RBSE Solutions for Class 11 Maths Chapter 9 Logarithms

RBSE Solutions for Class 11 Maths Chapter 9 Logarithms

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 = 1/25
Solution:
Logarithm form of 5^-2 = 1/25 is
log₅(1/25) = -2.

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

Question 6.
4^3/2 = 8
Solution:
Logarithm form of 4^3/2 = 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( 1/27) = -3
Solution:
Exponential form of log₃( 1/27) = -3 is 3^-3 = 1/27

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

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

Question 14.
If log₁₂₅P = 1/6 then find the value of P.
Solution:
log₁₂₅ P = 1/6
⇒ P = (125)^1/6
⇒ P = [(5)³]^1/6
⇒ P = 5^1/2 = √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 log_m mⁿ = n log_mm
and log_mm = 1
∴ log_m(m)_n = n x log_mm = n x 1 = n
⇒ log_m(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) (∵ log_mm = 1)
= 0 = R.H.S.
Hence Proved.

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:
= 9/4 = 2 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

Rajasthan Board RBSE Class 11 Maths Chapter 9 Logarithms Ex 9.3

Question 1.
Find the characteristic of logarithm of following numbers :
(i) 1270
(ii) 20.125
(iii) 7.985
(iv) 431.5
(v) 0.02
(vi) 0.02539
(vii) 70
(viii) 0.000287
(ix) 0.005
(x) 0.00003208
(xi) 0.000485
(xii) 0.007
(xiii) 0.0005309
Solution:
(i) Number 1270 is 4 digit number.
So, characteristic of its logarithm will be 4 – 1 = 3.

(ii) In 20.125, integral part is 20 which contains 2 digit.
So, characteristic of its logarithm will be 2 – 1 = 1.

(iii) In 7.985, integral part is 7 which contains 1 digit.
So, characteristic of its logarithm will be 1 – 1 = 0.

(iv) In 431.5, integral part is 431 which contains 3 digit.
So, characteristic of its logarithm will be 3 – 1 = 2.

(v) In 0.02, there are 1 zero between decimal point and first significant digit.
So, characteristic of its logarithm will be – (1 + 1) = – 2 or 2¯¯¯ .

(vi) In 0.02539, there are zero between decimal point and first significant digit.
So, characteristic of its logarithm will be – (1 + 1) = – 2 or 2¯¯¯.

(vii) Number 70 is 2 digit number.
So, characteristic of its logarithm will be 2 – 1 = 1.

(viii) In 0.000287, there are 3 zero between decimal point and first significant digit.
So, characteristic of its logarithm will be -(3 + 1) = -4 or 4¯¯¯.

(ix) In 0.005, there are 2 zero between decimal point and first significant digit.
So, characteristic of its logarithm will be – (2 + 1) = – 3 or 3¯¯¯.

(x) In 0.00003208, there are 4 zero between decimal point and first significant digit.
So, characteristic of its loga¬rithm will be – (4 + 1) = – 5 or 5¯¯¯ .

(xi) In 0.000485, there are 3 zero between decimal point and first significant digit.
So, characteristic of its logarithm will be – (3 + 1) = -4 or 4¯¯¯.

(xii) In 0.007, there are 2 zero between decimal point and first significant digit.
So, characteristic of its logarithm will be – (2 + 1) = – 3 or 3¯¯¯.

(xiii) In 0.0005309, there are 3 zero between decimal point and first significant digit.
So, characteristic of its logarithm will be – (3 + 1 ) = – 4 or 4¯¯¯.

Question 2.
Find the logarithm of the following numbers by using log table :
(i) 2813
(ii) 400
(iii) 27.28
(iv) 9
(v) 0.678
(vii) 0.08403
(viii) 0.000287
(ix) 1.234
(x) 0.00003258
(xi) 0.000125
(xii) 0.00003208
Solution:
(i) 2813
Characteristic : There are characteristic will be 4 – 1 = 3.
Mantissa : In log table, in first column, in front of 28 and under the column 1.
Find the number which is = 4487.
For fourth digit 3’s mean difference = 5
On adding = 4492
So, mantissa of log₁₀ 2813 = 0.4492
Thus, log₁₀ 2813 = Characteristic + Mantissa
= 3 + 0.4492 = 3.4492

(ii) 400
Characteristic : There are 3 digit in number 400. So, characteristic will be 3 – 1 = 2.
Mantissa : In log table, in first column, in front of 40 and under the column 0, find the number which is 6021.
So, mantissa of log₁₀ 400 = 0.6021
Thus, log₁₀ 400 = Characteristic + Mantissa
= 2 + 0.6021 =2.6021

(iii) 27.28
Characteristic : Here, integral part is 27 which contains 2 digit. So, characteristic of its logarithm will be 2 – 1 = 1.
Mantissa : In log table, in first column, in front of 27 (ignoring decimal point) and under the column 2, find the number which is = 4346
For fourth digit 8’s mean difference = 13
On adding = 4359
So, mantissa of log₁₀ 27.28 = 0.4359
Thus, log₁₀ 27.28 = Characteristic + Mantissa
= 1+ 0.04359 =1.4359

(iv) 9
Characteristic : There are only 1 digit in number 9. So, characteristic will be 1 – 1 = 0.
Mantissa : In log table, in first column, in front of 90 and under the column 0 find the number which is = 9542.
So mantissa of log10 9 = 0.9542
Thus, log₁₀ 9 = Characteristic + Mantissa
= 0 + 0.9542 = 0.9542

(v) 0-678
Characteristic : In 0.678, there are no zero between decimal point and first significant digit. So, characteristic of its logarithm will be – (0+ 1) = – 1 = 1
Mantissa : In log table, in first column, in front of 67 (ignoring decimal point) and under the column 8 find the num-ber which is 8312.
So, mantissa of log₁₀ 0.678 = 0.8312
Thus,
log₁₀ 0.678 = Characteristic + Mantissa
= 1¯¯¯ + 0.8312 = 1¯¯¯. 8312

(vi) 0.0035
Characteristic : In 0.0035, there are 2 zero between decimal point and first significant digit. So, characteristic of its logarithm will be
-(2+1) = -3 = 3¯¯¯
Mantissa : In log table, in first column, in front of 35 (ignoring decimal point) and under the column 0, find the number which is = 5441
So, mantissa of log₁₀0.0035 = 0.5441
Thus, log₁₀ 0.0035 = Characteristic + Mantissa
= 3¯¯¯ + 0.5441 = 3¯¯¯.5441

(vii) 0.08403
Characteristic : In 0.08403, there are 1 zero between decimal point and first significant digit. So, characteristic of its logarithm will be
-(1 + 1) = – 2 = 2
Mantissa : In log table, in first column, in front of 84 (ignoring decimal point) and under the column 0 find the num-ber, which is = 9243
For fourth digit 3 mean difference = 2
On adding = 9245
So, mantissa of log₁₀0.08403 = 0.9245
Thus,
log₁₀ 0.08403 = Characteristic + Mantissa
= 2¯¯¯ + 0.9245 = 2¯¯¯.9245.

(viii) 0.000287
Characteristic: In 0.000287, there are 3 zero between decimal point and first significant digit. So, characteristic of its logarithm will be
-(3 + 1) = -4 = 4¯¯¯
Mantissa : In log table, in first column, in front of 28 (ignoring decimal point) and under the column 7 find the number which is 4579.
So, mantissa of log₁₀ 0.000287 = 0.4579
Thus,
log₁₀ 0.000287 = Characteristic + Mantissa
= 4¯¯¯ + 0.4579
= 4¯¯¯.4579

(ix) 1.234
Characteristic = 1- 1 = 0
Mantissa = 0.0899 + 14
= 0.0913
log₁₀1.234 = Characteristic + Mantissa
= 0 + 0.0913
= 0.0913

(x) 0.00003258
Characteristic: In 0.00003258, there are 4 zero between decimal point and first significant digit. So characteristic of its logarithm will be
-(4 + 1) = -5 = 5
Mantissa : In log table, in first column, in front of 32 (ignoring decimal point) and under the column 5 find the num-ber, which is = 5119
For fourth digit 8 mean difference = 11
On adding = 5130
So, mantissa of log₁₀ 0.00003258 = 0-5130
Thus,
log₁₀ 0.00003258 = Characteristic + Mantissa
= 5¯¯¯ + 0.5130
= 5¯¯¯.5130

(xi) 0.000125
Characteristic : In 0.000125, there are 3 zero between decimal point and first significant digit. So, characteristic of its logarithm will be
– (3 + 1) = – 4 = 4¯¯¯

Mantissa : In log table, in first column, in front of 12 (ignoring decimal point) and under the column 5 find the num-ber which is = 0969
So, mantissa of log₁₀0.000125 = 0.0969
Thus,
log₁₀o 0.000125 = Characteristic + Mantissa
= 4¯¯¯ + 0.0969
= 4¯¯¯.0969

(xii) 0.00003208
Characteristic: In 0-00003208, there are 4 zero between decimal point and first significant digit. So, characteristic of its logarithm will be
-(4 + 1) = -5 = 5
Mantissa : In log table, in first column, in front of 32 (ignoring decimal point) and under the column 0 find the num-ber which is = 5051
For fourth digit 8’s mean difference = 11
On adding = 5062
So, mantissa of log₁₀0.0003208 = 0.5062
Thus,
log₁₀0.0003208 = Characteristic + Mantissa = 5¯¯¯ + 0.5062
= 5¯¯¯ .5062

Rajasthan Board RBSE Class 11 Maths Chapter 9 Logarithms Ex 9.4

Question 1.
Find the antilog of the following numbers :
(i) 1.3210
(ii) 2.4127
(iii) 0.084
(iv) 1¯¯¯ .301
(v) 3¯¯¯ .2462
(vi) 2¯¯¯ .0258
Solution:
(i) 1.3210
(a) Mantissa of given number is 0.3210.
(b) Common number of row of 0.32 and column of 1, is 2094.
(c) Characteristic of given number is 1. So antilog of number will be 2 digit number.
(d) The number obtained in step (b) will be written as 20.94.
Thus antilog 1.3210 = 20.94.

(ii) 2.4127
(a) Mantissa of given number is 0.4127.
(b) Common number of row of 0.41 and column of 2, is 2582.
(c) In the same line, in the mean difference column of head 7, number is 4.
(d) Sum of step (b) and (c) = 2586.
(e) Characteristic of given number is 2. So antilog of number will be 3 digit number.
(f) The number obtained in step (d) will be written as 258.6.
Thus antilog 2.4127 = 258.6.

(iii) 0.084
(a) Mantissa of given number is 0.084.
(b) Common number of row of 0.08 and column of 4, is 1213.
(c) Characteristic of given number is 0. So anitlog of number will be 1 digit number.
(d) The number obtained in step (b) will be written as 1.213.
Thus anitlog 0.084 = 1.213.

(iv) 1¯¯¯ .301
(a) Mantissa of given number is 0.301.
(b) Common number of row of 0.30 and column of 1, is 2000.
(c) Characteristic of given number is 1¯¯¯. Thus 1-1=0.
(d) The number obtained in step (b) will be written as 0.20000.
Thus antilog 1¯¯¯.301 = 0.2000.

(v) 3¯¯¯ .2462
(a) Mantissa of given number is 0.2462.
(b) Common number of row of 0.24 and column of 6, is 1762.
(c) In the same line, in the mean difference column of head 2, number is 1.
(d) Sum of step (b) and (c) = 1762 + 1 = 1763.
(e) Characteristic of given number is 3¯¯¯. Thus = 3 – 1 = 2.
(f) The number obtained in step (d) will be written by two zero of right side as 0.001763.
Thus Antilog 3¯¯¯.2462 = 0.001763.

(vi) 2¯¯¯ .0258
(a) Mantissa of given number is 0.0258.
(b) Common number of row of 0.02 and column of 5, is 1059.
(c) In the same line, in the mean difference column of head 8, number is 2.
(d) Sum of step (b) and (c) = 1059 + 2 = 1061.
(e) Characteristic of given number is = 2¯¯¯. Thus 2 – 1 = 1.
(f) The number obtained in step (d) will be written by one zero of right side as 0.01061.
Thus Antilog 2¯¯¯.466 = 292.4

Question 2.
Find the value of :
(i) antilog 3.1234
(ii) antilog 2¯¯¯.5821
(iii) antiiog 0.3
(iv) antilog 2.466
Solution:
(i) antilog 3.1234
(a) Mantissa of given number is 0.1234.
(b) In antilog table, common number of row of 0.12 and column of 3. is 1327.
(c) In the same line, in the mean difference column of head 4, number is 1.
(d) Sum of step (b) and (c) = 1327 + 1 = 1328.
(e) Characteristic of given number is 3. So, antilog of number will be 4 digit number.
(f) The number obtained in step (d) will be written as 1328.0.
Thus Antilog 3.1234 = 1328.0

(ii) antilog 2¯¯¯.5821
(a) Mantissa of given number is 0.5821.
(b) In antilog table, common number of row of .58 and column of 2, is 3819.
(c) In the same line, in the mean difference column of head l, number is 1.
(d) Sum of step (b) and (c) = 3819 + 1 = 3820.
(e) Characteristic of given number is 2¯¯¯. Thus 2 – 1 = 1.
(f) The number obtained in step (d) will be written by one zero of right side as 0.033820.
Thus, Antilog 2¯¯¯.5821 = 0.0382.

(iii) antilog 0.3
(a) Mantissa of given number is 0.3000.
(b) In antilog table, common number of row of 0.30 and column of 0 is 1995.
(c) Characteristic of given number is 0. So, antilog of number will be 1 digit number.
(d) The number obtained in step (b) will be written as 1.995.
Thus Antilog 0.3 = 1.995.

(iv) antilog 2.466
(a) Mantissa of given number is 0.466.
(b) In antilog table, common number of row of 0.46 and column of 6 is 2924.
(c) Characteristic of given number is 2. So antilog of number will be 3 digit number.
(d) The number obtained in step (b) will be written as 292.4.
Thus Antilog 2.466 = 292.4.

Question 3.
Find the value of x in the following :
(i) log x = 2¯¯¯.6727
(ii) log x = 0.452
Solution:
(i) log x = 2¯¯¯.6727
Taking anitlog on both sides
antilog (log x) = antilog 2¯¯¯.6727
⇒ x = antilog 2¯¯¯.6727
(a) Mantissa of given number is 0.6727.
(b) In antilog table, common number of row of 0.67 and column of 2, is 4699.
(c) In the same line, in the mean difference column of lead 7, number is 8.
(d) Sum of step (b) and (c) = 4699 + 8 = 4707.
(e) Characteristic of given number is 2¯¯¯. Thus 2 – 1 = 1.
(f) The number obtained in step (d) will be written by one zero of right side as 0.04707. Thus
antilog 2¯¯¯.6727 = 0.4707.

(ii) log x = 0.452
Taking antilog on both sides
antilog (log x) = antilog (0.452)
⇒ x = antilog (0.452)
(a) Mantissa of given number is 0.452.
(b) In antilog table, common number of row of 0.45 and column of 2, is 2831.
(c) Characteristic of given number is 0. So, antilog of number will be l digit number.
(d) The number obtained in step (b) will be written as 2.831. Thus
antilog 0.452 = 2.831
Hence, x = 2.831

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) 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) 1/logba
(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(a/b)
Comparing on both sides,
⇒ a – b = 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
= 8¯¯¯ + 0.6686 = 8¯¯¯ .6686
Hence,
log (0.06)⁶ = 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 1¯¯¯.5662
(ii) antilog 2.5662
(iii) antilog 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 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 2¯¯¯.5662 are same. So. antilog of two numbers will be same digit number.
Now, characteristic of which number is 2¯¯¯ then there will be one zero after decimal point in anitlog.
∴ Required antilog 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
⇒ 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( 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
= 2¯¯¯ + 0.4313 = 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( x/x–1) = log 3
On comparing,
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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