2. Find the continued product of 3.902, 597.16, and 0.0314728. Here, the 2 cancels the +2, and the 1 carried from the decimal part is set down. 3. Find the continued product of 3.586, 2.1046, 0.8372, and 0.0294. Ans. 0.1857615. DIVISION BY MEANS OF LOGARITHMS. 16. From the principle proved in Art. 6, we have the following RULE. Find the logarithms of the dividend and divisor, and subtract the latter from the former; then find the number corresponding to the resulting logarithm, and it will be the quotient required. The Here, 1 taken from I, gives 2 for a result. subtraction, as in this case, is always to be performed in the algebraic sense. 3. Divide 37.149 by 523.76. Ans. 0.0709274. The operation of division, particularly when combined with that of multiplication, can often be simplified by using the principle of 17. The ARITHMETICAL COMPLEMENT of a logarithm is the result obtained by subtracting it from 10. Thus, 8.130456 The arithmetical is the arithmetical complement of 1.869544. complement of a logarithm may be written out by commencing at the left hand and subtracting each figure from 9, until the last significant figure is reached, which must be The arithmetical complement is denoted by taken from 10. the symbol (a. c.). Let a and represent any two logarithms whatever, and a b their difference. and subtract it from, ab, without altering its value, wo But, 10-6 is, by definition, the arithmetical complement of b: hence, Equation (10) shows that the difference between two logarithms is equal to the first, plus the arithmetical complement of the second, minus 10. Hence, to divide one number by another by means of the arithmetical complement, we have the following RULE. Find the logarithm of the dividend, and the arithmetical complement of the logarithm of the divisor, add them toge ther, and diminish the sum by 10; the number correspond ing to the resulting logarithm will be the quotient required. Applying logarithms, the logarithm of the 4th term, is equal to the sum of the logarithms of the 2d and 3d terms, minus the logarithm of the 1st: Or, the arithmetical complement of the 1st term, plus the logarithm of the 2d term, plus the logarithm of the 3d term, minus 10, is equal to the logarithm of the 4th term. The operation of subtracting 10, is performed mentally. RAISING OF POWERS BY MEANS OF LOGARITHMS. 18. From Article 7, we have the following RULE. Find the logarithm of the number, and multiply it by the exponent of the power; then find the number corresponding to the resulting logarithm, and it will be the power required. 19. From the principle proved in Art. 8, we have the following RULE. Find the logarithm of the number, and divide it by the index of the root; then find the number corresponding to the resulting logarithm, and it will be the root required. EXAMPLES. 1. Find the cube root of 4096. The logarithm of 4096 is 3.612360, and one-third of this is 1.204120. The corresponding number is 16, which is the root sought. When the characteristic is negative and not divisible by the index, add to it the smallest negative number that will make it divisible, and then prefix the same number, with a plus sign, to the mantissa. 2. Find the 4th root of .00000081. The logarithm of .00000081 is 7.908485, which is equal to 81.908485, and one-fourth of this is 2.477121. The number corresponding to this logarithm is 09: hence, .03 is the root required. |