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RBSE Solutions for Class 11 Maths Chapter 9 Logarithms Ex 9.2

June 6, 2019 by Fazal Leave a Comment

Rajasthan Board RBSE Class 11 Maths Chapter 9 Logarithms Ex 9.2

Question 1.
Prove that:
log 630 = log 2 + 2 log 3 + log 5 + log 7.
Solution:
L.H.S. = log 630
= log (2 x 3 x 3 x 5 x 7)
= log (2 x 32 x 5 x 7)
= log 2 + log 32 + log 5 + log 7
= log 2 + 2 log 3 + log 5 + log 7
Hence Proved.

Question 2.
Prove that:
RBSE Solutions for Class 11 Maths Chapter 9 Logarithms
Solution:
RBSE Solutions for Class 11 Maths Chapter 9 Logarithms

Question 3.
Prove that:
log 10 + log 100 + log 1000 + log 10000 = 10
Solution:
L.H.S.
= log 10 + log 102 + log 103 + log 104
= log 10 + 2 log 10 + 3 log 10 + 4 log 10
= 10 log 10
= 10 x 1 (∵ log 10 = 1)
= 10
= R.H.S.
Hence Proved.

Question 4.
If log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451 and log 11 = 1.0414, then find the value of the following :
RBSE Solutions for Class 11 Maths Chapter 9 Logarithms
Solution:
(i) log 36 = log (2 x 2 x 3 x 3)
= log (22 x 32)
= log 22 + log 32
= 2 log 2 + 2 log 3
= 2(log 2 + log 3)
= 2(0.3010 + 0.4771)
= 2(0.7781)
= 1.5562

RBSE Solutions for Class 11 Maths Chapter 9 Logarithms
= log(2 x 3 x 7) – log 11
= log 2 + log 3 + log 7 – log 11
= 0.3010 + 0.4771 + 0.8451 – 1.0414
= 1.6232 – 1.0414
= 0.5818

RBSE Solutions for Class 11 Maths Chapter 9 Logarithms
= 5(log 11 – log 7)
= 5(1.0414 – 0.8451)
= 5(0.1963)
= 0.9815

(iv) log 70 = log (7 x 10)
= log 7 + log 10
= 0.8451 + 1
= 1.8451

RBSE Solutions for Class 11 Maths Chapter 9 Logarithms
= log 112 – log( 12 x 10)
= 2 log 11 – log 12 – log 10
= 2(1.0414)- log (2 x 2 x 3) – 1
= 2.0828 – 1 – 2 log 2 – log 3
= 1.0828 – 2(0.3010) – 0.4771
= 1.0828 – 0.6020 – 0.4771
= 1.0828 – 1.0791
= 0.0037

RBSE Solutions for Class 11 Maths Chapter 9 Logarithms

Question 5.
Find the value of x from following equation
logx4 + logx16 + logx64 = 12
Solution :
logx 4 + logx 16 + logx 64 = 12
⇒ logx 4 + logx 42 + log, 43 = 12
⇒ logx 4 + 2 logx 4 + 3 logx 4=12
⇒ 6 logx 4 = 12
⇒ logx 4 = 2
⇒ logx 2x = 2
⇒ 2 logx 2 = 2
⇒ logx 2 = 1
⇒ logx 2 = log2 2
On comparing, x = 2

Question 6.
Solve the equation :
log (x + 1) – log (x – 1) = 1
Solution:
log (x + 1) – log (x – 1) = 1
RBSE Solutions for Class 11 Maths Chapter 9 Logarithms

Question 7.
Find the value of 32-log34
Solution:
32 – log34 = 32 log33 – log34
= 3log3 32 – log34
= 3log3 9 – log3 4
= 3log3(9/4)
If log M = – 2, 1423 then to get its characteristic and mantissa is made positive as following:
= \(\frac { 9 }{ 4 }\) = 2 \(\frac { 1 }{ 4 }\) (∵ alogax = x)

Question 8.
Give the solution of following questions in one term :
(i) log 2 + 1
(ii) log 2x + 2 log x
Solution : (i) log 2 + 1
= log 2 + log 10 (∵ log1010 = 1)
= log (2 x 10)
= log 20

(ii) log 2x + 2 log x
= log 2x + log (x)2
= log (2x × x2)
= log2x3

Question 9.
Prove that:
(i) log53 . log34 . log25 = 2
(ii) logax × logby = logbx × loga y.
Solution:
(i) L.H.S. = log53 . log34. log25
RBSE Solutions for Class 11 Maths Chapter 9 Logarithms

RBSE Solutions for Class 11 Maths

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