Using antilogarithmic table find the value of the following.

i. antilog 3.072

ii. antilog 1.759

iii. antilog

iv. antilog

v. antilog 0.2732

vi. antilog

(i) Characteristic is 3

Mantissa is 0.072

From the antilog table antilog 0.072 = 1.180

Now as the characteristic is 3, therefore we will place the decimal after 3+1=4 numbers in 1180

∴ antilog 3.072 = 1180

(ii) Characteristic is 1

Mantissa is 0.759

From the antilog table antilog 0.759 = 5.741

Now as the characteristic is 1, therefore we will place the decimal after 1+1=2 numbers in 5741

∴ antilog 1.759 = 57.41

(iii) Characteristic is ̅1 = –1

Mantissa is 0.3826

From the antilog table antilog 0.382 = 2.410

Mean Value of 6 is 0.003

Thus, antilog 0.3826 = 2.410+0.003 = 2.413

Now as the characteristic is –1, therefore we will move decimal

–1+1=0 places left in 2.413

∴ antilog ̅1.3826 = 0.2413

(iv) Characteristic is ̅3 = –3

Mantissa is 0.6037

From the antilog table antilog 0.603 = 4.009

Mean Value of 7 is 0.006

Thus, antilog 0.6037 = 4.009+0.006 = 4.015

Now as the characteristic is –3,

therefore we will move decimal

–3+1=2 places left in 4.015

∴ antilog ̅3.6037 = 0.004015

(v) Characteristic is 0

Mantissa is 0.2732

From the antilog table antilog 0.273 = 1.875

Mean value 2 is 0.001

Thus, antilog 0.2732 = 1.875+0.001 = 1.876

Now as the characteristic is 0, therefore we will place the decimal after 0+1=1 numbers in 1876

∴ antilog 0.2732 = 1.876

(vi) Characteristic is ̅2 = –2

Mantissa is 0.1798

From the antilog table antilog 0.179 = 1.510

Mean Value of 8 is 0.003

Thus, antilog 0.1798 = 1.510+0.003 = 1.513

Now as the characteristic is –2, therefore we will move decimal

–2+1=1 places left in 1.513

∴ antilog ̅2.1798 = 0.01513