RBSE Solutions Class 11 Maths Chapter 9 Logarithms Exercise 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) \( \overline {2}.0258 \)
Answer:
(i) For 1.3210:
(a) The mantissa of the given number is 0.3210.
(b) From the antilog table, find the common number in row 0.32 and column 1, which is 2094.
(c) The characteristic of the given number is 1. This means the antilog will be a 2-digit number before the decimal point.
(d) The resulting number is 20.94.
(ii) For 2.4127:
(e) The characteristic of the given number is 2. This means the antilog will be a 3-digit number before the decimal point.
(f) Using the mantissa 0.4127 in the antilog table, the corresponding number found is 2586 (this step is inferred). The number obtained after placing the decimal is 258.6.
Thus, antilog 2.4127 = 258.6.
(iii) For 0.084:
(a) The mantissa of the given number is 0.084.
(b) From the antilog table, find the common number in row 0.08 and column 4, which is 1213.
(c) The characteristic of the given number is 0. This means the antilog will be a 1-digit number before the decimal point.
(d) The resulting number is 1.213.
Thus, antilog 0.084 = 1.213.
(iv) For \( \overline {1}.301 \):
(a) The mantissa of the given number is 0.301.
(b) From the antilog table, find the common number in row 0.30 and column 1, which is 2000.
(c) The characteristic of the given number is \( \overline {1} \). This means there will be 0 zeros after the decimal point before the first significant digit (as \( |-1| - 1 = 0 \)).
(d) The resulting number is 0.2000.
Thus, antilog \( \overline {1}.301 \) = 0.2000.
(v) For \( \overline {3}.2462 \):
(a) The mantissa of the given number is 0.2462.
(b) From the antilog table, find the common number in row 0.24 and column 6, which is 1762.
(c) In the same line, find the mean difference for column 2, which is 1.
(d) Sum of step (b) and (c) = 1762 + 1 = 1763.
(e) The characteristic of the given number is \( \overline {3} \). This means there will be 2 zeros after the decimal point before the first significant digit (as \( |-3| - 1 = 2 \)).
(f) The resulting number is 0.001763.
Thus, antilog \( \overline {3}.2462 \) = 0.001763.
(vi) For \( \overline {2}.0258 \):
(a) The mantissa of the given number is 0.0258.
(b) From the antilog table, find the common number in row 0.02 and column 5, which is 1059.
(c) In the same line, find the mean difference for column 8, which is 2.
(d) Sum of step (b) and (c) = 1059 + 2 = 1061.
(e) The characteristic of the given number is \( \overline {2} \). This means there will be 1 zero after the decimal point before the first significant digit (as \( |-2| - 1 = 1 \)).
The final number is 0.01061.
In simple words: To find the antilog, we look up the mantissa (the decimal part) in the antilog table. The characteristic (the whole number part) tells us where to put the decimal point in the final answer. If the characteristic is negative, like \( \overline{1} \) or \( \overline{2} \), it means we place zeros after the decimal point before the first main number.

🎯 Exam Tip: Remember that for a negative characteristic \( \overline{N} \), the number of zeros after the decimal and before the first non-zero digit is \( N - 1 \). For a positive characteristic \( N \), the number of digits before the decimal is \( N + 1 \).

Question 2. Find the antilog of:
(i) 3.1234
(ii) \( \overline {2}.5821 \)
(iii) 0.3
(iv) 2.466
Answer:
(i) For antilog 3.1234:
(a) The mantissa of the given number is 0.1234.
(b) In the antilog table, the common number in row 0.12 and column 3 is 1327.
(c) In the same line, the mean difference for column 4 is 1.
(d) Sum of step (b) and (c) = 1327 + 1 = 1328.
(e) The characteristic of the given number is 3. This means the antilog will be a 4-digit number before the decimal point.
(f) The resulting number is 1328.0.
Thus, Antilog 3.1234 = 1328.0.
(ii) For antilog \( \overline {2}.5821 \):
(a) The mantissa of the given number is 0.5821.
(b) In the antilog table, the common number in row 0.58 and column 2 is 3819.
(c) In the same line, the mean difference for column 1 is 1.
(d) Sum of step (b) and (c) = 3819 + 1 = 3820.
(e) The characteristic of the given number is \( \overline {2} \). This means there will be 1 zero after the decimal point before the first significant digit (as \( |-2| - 1 = 1 \)).
(f) The resulting number is 0.03820.
Thus, Antilog \( \overline {2}.5821 \) = 0.0382.
(iii) For antilog 0.3:
(a) The mantissa of the given number is 0.3000.
(b) In the antilog table, the common number in row 0.30 and column 0 is 1995.
(c) The characteristic of the given number is 0. This means the antilog will be a 1-digit number before the decimal point.
(d) The resulting number is 1.995.
Thus, Antilog 0.3 = 1.995.
(iv) For antilog 2.466:
(a) The mantissa of the given number is 0.466.
(b) In the antilog table, the common number in row 0.46 and column 6 is 2924.
(c) The characteristic of the given number is 2. This means the antilog will be a 3-digit number before the decimal point.
In simple words: Each step helps us build the antilog value. We find the numerical digits from the mantissa in the table, then use the characteristic to correctly place the decimal point.

🎯 Exam Tip: Pay close attention to the mean difference column when using antilog tables, as it adds a small correction to the main value.

Question 3. Solve for x:
(i) log x = \( \overline {2}.6727 \)
(ii) log x = 0.452
Answer:
(i) For log x = \( \overline {2}.6727 \):
To find x, we take the antilog of both sides:
Antilog (log x) = Antilog \( \overline {2}.6727 \)
\( \implies \) x = Antilog \( \overline {2}.6727 \)
(a) The mantissa of the given number is 0.6727.
(b) In the antilog table, the common number in row 0.67 and column 2 is 4699.
(c) In the same line, the mean difference for column 7 is 8.
(d) Sum of step (b) and (c) = 4699 + 8 = 4707.
(e) The characteristic of the given number is \( \overline {2} \). This means there will be 1 zero after the decimal point before the first significant digit (as \( |-2| - 1 = 1 \)).
(f) The resulting number is 0.04707.
Thus, Antilog \( \overline {2}.6727 \) = 0.04707.
(ii) For log x = 0.452:
To find x, we take the antilog of both sides:
Antilog (log x) = Antilog (0.452)
\( \implies \) x = Antilog (0.452)
(a) The mantissa of the given number is 0.452.
(b) In the antilog table, the common number in row 0.45 and column 2 is 2831.
(c) The characteristic of the given number is 0. This means the antilog will be a 1-digit number before the decimal point.
(d) The resulting number is 2.831.
Thus, Antilog 0.452 = 2.831.
Hence, x = 2.831.
In simple words: To find 'x' when 'log x' is given, you need to do the opposite of log, which is finding the antilog. You use the same steps as finding antilog for any number to get the value of 'x'.

🎯 Exam Tip: Always remember that \( x = \text{antilog}(y) \) is the inverse operation of \( \log x = y \). The process of finding antilog remains consistent whether it's directly asked or used to solve for x.