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Find the antilogarithm of each of the following: 2.7523 (ii) 3.7523 (iii) 5.7523 (iv) 0.7523 1.7523 (vi) 2.7523 (vii) 3.7523

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(i) The mantissa of `2.7523` is positive and is equal to ` 0.7523`. Now, in antilog table, look into the row starting ` 0.75`. In this row, look at the number in the column headed by 2. The number is 5649. Now in the same row move in the column of mean differences and look at the number in the column headed by 3. The number there is 4. Add this number to 5649 to get 5653.
The characteristic is 2. So, the decimal point is put after three digits to get `565.3`.Hence, antilog `(2.7523)=565.3`
(ii) The mantissa of `3.7523` is the same as the mantissa of the number in Step (i), but the characteristic is 3. Hence, antilog ` (3.7523) = 5653.0`
(iii) The mantissa of ` 5.7523` is the same as the mantissa of the number is Step (i), but the characteristic is 5. Hence, antilog ` (5.7523) = 565300.0`.
(iv) Proceeding as above, we have antilog `(0.7523)=5.653`.
(v) In this case, the characteristic is ` bar(1), i.e., - 1`. Hence,
antilog ` (bar(1).7523) = 0.5643`
(vi) In this case, the characteristic is ` bar(2), i.e., - 2`. So, we write one zero on the right side of the decimal point. Hence,
antilog ` (bar(2).7623)=0.05653`
(vii) Proceeding as above, antilog ` (bar(3).7523)= 0.005623`.
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