RBSE Solutions for Class 11 Maths Chapter 9 Logarithms Ex 9.1

Rajasthan Board RBSE Class 11 Maths Chapter 9 Logarithms Ex 9.1

(Q. 1 to 6) Write the following in logarithm form :

Question 1.
26 = 64
Solution:
Logarithm form of 26 = 64 is
log2 64 = 6

Question 2.
104 = 10000
Solution:
Logarithm form of 104 = 10000 is
log1010000 = 4

Question 3.
210 = 1024
Solution:
Logarithm form of 210= 1024 is
log2 1024 = 10

Question 4.
5-2 = \(\frac { 1 }{ 25 }\)
Solution:
Logarithm form of 5-2 = \(\frac { 1 }{ 25 }\) is
log5(\(\frac { 1 }{ 25 }\)) = -2.

Question 5.
10-3 = 0.001
Solution:
Logarithm form of 10-3 = 0.001 is
log100.001 = – 3

Question 6.
43/2 = 8
Solution:
Logarithm form of 43/2 = 8 is
log4 8 = 3/2

(Q. 7 to 12) Write the following in the power form :

Question 7.
log5 25 = 2
Solution:
Exponential (power) form of log5 25 = 2 is 52 = 25

Question 8.
log3 729 = 6
Solution:
Exponential (power) form of log3 729 = 6 is 36 = 729.

Question 9.
log10 0.001 = – 3
Solution:
Exponential (power) form of log10 0.001 = – 3 is
10-3 = 0.001

Question 10.
log10 0.1 = – 1
Solution:
Exponential form of log10 0.1 = – 1 is 10-1 = 0.1

Question 11.
log3( \(\frac { 1 }{ 27 }\)) = -3
Solution:
Exponential form of log3( \(\frac { 1 }{ 27 }\)) = -3 is 3-3 = \(\frac { 1 }{ 27 }\)

Question 12.
log√24 = 4
Solution:
Exponential (power) form of log√24 = 4 is (√2)4 = 4

Question 13.
If log81x = \(\frac { 3 }{ 2 }\), then find the value of x.
Solution:
log81x = \(\frac { 3 }{ 2 }\)
⇒ x = (81)3/2
= (9²)3/2 = (9)3 = 729
Hence, the value of x is 729.

Question 14.
If log125P = \(\frac { 1 }{ 6 }\) then find the value of P.
Solution:
log125 P = \(\frac { 1 }{ 6 }\)
⇒ P = (125)1/6
⇒ P = [(5)3]1/6
⇒ P = 51/2 = √5
Hence, the value of P is √5.

Question 15.
If log4 m = 1.5, then find the value of m.
Solution:
log4 m = 1.5
⇒ m = (4)1.5
⇒ m = (22 )1.5
⇒ m = 23 = 8
Hence, the value of m is 8.

Question 16.
Prove that :
log4[log2{log2 (log3 81)}] = 0
Solution:
We know that logm mn = n logmm
and logmm = 1
∴ logm(m)n = n x logmm = n x 1 = n
⇒ logm(m)n = n …(i)
L.H.S. = log4[log2{log2(log3 81)}]
= log4[log2 {log2(log334)}] (∵ 81 = 34)
= log4[log2 {(log24)} [According to equal (i)]
= log4{log2(log222)} (∵ 4 = 22)
= log4(log22) [According to equal (i)]
= log4(1) (∵ logmm = 1)
= 0 = R.H.S.
Hence Proved.

RBSE Solutions for Class 11 Maths