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0.00467 × 348.1=3.66932+2.54171=.21102=1.626.

0.0861 × 0.00843=2.93500+3.92583-4.86083=

0.0007258.

Example 2. Article 39.

36.4×7.82 × 68.91 × 0.3246 = 1.56110 + 0.89321 + 1.83828+1.58500-3.87759-7544, the answer.

Example 3,

0.00629×2.647×0.082 × 278.8×0.00063= 3.79365 + 0.42275+2.91381+2.44529+4.79934-4.37984=

0.0002898, the answer.

Division by Logarithms. Art. 42.

Natural number, 6.832 Log. 0.83455

Natural number .0362 Log. 2.55871

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23x=7, x=.03, and 188.73=the answer.

Natural number, 0.0985 Log. 2.99344
Natural number, 0.007241 Log. 3.85980
1.13364

13.6

Natural number, 0.0621 Log. 2.79309
Natural number, 3.68 Log. 0.56585

0.01687 2.22724

Involution by Logarithms. Art. 46.

Example 5. What is the 7th power of 0.9061?
Root, 0.9061 Log. 1.9571761304

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Example 6. What is the 5th power of 0.9344?

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Evolution by Logarithms. Art. 51.

Example 2. What is the 8th power of the 9th root of 654 ? Given number, 654 Log. 2.8155777483

Multiplying by

8

Divid ing by

9)22.5246219864

Power required 318.3

2.5027357762

COMPOUND INTEREST.

Reduction of the formulæ.

1. By article 61, A=nxlog. a+log. P
2. Transposing, —log. P=−A+n×!og. a
3. Changing signs, log. P=A-nx log.a

4. Transposing in the 1st, -n log. a=−A+log. P
5. Changing signs, n× log. a=A-log. P

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By the above formulæ, the several examples which follow may be solved, in which P represents the principal, a the amount of one dollar for one year, n any number of years, and A the amount of the given principal for n years.

Eleven examples under article 61.

Example 3. The amount of $1 for one year=1.06, whose 0.0253059

=

log.
Multiplying by 10 years,

10

Product of log. of 1,06 into 10 yrs. 0.2530590
Logarithm of $1000=

Natural number $1790.84

3.0000000

3.2530590

The 84 cents is obtained by the rule for finding the natural number in Art. 34. By a comparison of the answer here given, with that of Day, they will be found not

perfectly to agree, which is the consequence of different tables being employed in the operation.

Example 4. The amount of $1 for one year is 1.04,

whose logarithm is

0.017033

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Diff. between 0.357693 & 3.215638=2.857945

The natural number belonging to 2.857945 is 721.

Example 5. The amount of $1 for one year is 1.06,

whose logarithm is

Multiplying by 4 years,

0.0253059

4

0.1012236

Logarithm of $202=

2.3053514

Difference between 0.1012236 and

2.3053514

2.2041278

The natural number belonging to 2.2041278 is $160.00.

Example 6. Logarithm of $400= 2.602060 Logarithm of 569.3=

2.755341

0.017031

Diff. between 2.602060 and 2.755341=0.153281
Dividing 0.153281 by 9,

The natural number belonging to the logarithm 0.017031 is 1.04; therefore 4 per cent is the rate of interest sought.

Example 7. Logarithm of $500, the principal 2.698970 Logarithm of $900, the amount=

Diff. between 2.698970 and 2.954243

2.954243

0.255273

The log. of 1.05, the amount of $1 for one year is .021189, and dividing.255273 by this we obtain 12 years, the answer.

Example 8. Log. of $10,000, the principal 4.000000 Log. of 16.288, the amount,=

Diff. between 4.000000 and 4.211291=

4.211921

.211921

The log. of 1.05 the amount of $1 for one year is .021189, and .211921 divided by this, the quotient is 10 years, the

answer.

Example 9. Log. of $20,000, the amount=4.301030

Log. of $11,106, the principal=

Diff. between 4.301030 and 4.045714=

Dividing by 15 years,

4.045714

0.255316

0.017021

The natural number belonging to 0.117021 is 1.04, and therefore 4 is the rate per cent.

Example 10. Log. of 1.06, the amount of $1 for one year =

Multiplying by 8, the number of years,

Log. of $3188, the amount=

0.0253059

8

0.2024472

3.5035183

Diff. between 0.2024472 and 3.5035183=3.30107002
The natural number belonging to 3.30107002 is $2000.

Example 11. Log. of 1.03, the amount of $1 for half a

year =

Multiplying by 20, the No. of half years,

Product of 20 into .012937=

Logarithm of $1200, the principal=

The sum of .256740 and 3.079181=

.012837

20

.256740

3 079181

3.335921

The natural number belonging to 3.335921 is 2167.3.

Example 12. Assume the principal $650, whose logarithm is

Logarithm of $1300, the amount=

Diff. between 2.812913 and 3.113943=

2.812913

3.113943

0.301030

The logarithm of the amount of $1 for one year, is 0.025305, and 0 301030 divided by 0.025305 is 11.89, with a remainder 15355, so that the quotient is very nearly 11.9, the answer.

Example 13. Log. of 1.06, amount of $1 for one year is,

Multiply by 281,

0.0253059

28.25

Logarithm of $5000,=

Sum of 0.714881675 and 3,698970=

0.714891475

3.698970

4.413861675

The natural No. belonging to 4.413861675, is 25943.532.

INCREASE OF POPULATION. Art. 61. b.

The natural increase of population in a country, may be calculated in the same manner as compound interest, making use of the same formula which were there employed, letting Pthe population at the beginning of the year, a=the rate of increase, n=any number of years, and A= the amount of the population at the end of n years.

Example 2. By the formula, the amount, or the number

of inhabitants at the end of 100 years=100×log.

of 5,000,000.

31

30

+log.

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Example 3. By the formula, the population at the com

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Natural number of 7,301,044=20,000,000, the answer.

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