { 8 }\) .6686

Hence,

log (0.06)^{6} = \(\overline { 8 }\) .6686 or -7.3314

Question 24.

Prove that:

Solution:

Question 25.

in the sum and difference of logarithm.

Solution:

= log (11)^{3} – log{(5)^{7} x (7)^{5}}

= 3 log 11 – log (5)^{7} – log (7)^{5}

= 3 log 11 – 7 log 5 – 5 log 7

Question 26.

(a) If antilog 1.5662 = 36.83, then find the value of the following :

(i) antilog \(\overline { 1 }\).5662

(ii) antilog 2.5662

(iii) antilog \(\overline { 2 }\).5662

(b) Find the value of antilog (log x).

Solution:

(a) Given, antilog 1.5662 = 36.83

(i) ∵ Mantissa of 1.5662 and j.5662 are same. So, anitlog of two number will be same digit numbers.

Now, characteristics of which number is 1 then there will be no zero after decimal point in antilog.

∴ Required antilog \(\overline { 1 }\).5662 = 0.3683.

(ii) ∵ Mantissas of 1.5662 and 2.5662 are same. So, antilog of two numbers will be same digit number.

Now, characteristic of which number is 2 then decimal point will be after 3 digit in antilog.

∴ Required anitlog 2.5662 = 368.3.

(iii) ∵ Mantìssa of 1.5662 and \(\overline { 2 }\).5662 are same. So. antilog of two numbers will be same digit number.

Now, characteristic of which number is \(\overline { 2 }\) then there will be one zero after decimal point in anitlog.

∴ Required antilog \(\overline { 2 }\).5662 = 0.03683.

(b) antilog and log are opposite to each other.

∴ antilog (log x) x.

Question 27.

Find (17)^{1/2} whereas log17 = 1.2304 and antilog 0.6152 = 4.123.

Solution:

Let(17)^{1/2} = x

Taking log on both sides,

log (17)^{1/2} = logx

⇒ log^{1/2} = logx

⇒ \(\frac { 1 }{ 2 }\) log17 = log x

⇒ log x = 0.6152

Taking antilog on both sides,

antilog (log x) = antilog 0.6152

⇒ x = 4.123

Hence, (17)^{1/2} = 4.123

Question 28.

l0g_{10} = 0.4771, then find log_{10} 0.027.

Solution:

Given,

log_{10}3 = 0.4771

log_{10} 0.027 = log( \(\frac { 27 }{ 1000 }\))

= log_{10} (27 x 10^{-3})

= log_{10} (3^{3}x 10^{-3})

= log_{10}(3)^{3} + log_{10} (10)^{-3}

= 3 log_{10}3 – 3 log_{10}10

= 3 x 0.4771 – 3 x 1

= 1.4313 – 3

= – 1.5687

= -2 + 0.4313

= \(\overline { 2 }\) + 0.4313 = \(\overline { 2 }\).4313

Hence, log_{10} 0.027 = latex]\overline { 2 }[/latex].4313 or – 1.5687

Question 29.

By using logarithm, find the value of

Solution:

Taking log on both sides,

⇒ log (520.4 x 8.065) – log 97.53 = logx

⇒ log 520.4 + log 8.065 – log 97.53 = logx …..(i)

⇒ 2.7163 + 0.9066 – 1.9881 = logx

⇒ 3.6229 – 1.9881 = logx

⇒ 1.6348 = logx

∴ x = antilog 1.6348 = 43.13

Question 30.

If log x – log (x – 1) = log 3, then find the value of x.

Solution:

log x – log (x – 1) = log 3

⇒ log( \(\frac { x }{ x – 1 }\)) = log 3

On comparing,

\(\frac { x }{ x – 1 }\) = 3

⇒ x = 3 (x – 1)

⇒ x = 3x – 3

⇒ 3x – x = 3

⇒ 2x = 3

⇒ x = 3/2

Hence, the value of x is 3/2.

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