CA CMA DISA(ICAI) B.COM
3956 Points
Joined November 2010
II Method: Using ex/e-x table
Let ex = y
then, ln(y) =x
Example 2 :
Find natural log of (i) 0.2567(ii) 0.3570 (iii)0.2466 (iv) 0.1979 and (v) 4.0960
Answers
(i) Consulting the table of values of ex/ e-x, we find :
0.2567 = e-1.36
Hence, natural log of 0.2567 = -1.36
(ii) Consulting the table of values of ex/e-x, we find :
0.3570 = e-1.03
Hence, natural log of 0.3570 = -1.03
(iii) Consulting the table of values of ex/e-x, we find :
0.2466 = e-1.40
Hence, natural log of 0.2466 = -1.40
(iv) Consulting the table of values of ex/e-x, we find :
0.1979 = e-1.62
Hence, natural log of 0.1979 = -1.62
(v) Consulting the table of values of ex/e-x, we find :
4.0960 = e1.41
Hence, natural log of 4.0960 = 1.41
Example 3 : Find the natural log of (a) 0.75 (b) 1.24 using ex/e-x table
Answer
(a) ln(0.75) = ?
e-0.28 = 0.7558
e-0.29 = 0.7483
0.7558 = e-0.28
0.7483 = e-0.29
For 0.0075 ↓ in LHS, e’s power ↓ by 0.01
For 1↓ in LHS, e’s power ↓ by 0.01/0.0075
For 0.0058 ↓ in LHS, e’s power ↓ by (0.01/0.0075)x0.0058 i.e. by 0.0077
= 0.7558 – 0.0058 = e-0.28-0.0077
0.75 = e-0.287686
ln 0.75 = -0.287686
(b) ln(1.24) = ?
e0.21 = 1.2337
e0.22 = 1.2461
1.2337 = e0.21
1.2461 = e0.22
For 0.0124 ↑ in LHS, e’s power ↑ by 0.01
For 1 ↑ in RHS, e’s power ↑ by 0.01/0.0124
For 0.0063 ↑ in RHS, e’s power ↑ by
(0.01/0.0124)x0.0063 i.e. 0.0050806
1.2337 +0.0063 = e0.21+0.0050806
= 1.24 = e0.215097
ln 1.24 = 0.215097