RBSE Solutions for Class 11 Maths Chapter 9 Logarithms Miscellaneous Exercise

Rajasthan Board RBSE Class 11 Maths Chapter 9 Logarithms Miscellaneous Exercise

Question 1.
log_√2x = 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_√2x= 4
Then (√2)⁴ = x
⇒ x = (√2)⁴ = (2^1/2 )⁴ = 2² = 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_an = 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 = (3²)²
On comparing, x = 3² = 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_ba.log_cb.log_ac is :
(A) 0
(B) log abc
(C) 1
(D) log (a^a b^b c^c)
Solution:
log_bo.log_c6.log_ac

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_a0 = – ∞
If a > 1
Hence, option (A) is correct.

Question 7.
If a < 0 then value of log_a0 is :
(A) – ∞
(B) ∞
(C) 0
(D) 1
Solution:
log 0 = ∞
If a < 0
Hence, option (B) is correct.

Question 8.
Other form of Iog_ab :
(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 log₂7 is :
(A) Integer
(B) Rational
(C) Irrational
(D) Prime
Solution:
log₂ 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 = log₃ 5 and b = log₇ 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 log₂ x + log₂ (x – 1) = 1, then find the value of x.
Solution:
log₂x + log₂(x – 1) = 1
⇒ log₂ x(x – 1) = 1
⇒ x(x – 1) = 2¹
⇒ x² – x – 2 = 0
⇒ x² – 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 – b² = a
⇒ ab – a = b²
⇒ a(b – 1) = b²
⇒ a = b²/b – 1
Hence, value of a in terms of b is b²/b – 1.

Question 13.
If , then find relation among a,b and c.
Solution:
According to question,

⇒ log_xa + log_xc= 2 log_xb
⇒ log_x(ac) = log_x(b)²
On comparing, ac = b² or b² = 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 10²)
= log 2 + log 10²
= 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 5² + log 4) – (log 3 + log 2)
= log (6 x 5² x 4) – log 3 x 2

= log (5² 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:
log₁₀ tan 1°. log₁₀ tan 2°……. log₁₀ tan 89° = 0
Solution :
L.H.S.
= log₁₀ tan 1°. log₁₀ tan 2°…. log₁₀ tan 45° log₁₀ tan 89°
= log₁₀ tan 1°. log₁₀ tan 2°…. log₁₀ (1)…. log₁₀ tan 89°
= log₁₀tan 1°- log₁₀ tan 2°…. x 0 x…. log₁₀ tan 89°
= 0
= R.H.S. Hence Proved.

Question 20.
Prove that:
log₃4 . log₄5 . log₅ 6 . log₆7 . log₇8. log₈ 9 = 2
Solution:
L.H.S.
= (log₃4 – log₄5)- (log₅6 . log₆7) (log₇8- log₈9)
= (log₃5 x log₅ 7) x log₇ 9
= log₃7 x log₇9
= log₃9 = log₃3²
= 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.
l0g₁₀ = 0.4771, then find log₁₀ 0.027.
Solution:
Given,
log₁₀3 = 0.4771
log₁₀ 0.027 = log( \(\frac { 27 }{ 1000 }\))
= log₁₀ (27 x 10^-3)
= log₁₀ (3³x 10^-3)
= log₁₀(3)³ + log₁₀ (10)^-3
= 3 log₁₀3 – 3 log₁₀10
= 3 x 0.4771 – 3 x 1
= 1.4313 – 3
= – 1.5687
= -2 + 0.4313
= \(\overline { 2 }\) + 0.4313 = \(\overline { 2 }\).4313
Hence, log₁₀ 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.

RBSE Solutions for Class 11 Maths