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412. Division by Logarithms. According to Art. 397, to find the quotient of two numbers we have the following
From the logarithm of the dividend subtract the logarithm of (the divisor; the difference will be the logarithm of the quotient.
The word difference is here to be understood in its algebraic sense; the decimal part of the logarithm being invariably positive, while the characteristic may be either positive or negative.
Ex. 1. Find the quotient of 888.7 divided by 42.24.
413. Involution by Logarithms. — According to Art. 398, to involve a number to any power we have the following
RULE. Multiply the logarithm of the number by the exponent of the power required.
It should be remembered that what is carried from the decimal part of the logarithm is positive, whether the characteristic be positive or negative.
Ex. 1. Find the fifth power of 2.846.
5 The fifth power is 186.65, whose log. is 2.2710. Ex. 2. Find the cube of .07654.
The logarithm of .07654 is 2.8839.
414. Evolution by Logarithms. — According to Art. 399, to extract any root of a number we have the following
Divide the logarithm of the number by the index of the root required.
Ex. 1. Find the cube root of 482.4.
Dividing by 3, we bave 0.8945, which corresponds to 7.843, which is therefore the root required. Ex. 2. Find the 100th root of 365.
Ans. 1.061. When the characteristic of the logarithm is negative, and is not divisible by the given divisor, we may increase the characteristic by any number which will make it exactly divisible, provided we prefix an equal positive number to the decimal part of the logarithm.
Ex. 3. Find the seventh root of 0.005846.
7+4.7669. Dividing by 7, we have 1.6810, which is the logarithm of .4797, which is therefore the root required.
Ex. 4. Find the 10th root of 0.007815.
415. Proportion by Logarithms.- The fourth term of a proportion is found by multiplying together the second and third terms and dividing by the first. Hence, to find the fourth term of a proportion by logarithms, we have the following
RULE. Add the logarithms of the second and third terms, and from their sum subtract the logarithm of the first term. Ex. 1. Find a fourth proportional to 72.34, 2.519, and 357.5.
Ans. 12.45. Ex. 2. Find a fourth proportional to 43.17, 275, and 5.762. Ex. 3. Find a fourth proportional to 5.745, 781.2, and 54.27.
Exponential Equations. 416. An exponential equation is one in which the unknown quantity occurs as an exponent. Thus,
a*=0 is an exponential equation, from which, when a and b are known, the value of x may be found. If a=2 and b=8, the equation becomes
2+=8, in which the value of x is evidently 3, since 23=8. If a=16 and b=2, the equation becomes
16+=2, in which the value of x is evidently 1, since 16*=2.
417. Solution by Logarithms.—When b is not an exact power or root of a, the equation is most readily solved by means of logarithms. Taking the logarithm of each member of the equation ap=b, we have
log. a=log. 6, whence
log. a Ex. 1. Solve the equation 3*=20. log. 20 1.3010
=2.727 nearly. log. 3
.4771 Ex. 2. Solve the equation 5x=12.
2 * .
Ex. 3. Solve the equation ()
Ex. 4. Solve the equation 10x=7.
Ex. 5. Solve the equation 12+=3.
Ex. 6. Solve the equation 12*=7.
Compound Interest. 418. Interest is money paid for the use of money. When the interest, as soon as it becomes due, is added to the principal, and interest is charged upon the whole, it is called compound interest.
419. To find the amount of a given sum in any time at compound interest. It is evident that $1.00 at 5 per cent. interest becomes at the end of the year a principal of $1.05; and, since the amount at the end of each year must be proportioned to the principal at the beginning of the year, the amount at the end of two years will be given by the proportion
1.00:1.05:: 1.05:(1.05). The sum (1.05) must now be considered as the principal, and the amount at the end of three years will be given by the proportion
1.00:1.05:: (1.05)2: (1.05)”. In the same manner, we find that the amount of $1.00 for n years at 5 per cent. compound interest is (1.05)". For the same reason, the amount for n years at 6 per
cent. is (1.06)". It is also evident that the amount of P dollars for a given time must be P times the amount of one dollar. Hence, if we put
P to represent the principal,
A the amount of the given principal for n years, we shall have
A=P(1+r)" This equation contains four quantities, A, P, n, r, any three of which being given, the fourth may be found. The computation is most readily performed by means of logarithms. Taking the logarithms of both members of the preceding equation and reducing, we find
log. A=log. P+nx log. (1+r),
log. A-log. P
Ex. 1. How much would 500 dollars amount to in five years at 6 per cent. compound interest ? The log. of 1.06 is
0.1265 The log. of 500 is
2.6990 The amount is $669.10, whose log. is 2.8255.
Ex. 2. What principal at 6 per cent. compound interest will amount to 500 dollars in seven years ? Ans. $332.60.
Ex. 3. At what rate per cent. must 500 dollars be put out at compound interest so that it may amount to $680.30 in seven years?
Ans. 44 per cent. Ex. 4. In what time will 500 dollars amount to 900 dollars at 6 per cent. compound interest?
Ans. 1011 years. Ex.5. How much would 400 dollars amount to in nine years at 5 per cent. compound interest?
Ex. 6. What principal at 5 per cent. compound interest will amount to 400 dollars in eight years?
Ex. 7. At what rate per cent. must 400 dollars be put out at compound interest so that it may amount to $620.70 in nine years?
Ex. 8. In what time will a sum of money double at 6 per cent. compound interest?
Ex. 9. In what time will a sum of money double at 5 per cent. compound interest ?
Annuities. 420. An annuity is a sum of money stipulated to be paid annually, and to continue for a given number of years, for life, or forever.
421. To find the amount of an annuity left unpaid for any number of years, allowing compound interest.
Let a denote the annuity, n the number of years, r the interest of one dollar for one year, and A the required amount. The amount due at the end of the first
year At the end of the second year the amount of the first an