RBSE Solutions for Class 11 Maths Chapter 9 Logarithms Ex 9.2

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:

Solution:

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 :

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

= 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

= 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

= 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

Question 5.
Find the value of x from following equation
log_x4 + log_x16 + log_x64 = 12
Solution :
log_x 4 + log_x 16 + log_x 64 = 12
⇒ log_x 4 + log_x 42 + log, 43 = 12
⇒ log_x 4 + 2 log_x 4 + 3 log_x 4=12
⇒ 6 log_x 4 = 12
⇒ log_x 4 = 2
⇒ log_x 2^x = 2
⇒ 2 log_x 2 = 2
⇒ log_x 2 = 1
⇒ log_x 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

Question 7.
Find the value of 3^2-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 \(\frac { 1 }{ 4 }\) (∵ a^log_ax = 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) log_ax × log_by = log_bx × log_a y.
Solution:
(i) L.H.S. = log₅3 . log₃4. log₂5

RBSE Solutions for Class 11 Maths