Page images
PDF
EPUB

(6) Suppose a logarithm, as 3.7531, is given to find the corresponding number.

The exact mantissa, 7531, is not found in the table, therefore take out the next larger, 7536, and the next smaller, •7528, and retain the characteristic in arranging the work. Thus, the number corresponding to 3.7536 is 5670

66 3.7528 is 5660

[blocks in formation]

Now the logarithm 3.7531 is .0003 greater than the logarithm 3.7528, and a difference in logarithms of .0008 corresponds to a difference in numbers of 10; therefore we should increase the number corresponding to the logarithm 3.7528 by .0003 3

of 10. -0008 8 Thus the number corresponding to the logarithm

3.7531=5660

3.7 correction

or

=5663.7 (c) Suppose a logarithm, as 8•8225 – 10 or 2.8225, is given to find the corresponding number.

Take out the mantissæ as in the previous example.
The number corresponding to 2.8228 is .0665 [Art. 409.]

“ 2.8222 is 0664

[merged small][ocr errors]

Now the logarithm 2.8225 is .0003 greater than the logarithm 2.8222, and a difference in logarithms of .0006 corresponds to a difference in numbers of .0001; therefore we should increase

.0003 the number corresponding to the logarithm 2.8222 by 3

·0006 of .0001. 6 Thus the number corresponding to the logarithm 2.8222=.0664

2.8225=.0664 Correction, •00005

or

66

= .06645

EXAMPLES XL. b. Find the common logarithms of the following: 1. 50. 2. 203.

3. 5•3. 4. 673. 5. 341.

6. 0·045. 7. 5265. 8. 12345.

9. 0·010203. 10. 354.076.

Find the numbers corresponding to the following common logarithms : 11. 1.8156. 12. 2:1439.

13. 4.0022. 14. 1.9131. 15. 3.8441.

16. 7.4879–10.

ARITHMETICAL COMPLEMENT.

419. The logarithm of the reciprocal of a number is called the Arithmetical Complement or Cologarithm of that number.

Thus colog 210= log zio= log 1 – log 210.

Since log 1=0, we write it in the form 10-10 and then subtract log 210, which gives

colog 210=(10-2 3222) - 10

=7.6778-10. Hence

RULE. To find the cologarithm of a number, subtract the logarithm of the number from 10 and write 10 after the result.

420. The advantage gained by the use of cologarithms is the substitution of addition for subtraction.

4.26 Example. Find by use of logarithms the value of

7.42 x .058 4.26

1 1 =4.26 x

Х ; 7.42 x .058

7.42 058 4.26

1

1 log

- log 4.26+ log + log
7.42 x .058

7:42 058
= log 4.26+ colog 7:42+ colog •058
= •6294+ (9.1296 - 10)+1.2366

=10.9956 – 10.
The number corresponding to this logarithm is 9.9.

The student should carefully notice the operations involved in finding colog 0.058.

1
colog •058= log =10-[log .058] -10 (Art. 419).

•058
=10— [8.7634-10] -10 (Art. 409.)
=10-8.7634 +10-10
=1•2366.

EXPONENTIAL EQUATIONS.
421. Equations in which the unknown quantity occurs as
an exponent are called Exponential Equations, and are readily
solved by the aid of logarithms.

Example. . Find the value of 2 in 157=28.
Taking the logarithms of both sides of the equation, we have

log 15x = log 28;
.. x log 15= log 28.

1.4472

=1.2305+ 1.1761

X

log 28
log 15

EXAMPLES XL. C. Find by use of logarithms : 24.051 x .02456

145•206 (-7.564) 1.

2.* •006705 x .0203

448.1 % (-2406) (-47-85) 3. (742.8024).

4. (- .0012045) 8. 4.8 x V.002 x 442-6

V9.8149 x 80.80008 5.

6. (18)2 x 78 x (3-4562)

V8283 X (-0006412) 7. 845692:1 x.845856.

8. •00010101 x (7117:1). (285.42)104 x (5.672) 9.

12.876 x V.068 x (·005157)2

10. V V20x V-02 x 124.89

29-029 x (5281)*x (-40° 11. 34+2=405. 12. 105–3x = 27--22. 13. 123.–4 x 187—2x=1458. 14. 2* x 6x-2=524 x 71-4.

15. 2x+y=6y,

3x=3 x 2y+1. 16. 31-4-v=4-1,

222-1=33y-.

3

* Treat negative quantities occurring in logarithmic work as positive. When the numerical result is obtained, determine its sign by the rules for multiplication and division already given.

[ocr errors]

CHAPTER XLI.

INTEREST AND ANNUITIES.

422. QUEStions involving Simple Interest are easily solved by the ordinary rules of Arithmetic; but in Compound Interest the calculations are often extremely laborious. We shall now show how these arithmetical calculations may be simplified by the aid of logarithms. Instead of taking as the rate of interest the interest on $100 for one year, it will be found more convenient to take the interest on $1 for one year. If this be denoted by $r, and the amount of $1 for 1 year by $R, we have R=1+r.

423. To find the interest and amount of a given sum in a given time at compound interest.

Let P denote the principal, R the amount of $1 in one year, n the number of years, I the interest, and M the amount.

The amount of Pat the end of the first year is PR; and, since this is the principal for the second year, the amount at the end of the second year is PRXR or PR2. Similarly the amount at the end of the third year is PR8, and so on; hence the amount in n years is PR; that is,

M=PR"; and therefore

I=P(R"-1). Example. Find the amount of $100 in a hundred years, allowing compound interest at the rate of 5 per cent., payable quarterly ; having given log 2=:3010300, log 3=.4771213, log 14.3906=1:15808.

1 5 The amount of $1 in a quarter of a year is $(1+

or $

81

4 100 The number of payments is 400. If M be the amount, we have

M=100

; 80

80

[ocr errors]

log M=log 100+400(log 81 – log 80)

=2+400(4 log 3–1–3 log 2)

=2+400(.0053952)=4:15808; whence

M=1439.6.
Thus the amount is $14390.60.
Note. At simple interest the amount is $600.

424. To find the present value and discount of a given sum due in a given time, allowing compound interest.

Let P be the given sum, V the present value, D the discount, R the amount of $1 for one year, n the number of years.

Since V is the sum which, put out to interest at the present time, will in n years amount to P, we have

P=VR";

.:. V=PR-n, and

D=P-V=P(1-R-n).

ANNUITIES.

425. An annuity is a fixed sum paid periodically under certain stated conditions; the payment may be made either once a year or at more frequent intervals. Unless it is otherwise stated we shall suppose the payments annual.

426. To find the amount of an annuity left unpaid for a given number of years, allowing compound interest.

Let A be the annuity, R the amount of $1 for one year, n the number of years, M the amount.

At the end of the first year A is due, and the amount of this sum in the remaining n–1 years is A Rn-1; at the end of the second year another A is due, and the amount of this sum in the remaining n -2

years

is ARN-2; and so on.
... M=ARn-l+ ARn-2+ + AR2+ AR+A
=A(1+R+R2+

to n terms)
Rn-1
=A

R-1 427. To find the present value of an annuity to continue for a given number of years, allowing compound interest.

Let A be the annuity, R the amount of $1 in one year, in the number of years, V the required present value.

« PreviousContinue »