478. As the numerator is divided by the denominator, we should (473) subtract the log. of the denominator from that of the numerator. But as this is impossible, the denominator being the greater, we subtract the log. of the numerator from that of the denominator, and prefix to the remainder the sign -, which shows by how much the subtraction is impossible. Wherefore, operating as above, the log. of the fraction is - 1,411245.

232

479. To multiply by a fraction we should (472) add its log. ; but, as this is negative, its addition consists (98) in merely performing the subtraction indicated by its sign. Thus, if we have 6264 × 2, we write

Having performed the subtraction indicated by the sign, we have 2,385607. Seeking this in the table, we find that it differs from the log. of 243 by only a unit of the last decimal order. Now as the logarithms of the table are only true to about half a unit of the last decimal order, we take 243 for the true result, which it is in effect.

480. To divide by a fraction we should (473) subtract its log. But to subtract a negative number is (96) to add it as a positive number. Wherefore, if we have 63÷, we write l. 63............ 1,799341

and (96) having performed the subtraction of the negative log. by adding it as a positive number, we have 3,210586. Seeking this in the table, we find its number 1624, which is the true quotient.

481. From the series of numbers 10, 100, 1000, 10000, &c. and their logarithms 1, 2, 3, 4, &c., it is evident that to add 1, 2, 3, &c. units to the index of a logarithm is to multiply the

corresponding number by 10, 100, 1000, &c.; and vice versa, to subtract from the index 1, 2, 3, &c. units, is to divide the corresponding number by 10, 100, 1000, &c.

482. Let n and n+1 be any two consecutive numbers, and d the difference of their logarithms. Then (473)

= 7 (" + 1 ) = 7(1 + 1 ); that is, d approaches nearer to

n

n

7. 1 or 0, according to the increase of n.

By inspecting the column marked Diff., the student will see that near the end of the table several numbers in succession have nearly the same diff. between their logarithms; that is, their logarithms successively differ by only about half a unit of the fourth decimal order.

Now suppose the difference between two consecutive logarithms to be always the same, and put q: this difference. Then, as the diff. between any two consecutive numbers of the table is a unit, if a, b, c, d, &c. be consecutive numbers, and A, B, C, D, &c. their respective logarithms, the diff. between a and b is 1 unit, between a and c, 2 units, between a and d, 3 units, &c. Also the diff. between A and B is q, between A and C is 2q, between A and D is 3q, &c. Now as the equimultiples of any two numbers (165) have the same ratio to each other that the numbers have, it is plain that any two of the numeral differences are to one another as the differences of the logarithms of the corresponding numbers. Thus-1:2::g : 2q; 1:3::q: 39; 2: 3 :: 29: 3q, &c.

483. This proportion, though not exact, is near enough, when the numbers are large for ordinary purposes, and is of great utility in finding the logarithms of numbers which exceed the limit of the table, as well as in finding the numbers from their logarithms.

Suppose, for example, we would find the log. of 458264; we separate by a comma two figures on the right, that the part on the left may be found in the table; thus, 4582,64. We then find the log. of 4582, which is 3,661055. Opposite to this, in the column marked Diff., we find 95. We then say 1:,64 ::95 60,80, or 61. Hence, we have 3,661055+ 61 3,661116 for the log. of 4582,64. But as this number is 100 times too little, we add 2 units (481) to the index of its log., and have 5,661116 for the required logarithm.

=

Again, if we would find the number corresponding to the log. 5,661116, we first suppose 2 units to be taken from its

index 5; and seeking in the table the log. next inferior to 3,661116, we find 3,661055, and its corresponding number, 4582. Then, taking the diff. between the log. found and the given one, which is 61, we say 95: 61::1:,64. This we append to the number 4582, which thus becomes 4582,64. Then, as this number is 100 times too little, (481,) we remove the comma, and have 458264 for the number sought..

484. From the above (476 and 481) we easily see that the log. of a decimal number is the log. of the same number considered as integral, having its index diminished by as many units as there are decimals in the number; or having for its index as many units, less one, as there are figures on the left of the comma. Thus, as the log. of 785398 is 5,895090, to have the log. of 785,398, as there are 3 decimals, we simply diminish the index 5 by 3 units, and have 2,895090 for the required logarithm. Also, it is easy to see that in this case the number of units in the index will always be one less than the number of figures remaining on the left of the comma. Hence the log. of 7,85398 is 0,895090.

-.

485. When there are decimals only, we find the log. as before, and subtract it from as many units as there are decimals, prefixing to the remainder the sign Thus, to have the log. of ,785398, as there are 6 decimals, we subtract the log. of 785398 from 6, and prefixing the sign, we have -0,104910 for the required logarithm.

486. Suppose that the given log. is such that while its index exceeds, its decimal part is found in the table; we write the corresponding number found in the table, and subtract the index of its log. from that of the given log. We then, on the right of the number found, place as many ciphers as there are units remaining in the index of the given log. Thus, if we would have the number corresponding to the log. 7,146438, we find the number corresponding to the decimal part, with 3 for its index, to be 1401. Then, as 7-34, we place 4 ciphers on the right of 1401, and have 14010000 for the required number.

487. When only the first figures of the decimal are found, we may proceed thus: If we would find the number corresponding to the log. 5,263584, we find in the table the next inferior decimal,263399, supposing its index to be 3 units; hence its corresponding number is 1834. Then with the diff. be

tween these two logarithms, which is 185, and the tabular diff. 237, we say

consequently, the corresponding number is 1834185, nearly. Now, as the proposed log. (481) belongs to a number 100 times greater, the true number is nearly 18340019500; that is, 18347814, or 183478,06-.

488. As the proportion here used only approximates to the truth when the numbers are large, if the proposed log. is below that of 1500, we add to its index as many units as we can without passing the limit of the table. Having found the corresponding number, we separate as decimals, by a comma, as many figures as we have added units to the index. These decimals will often suffice, but if they do not, we find more, as above, by proportion, and annex them.

For ex. If we would find the number corresponding to 0,624569, as this is between the logarithms of 4 and 5, and much under that of 1500, we seek it with 3 units for index, and find that is between 4212 and 4213; that is, it is 4212 and a fraction; therefore, pointing off 3 figures for decimals, we have 4,212 + for the required number. But if we would have more decimals, we take the diff. between the log. of 4,212 and the given log., which is 81; also the tabular diff., which is 103, and say 103:81::1: ,786+, or,79 —, which we annex to 4,212, and have 4,21279 for the required number. The logarithms of the table being only true to about half a decimal unit of the sixth order, it is not safe to push this second approximation to more than 2 places of figures.

81 103

489. When the log. is negative, we subtract it from (or rather (98) add it to) a number of units, such that the remainder may not exceed the limit of the table; and having found the number answering to this remainder, we separate on the right, by a comma, as many figures as there are units in the number, from which we subtract the log. For ex., if we have the log. -1,724685, we say, 41,724685 2,275315, which answers all but the last decimal to the log. of 1885. Now the sum of any two logarithms is the log. of their product. But 4 is the log. of 1000, and 1,724685 the log. of a fraction. Therefore, 2,275315 is the log. of the fraction multiplied by 10000; consequently, as 1885 is 10000 times too great, we place a comma on the left, and have,1885 for the fraction sought, within a ten-thousandth.

Use of the Table.

490. The student will, by the following examples, obtain some idea of the vast utility of logarithms, especially if he proves them by the usual methods.

1. Required the quotient of 3725÷2957, within a thousandth:

Seeking the log. of the remainder with an index of 3 units, we have 1259+; and as this is 1000 times too great, (481,) we point off 3 figures, and have 1,259+ for the required number. 2. Required the square root of 143, within a ten-thousandth. 7. 143=2,155336; dividing by 2, we have 1,077668, which (488) gives 11,9582 + for the required number.

3. Required the square root of 117.

7. 119-7. 1172,075547-2,068186

0,007361; di viding by 2, we have -0,003680; adding 4, we have 8,996320, which (488) gives the root,991561, true to a millionth.

Hence,

4. Required the square root of. 7.7-7.30,845098-0,4771210,367977. 0,367977. Taking one-half, we have 0,183989 ; adding 4, we have 3,816011, which (488) gives the root,654653, true to a millionth.

7. 3/

5. Required the cube root of 91, within a thousandth. 7. 911,959041; dividing by 3, we have 0,653014 which (488) gives 4,498 for the required root. The true, root is 4,4979+.

6. Required the seventh root of 13, within a hundred-thousandth.

7. 13=1,113943; dividing by 7, we have 0,159135 – which gives the root 1,44256.

7. Required the fifth root of, within a hundred-thousandth.

=

7. 11-7.7 = 1,0413930,845098 0,196295. Hence, (478) 7. 70,196295; dividing by 5, we have

0,039259; adding 4 units, we have 3,960741, which (488) gives the root,91356.