Multiplying by 3 is the same as adding the number twice to itself. The characteristic becomes 28, but omitting two of the tens or 20, it becomes 8, which shows it to be the logarithm of a fraction whose first place is hundredths. If it is required to find the 3d root of a fraction, it is easy to see, that having taken out the logarithm of the fraction, it will be necessary to add two tens to the characteristic, for it is then considered the third power of some other fraction, and in raising the fraction to that power, two tens would be subtracted. In the last example the logarithm of the power is 8.732476, but in order to take its 3d root, it will be necessary to add the two tens which were omitted. For the second root one ten must be previously added, and for the fourth root, three tens, &c. 4. What is the 3d root of .027 f log. .027 - - - - 8.431364 or considered as a 3d power 28.431364 (3 log. .3. Ans. - - - 9.477121 5. What is the 2d root of .0016: log. .0016 - - 7.204120 or considered a second power 17.204120 (2 log. .04. Alms. 8.602060 In dividing a whole number by a fraction, if 10 be added to the characteristic of the dividend, it cancels the 10 supposed to be added to the divisor. If both are fractions the ten in the one cancels it in the other; and if the dividend only is a fraction, the answer will of course be a less fraction. Consequently in division the results will require no alteration. Here in subtracting I suppose 10 to be added to the first characteristic, and say 8 from 11, &c. 7. Divide .2172 by .006. log. .2172 - - - 9.336860 In taking the arithmetical complement, the logarithm of the number may be subtracted immediately from 10. The logarithm of 2 being .301030, its arithmetical complement is 1.698970. Adding 10 it becomes 9.698970. It would be the same if subtracted immediately from 10 thus 10 — .301030 – 9.698970. 8. It is required to find the value of a in the following expression : Observe that the 2d power of 38 is found by multiplying the logarithm of 38 by 2, the 3d power by multiplying it by 3, &c. which will give the logarithm of the result. Hence we have the following equation; the logarithm of 38 being 1.579784 and that of 583 being 2.765669. The value of a is found by dividing one logarithm by the other in the same manner as other numbers. It might be done by logarithms if the tables were sufficiently extensive to take out the numbers. By a table with six places an answer correct to four decimal places may be obtained. In taking out the logarithms the right hand figure may be omitted without affecting the result in the first four decimals. log. 2.76567 - - 0.441794 log. 1.57978 - - , (), 198588 log, c = 1.75064 + - 0.243206 13. What is the value of a in the equation 1537% = 52: This gives first 1537 = 52*. * This may now be solved like the last. LII. Questions relating to Compound Interest. It is required to find what any given principal p will amount to in a number n of years, at a given rate per cent. r, at compound interest. Suppose first, that the principal is $1, or £1, or one unit of money of any kind. The interest of 1 for one year is #. or simply r, if r is considered a decimal. The amount of 1 for one year then, will be 1 + r. The amount of p dollars will be p (1 + r.). For the second year, p (1 + r.) will be the principal, and the amount of 1 being (1 + r.), the amount of p (1 + r.) will be p (1 + r.) (1 + r.) or p (1 + r.)”. For the third year p (1 + r.)" being the principal, the amount will be p (1 + r.)” (1 + r.) or p (1 + r.)". For n years then, the amount will be p (1 + r.)". Putting .4 for the amount, we have .4 = p (1 + r.)". This equation contains four quantities, off, p, r, and n, any three of which being given, the other may be found. Logarithms will save much labour in calculations of this kind. Examples. 1. What will $753.37 amount to in 53 years, at 6 per cent compound interest ? Here p = 753.37, r =.06, and n = 5}. log. 1 + r = 1.06 - - 0.025.306 2. What principal put at interest will amount to $5000 in 13 years at 5 per cent. compound interest ? 3. At what rate per cent. must $378.57 be put at compound interest, that it may amount to $500 in 5 years Solving the equation 4 = p (1 + r.)" making r the unknown quantity, it becomes 4. In what time will $284.37 amount to 750 at 7 per cent.” Making n the unknown quantity, the equation.A = p (1 + 1)" becomes |