8 5 Consequently the log. of (x =)10 Substituting for M the number last found, we obtain the log. of 2.30102999599, and thence the log. of 4, 8, 16, 32, &c. may be had by simple multiplication. Also the log. of 5= log. of 10- log. of 2.69897000401; whence be found the logarithms of all the powers of 5. may The numerical value of M being thus ascertained, if we make p denote a number whose log. is known, we have log. p+1=log. p+ log+1log.p+ 2M A B C X + 2p+1 3.(2p+1)* 5.(2p+1) 7.(2p+1) + &c. Where A, B, C, &c. denote the preceding term exclusive of the divisor, 3, 5, &c., and thus the logarithms of all the prime numbers may be computed. But as the above series, when the number p is small, converges but slowly, and therefore, requires a considerable number of terms to be used, the labour may be abridged by proper expedients, a few of which are subjoined. Let the log. of 3 be required; put p=80, then To which add log. of 80=*1+3 log. of 2=1.9030899880 Hence, log. of 81, or 4 log. of 3= And log. 3=.4771212549. 1.9084850199 Again, let the log. of 7 be required; put p=27, then Adding log. of 33, or of 27=1.4313637647 Log. of 28=1.4471580318 Subtracting log. of 4, .6020599920 Log. of 7.8450980398 As the logarithms of 10 and its powers are wholly integral, it is manifest the logarithm of a number is changed only in the integral part, by varying the position of the decimal point in the number itself. In general, suppose x+2 to denote a prime number whose logarithm is required, those of the inferior even numbers being known. Since = -3x+2 where both terms (x-1)3× x+2 are divisible by 4, because x-1, and x+1 are even numbers, it follows that this fraction in its lowest terms, may be expressed by P+1 p and therefore the log. of (x−1)2.(x+2) (x+1)3.(x-2) be obtained from the formulæ above given; to which adding 2 log. (x+1)+ log. (x-2)-2 log. (x-1) the result will be the log. of x+2. Let the log of 13 be required. Here x=11, and our fraction becomes In this manner the logarithms may be derived from those already obtained, to any proposed extent. The differential method may be advantageously applied, to the completion of a logarithmic table. But the labour of those computations, being already finished, and not likely to be renewed, a further elucidation of the subject is deemed unnecessary. SECTION X. SURDS. 104. It has been remarked, (note, art. 36.) that the even roots of negative quantities are impossible; hence whenever, in the solution of a problem, the square root of a negative quantity appears in the result, such result is impossible, or imaginary. But instead of abandoning, as hopeless, every example in which such expressions appear, they are found conformable to the general principles of the science, and sometimes connected with the most refined analytical processes. Impossible roots may be introduced into a problem in two ways, quite distinct from each other. First, by admitting incompatible assumptions into the data of the problem; in which case the impossible root serves to detect that incompatibility; and its appearance or disappearance marks the limits of the problem. Thus in the equation x2-2ax-b, we have (by art. 62,) x=a±√/a2-b as the general expression of the value of x. If now ao>b,* a2-b is positive, and therefore ✔ab, and consequently x, a possible quantity. But if a2 <b, a3—b is negative; and therefore, ab impossible; whence x is, in this case, an imaginary quantity. Here the terms of the original equation are compatible, whenever a is equal to, or greater than b; but incompatible when a2 <b; for x-2ax-b is equivalent to 2α-x.x=b, but 2a-x.x cannot be greater than a2. Put x=a±c, then 2a-x=ac, and 2a-x.x—α3—c3. When c=0, x=a, or 2a―x.x=(b=)a2; this is, therefore, the limit of b. *a>b is read a is greater than b, and ab, a is less than 6. Multiplying by x, x2+1=2ax, or x2-2ax=-1. Whence, x=α±aa—1, which value of x is imaginary when a<1. Hence the least possible sum of a number and its reciprocal is 2.* 105. Impossible roots may be sometimes obtained, when the data are compatible, by the admission of inconsistent suppositions into the solution. Suppose we have x2+nya, and xy=b, to find x and y; a, b, and n being any given numbers. To and from the first equation add and subtract 2✅n times the second, whence, x+2xy✓n+ny®=a+2b✔✅n, and x2-2xy✓n+ny2=a—2b/n; and by evolution, x+y√n=✔✅a+2b√n, (M.) x—y✔✅n=✅a—2b✅n, (N.) ✔a+2b√n+a-2b/n 2 *These principles may be applied to determine the maxima and minima of geometrical quantities. Let b= the base, a= the altitude of a plane triangle; x= altitude of its inscribed rectangle, = the area of the rectangle, which put =c; X= =2, this is, therefore, the greatest posssible value ab of x; and the greatest rectangle is 14 the triangle. |