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) \(\overline {1 }\) .301
(v) \(\overline { 3 }\) .2462
(vi) \(\overline { 2 }\) .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) \(\overline {1 }\) .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 \(\overline { 1 }\). Thus 1-1=0.
(d) The number obtained in step (b) will be written as 0.20000.
Thus antilog \(\overline { 1 }\).301 = 0.2000.
(v) \(\overline { 3 }\) .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 \(\overline { 3 }\). 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 \(\overline { 3 }\).2462 = 0.001763.
(vi) \(\overline { 2 }\) .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 = \(\overline { 2 }\). 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 \(\overline { 2 }\).466 = 292.4
Question 2.
Find the value of :
(i) antilog 3.1234
(ii) antilog \(\overline { 2 }\).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 \(\overline { 2 }\).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 \(\overline { 2 }\). 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 \(\overline { 2 }\).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 = \(\overline { 2 }\).6727
(ii) log x = 0.452
Solution:
(i) log x = \(\overline { 2 }\).6727
Taking anitlog on both sides
antilog (log x) = antilog \(\overline { 2 }\).6727
⇒ x = antilog \(\overline { 2 }\).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 \(\overline { 2 }\). 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 \(\overline { 2 }\).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