(3) To multiply 3.7482569 by 5. 3.7,482569 5 12.7412845 the 3 carried on from the last multiplication of the decimal places being added to -15, and thus giving -12 as a result. (4) To divide 14-2456736 by 4. Increase the negative characteristic so that it may be exactly divisible by 4, making a proper compensation, thus, 1. Add 3-1651553, 47505855, 66879746, 26150026. 3. Add 2:5324716, 36650657, 5·8905196, 3156215. 5. From 2.352678 take 5:428619. 6. From 5-349162 take 3·624329. 7. Multiply 24596721 by 3. 9. Multiply 9-2843617 by 7. 10. Divide 6:3725409 by 3. 11. Divide 14:432962 by 6. 12. Divide 453627188 by 9. 463. We shall now explain how a system of logarithms calculated to a base a may be transformed into another system of which the base is b. Let m be a number of which the logarithm in the first system is x and in the second y. Hence if we multiply the logarithm of any number in the 1 logab' system of which the base is a by we shall obtain the logarithm of the same number in the system of which the base is b 1 This constant multiplier is called THE MODULUS of the log.b system of which the base is b with reference to the system of which the base is a. 464. The common system of logarithms is used in all numerical calculations, but there is another system, which we must notice, employed by the discoverer of logarithms, Baron Napier, and hence called The Napierian System. The base of this system, denoted by the symbol e, is the number which is the sum of the series of which sum the first eight digits are 2.7182818. 465. Our common logarithms are formed from the Logarithms of the Napierian System by multiplying each of the S. A. 22 latter by a common multiplier called The Modulus of the Common System. This modulus is, in accordance with the conclusion of That is, if and N be the logarithms of the same number in the common and Napierian systems respectively, and so the modulus of the common system is 43429448. 467. The following are simple examples of the method of applying the principles explained in this Chapter. Ex. (1) Given log 2=3010300, log 3=4771213 and log 7=8450980, find log 42. Since 42=2×3×7 log 42=log 2+log 3+log 7 ='3010300+4771213 +8450980 = 1.6232493. Ex. (2) Given log 2=3010300 and log 3=4771213, find the logarithms of 64, 81 and 96. .. log 96=5 log 2+log 3=1·5051500+*4771213=1·9822713. Ex. (3) /(6.25). Given log 5=6989700, find the logarithm of 1. log 2500. EXAMPLES.-CLXI. Given log 2=3010300, find log 128, log 125 and 2. Given log 2=3010300 and log 7=8450980, find the logarithms of 50, 005 and 196. 3. Given log 2=3010300, and log 3=4771213, find the logarithms of 6, 27, 54 and 576. 4. Given log 2='3010300, log 3='4771213, log 7='8450980, find log 60, log 03, log 1'05, and log '0000432. and and 5. Given log2=3010300, log 18=1′2552725 and log 21=1-3222193, find log 00075 and log 31*5. 6. Given log 5='6989700, find the logarithms of 2, '064, 1 26914 520 7. Given log2=3010300, find the logarithms of 5,125 59015 8. What are the logarithms of 01, 1 and 100 to the base 10? What to the base '01 ? 9. What is the characteristic of log 1593, (1) to base 10, (2) to base 12? 11. Given log4='6020600, log 1·04=·0170333 : (a) Find the logarithms of 2, 25, 83-2, (625) 1 (b) How many digits are there in the integral part of (1.04)GOOD? 12. Given log 25=1.3979400, log 103='0128372: 1 (a) Find the logarithms of 5, 4, 51′5, ('064)100. (b) How many digits are there in the integral part of (103) co? 13. Having given log 3=4771213, log 7='8450980, |