Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 2.31

Example 40 : Given log 10 2 = 0.30103, find log 10 (1000/256)
solution:

10 10 10
log^1000 log 1000 log 256
256 = −^
=log 10 log 2 10 3 − 10 9
= 3 log 10 10 – 9 log 10 2 = 3 – 9 (0.30103) = 3 – 2.70927 = 0.29073.
Example 41 : Find the value of : (i) 0.8176 × 13.64, (ii) (789.45)1/8
solution:
(i) Let x= 0.8176 × 13.64; taking log on both sides.
Log x = log (0.8176 ×13.64) = log 0.8176 + log 13.64
= 1.9125 1.1348 1 0.9125 1 0.1348+ = − + + +
= 0.9125 + 0.1348 = 1.0473
∴ x = antilog 1.0473 = 11.15.

(ii) Let x = (789.45)1/8 or, logx log 789.45=^18 ( )= 81 (2.8973 0.3622)=
∴ x = antilog 0.3622 = 2.302.
Note. Procedure of finding the mantissa of 5 significant figures will be again clear from the following example:
Mantissa of 7894 (see above) is 0.8973 and for the fifth digit (i.e. for digit 5), the corresponding number in the
mean difference table is the digit 2, which is less than 5 ; so 0 is to be added to the mantissa 0.8973.
It again, the corresponding mean difference number is
5 to 14, carry 1
15 to 24, carry 2 and so on.
Example 42 : Find, from tables, the antilogarithm of – 2.7080
solution:

• 2.7080 = 3 – 2.7080 – 3 = .2920 –3 = 3.2920

∴ antilog (– 2.7080) = antilog 3.2920 = 0.001959

SELF EXAMINATION QUESTIONS

1. Find the number of zeros between the decimal point and the first significant figures in :

(i) (0.0011)^20 (ii)

1 100

11

(iii) (12.4)–15 [Ans. 59 ; 104 ; 16]

2. Given log 8 = 0.931, log 9 = 0.9542 ; find the value of log 60 correct to 4 decimal 4 places.
   [Ans. 0.7781]


3. Given log 2 = 0.30103, log 3 = 0.47712 ; find the value of :

(i) log 4500 (ii) log 0.015 (iii) log 0.1875. [Ans. 3.65321 ; 2.17609 ; (iii) 1.27300