238. The above laws may be thus expressed:
I. Multiplying powers is performed by adding indices.
II. Dividing powers is performed by subtracting indices.

III. Power-raising of a power is performed by multiplying indices.

Root-taking of a power is performed by dividing indices.

Extension of Index Notation. 239. Each of the two inverse laws, viz., II. and IV., are proved under a condition. This condition is equivalent to the statement that the index on the right-hand side shall be positive (Law II.) and integral (Law IV.). Otherwise the right-hand side would be meaningless, for we have only defined the symbol mm, when m is a positive integer.

It is convenient, however, to be able to use these formulæ for any case whatever.

To do this we have only to interpret negative or fractional indices so that Laws II. and IV. shall hold universally. That is, for all positive integral values of m and n,

2cm-n shall equal xm =- , according to Law II., and zemin shall equal </(x"), according to Law IV.

In the formula of Law II., then
First, let n = = m,

2cm-m must equal 20" - 20",
i.e. co must equal 1..

(1). Secondly, let m = 0, then 200-n must equal 2° = 2c",

1 i.e. -" must equal



Thirdly, in the formula of Law IV., let m be not divisible by

n, then


cñ must equal m/(x).


240. The interpretations (1), (2), (3) of the last article, which are of the greatest importance, thus enable us to use the four index Laws for all positive integral values of m and n. It only remains to show that, with the interpretation of negative and fractional indices given in the last article, the same laws hold for all values of m and n whatever. This may be shown by use of the preceding articles. Thus, e.g. To prove


= gőtå.
at means Vaca; :- (26)* = xa ; (neð jöd = vad (Law III.),
:. b = bd/(avad).

Similarly xå

od/ (izbe), :: x x xå – balj (zout) x bd (sobe) = bed](zad x zobe) by Art. 233, (3)

d'avad+bc) (Law I.)





ad+bc =3


Example 1. Write down a series of powers of x beginning with x3, in which each power is derived from the preceding by dividing by x. X3

20 1





1= 20;

[ocr errors]
[ocr errors]
[ocr errors]
[blocks in formation]

Thus we see that negative indices simply carry out the rule that to divide by a is equivalent to subtracting 1 from the index of x. Example 2. Interpret x3, x3, xx.

(.22)2 must equal x2x, i.e. x, i.e. x.

[ocr errors][merged small][merged small]
[ocr errors][merged small]

241. The equations

2cm = P and a = m/P are two different ways of stating the same relation between a, m, and P. In the first, P stands by itself; in the second, x stands by itself.

There is a third way of expressing this same relation, in which m stands by itself. Thus

If 3cm = P, m is called the logarithm of P to the base x; and we write m = logą P (the base being written below).

DEF. A logarithm of a given number to any base is the index of that power of the base, which is equal to the given number.

It should be remembered, then, that a logarithm is simply an index made to stand by itself. Hence the laws of logarithms are simply the laws of indices expressed in different notation. The equations

2cm P and m= logo P have the same meaning: and either one of them may be substituted for the other.

The student should examine the truth of the statement

[ocr errors]

zlog, P= P.

Example 1. What is the logarithm of 8 to the base 2 ?

This means to what power must 2 be raised to give 8. Now 23=8; .: 3 is the logarithm of 8 to the base 2; or 3=log2 8.

Example 2. Write down the values of log; 81, log10 10000, log. 64. Since 34=81,

.: 4=log: 81. Since 104 = 10000, .: 4=log10 10000. Since 43=64,

.:3=log, 64.

242. In using the relation acm P or m = logą P, we shall assume for the present that w and P (i.e. the base and power) are both positive, and that x is not equal to 1.

243. To show that there is only one positive or negative* value of the logarithm of a positive power to a positive base (not equal to 1).

If possible let m and n be two different logarithms of P to the base x: so that

P = c =c. Then, dividing by ac”, men=1.

Now m and n being positive or negative, m n must be positive, negative, or zero.

But, if is not equal to 1, no positive or negative power of a can be equal to 1. For

If x > 1, positive powers of x are > 1; negative powers are <l.
If a < 1, positive powers of x are <l; negative powers are > 1.

.. m-n must be zero. .. m= n.

i.e. the two logarithms supposed to be different are not different.

Cor. In the same manner it may be shown that, if x > 1, the logarithms of numbers to the base x increase as the numbers increase : but, if x < 1, the logarithms of numbers to the base x decrease as the numbers increase.

244. In order to prove any formula involving logarithms, the student has simply to translate from the logarithmic language with which he is unfamiliar into the index language with which he is more familiar. Thus he has only to remember that the two equations

2cm = P and m=logx P are precisely equivalent statements in the two different languages of indices and of logarithms.

h * Whether there are logarithms which are neither positive or negative is not here discussed.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small]

The Logarithmic Laws involving the same base. 246. Law I. log. (m x n) = log2 m + loga n.

Law II. loga (m = n) = logą m – logan.
Law III. log (m") - (log, m) x n.
Law IV. loga (m/m)

log.c (/m) = (logam) = n. These laws are respectively equivalent to those of Art. 238, but expressed in the language of logarithms.

= n.

247. To


logą (m x n) = logc m + loga n. Law I.

logx (m = n) = log. m - log, n. Law II.
Let logo m = a; this means aca = m.
Let logo n=b; this means ab
Now m x n (i.e. aca x ab) = x2+0 by Index Law I.
And mon (i.e. 2Q = ab) = x2-0 by Index Law II.
Expressing these equations in logarithmic language

logą (m n) = a + b, i.e. logx m + log. n.
loge (m = n) = a b, i.e. logx m – logo n.

« ForrigeFortsett »