Napier, the inventor of logarithms, and assign to M the simplest possible value. This value will therefore be unity; and we have log. (P+1)=log. P+2{2p+1 + 3(2P2+1)3 + 5(2P2+1)+ ··.} Expounding P successively by by 1, 2, 3, 4, &c., we find log 2= = log. 3 log. 2+2 log. 42 log. 2. 1 1 1 In this manner the Napierian logarithms of all numbers may be computed. and by means of this equation we can pass from one system of logs. to another, by multiplying x, the log. of any number in the system whose base is a, by the reciprocal of log. b in the same system; and thus we shall obtain the log. of the same number in the system whose base is b. Let the two systems be the Napierian and the common, in which the base of the former is 2.718281828... and the base of the latter is b=10, the base of our common system of arithmetic; then we have b=10, and a==2.718281828... and consequently if N denote any number, we shall and the modulus of the common system is, therefore, M= 1 2.3025851 ='43429448 ... 2 M=86858896 Hence, to construct a table of common logarithms, we have log. 8=log. 2+-8688896 ( — +353 + 1/4 + . . . ) = } 4771213 = €020600 =6389700 =7781513 1 13 18+ 3.133 + 5-138+. = 2M (n + ƒn3 + ƒn3 + ƒn2 + . . .) log. 10= .. log. P=2M -n =P; then 1+n=P (1—n) or n= P-1 P+1 3 1 &c. P = { + ( P = 1 ) + ( P = 1 ) ' + ... } and thus we have a series for computing the logs. of all numbers, without knowing the log. of the previous number. EXAMPLES IN LOGARITHMS. (1.) Given the log. of 2=0·301C300, to find the logs. of 25 and 0125. .. log. 0125=log. 1-log. 10-3 log. 2-1-3 log. 2=2-0969100 (2.) Calculate the common logarithm of 17. Ans. 1-2304489. (3.) Given the logs. of 2 and 3 to find the logarithm of 22.5. Ans. 1+2 log. 3—2 log. 2. (4.) Having given the logs. of 3 and 21, to find the logarithm of 83349. Ans. 6+2 log. 3+3 log. 21. ON EXPONENTIAL EQUATIONS. 209. An exponential equation is an equation in which the unknown appears in the form of an exponent or index; thus, the following are exponential equations: a2 = b, x2 = a, ab* = c, x12 = a, &c. When the equation is of the form a=b, or ab*= c, the value of x is readily obtained by logarithms, as we have already seen in Art. 201. But if the equation be of the form = a, the value of a may be obtained by the rule of double position, as in the following example: Ex. Given x = 100, to find an approximate value of x. The value of x is evidently between 3 and 4, since 33 = 27 and 41=256; hence, taking the logs. of both sides of the equation, we have Then, as the difference of the results is to the difference of the assumed numbers, so is the least error to a correction of the assumed number corresponding to the least error; that is, 0984511002689: 00273; hence 36·002733·59727, nearly Again, by forming the value of x for x = 3.5972, we find the error to be -·0000841, and for x= = 3.5973, the error is +0000149; hence, as 000099 00010000149: 0000151; therefore x = 3·5973 — ·0000151 = 3·5972849, the value nearly. • In equations of this kind, the following method may be adopted:-Leta; then ≈ log. = log. a; put log. =y, and log. a=b; then ry=b, and log. +log. y=log. b; hence y+log. y=log. b. Now, y may be found by double position, as above, and then becomes known. When a is less than unity, put x= and ; then we have by=y.. y log. b=log. y, and if log. b=c, and log. y=; then cy=z, and log. c+log. y=log. z, or log. c+x=log. z. Hence z may be found by the preceding method, and then y and a become known. a=/; INTEREST AND ANNUITIES. THE solution of all questions connected with interest and annuities may be greatly facilitated by the employment of algebraical formulæ. In treating of this subject we may employ the following notation: interest of £1 for one year. interest of p pounds for t years. amount of p pounds for t years at the rate of interest denoted by r. the number of years that p is put out to interest. SIMPLE INTEREST. PROBLEM I. To find the interest of a sum p for t years at the rate r. Since the interest of one pound for one year is r, the interest of p pounds for one year must be p times as much, or pr; and for t years t times as much as for one year, consequently, i = ptr........ .....( 1 ) PROBLEM IL-To find the amount of a sum p laid out for t years at simple interest at the rate r. The amount must evidently be equal to the principal together with the interest upon that principal for the given time, Required the interest of £973. 15s. for 24 years at 43 per cent. per annum. It will be found convenient to reduce broken sums of money and periods of time to decimals of a pound and of a year, respectively. By the formula (1) we have The amount of the above sum at the end of the given time will be PRESENT VALUE AND DISCOUNT AT SIMPLE INTEREST. The present value of any sum s due t years hence is the principal which in the time t will amount to s. The discount upon any sum due t years hence is the difference between that sum and its present value. PROBLEM III. To find the present value of s pounds due t years hence, simple interest being calculated at the rate r. By formula (2) we find the amount of a sum p at the end of t years to be 8 = pptr Consequently p will represent the present value of the sum s due t years hence, and we shall have PROBLEM IV. To find the discount on s pounds due t years hence, at the rate r, simple interest. ✶r is the interest of L1 for one year. To find the value of r when interest is calculated at the rate of L44, or L4.75 per cent, per annum, we have the following proportion. |