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which logarithms, in this case, differ only in their indices, the decimal or positive part, being the same in them all.

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519. And as the indices, or the integral parts of the logarithms of any numbers whatever, in this system, can always be thus readily found, from the simple consideration of the rule above-mentioned, they are generally omitted in the tables, being left to be supplied by the operator, as occasion requires.

520. It may here, also, be farther added, that, when the logarithm of a given number, in any particular system, is known, it will be easy to find the logarithm of the same number in any other system, by means of the equations, aa=n, ex'=n, which give

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x= log. n, x'= 1. n

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(1) (2). Where log. denotes the logarithm of n, in the system of which a is the base, and 1. its logarithm in the system of which e is the base.

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521. Whence axex, or axe, and ex =a, we shall have,

x

for the base a, =log. e, and for the base e,

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x=x' log. e, x'=x.l.a

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. (4).

(3)... Whence, if the values of x and x', in equations (1), (2), be substituted for x and x' in equations (3), (4), we shall have,

log. n= log. eXl.n, and l.n=

1

1

X log. n; or l.n=l.ax

log. e

log. n, and log. n=xl.n. where log. e, or its equal

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presses the constant ratio which the logarithms of n have to each other in the systems to which they belong.

522. But the only system of these numbers, deserving of notice, except that above described, is the one that furnishes what have been usually called hyperbolic or Neperian logarithms, the base of which is 2.718281828459

523. Hence, in comparing this with the common or tabular logarithms, we shall have, by putting a in the latter of the above formulæ 10, the expression

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Where log., in this case, denotes the common logarithm of the number n, and 7. its Neperian logarithm; the constant

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ing what is usually called the modulus of the common or tabular system of logarithms.

524. It may not be improper to observe, that the logarithms of negative quantities, are imaginary; as has been clearly proved, by LACROIX, after the manner of EULER, in his Traité du Calcul Differentiel et Integral; and also, by SUREMAIN-MISSERY in his Théorie Purement Algébrique des Quantités Imaginaires. See, for farther details upon the properties and calculation of logarithms, GARNIER's d'Algebre, or BONNYCASTLE's Treatise on Algebra in two vols. 8vo.

§ II. APPLICATION OF LOGARITHMS TO THE SOLUTION OF EXPO

NENTIAL EQUATIONS.

525. EXPONENTIAL EQUATIONS are such as contain quantities with unknown or variable indices: Thus, a=b,a=c,

y

ad, &c. are exponential equations.

526. An equation involving quantities of the form x, where the root and the index are both variable, or unknown, seldom occur in practice, we shall only point out the method of solv

ing equations involving quantities of the form a, the base a is constant or invariable.

x

ab,

where

527. It is proper to observe that an exponential of the

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form ab, means, a to the power of b*, and not ab to the power of x.

Ex. 1. Find the value of x in the equation a*=b.

Taking the logarithm of the equation ab, we have xX log. b

log. a=log. b; .'. x=; ; thus, let a=5, b=100; then in log. a'

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Б

α

Ex. 2. It is required to find the value of x in the equation

=C.

Assume by, then ac, and yX log. a= log. c; .. y=

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log.c
log.a

(which let)=d. Take the loga

rithm of the equation bd, then, by (Ex. 1.), x=

log, d

log.b

x

Thus, let a=9, b=3, c=1000; then in the equation 93 = log.c log. 1000

1000,

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3.14(d); and a=

log. d
log. b

log. 3.14

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log.3

.4771213

Ex. 3. Make such a separation of the quantities in the equa

tion (a3-6)=a+b, as to show, that

Taking the logarithm, we have

X log. (a+b) 1-xlog. (ab)"

a Xlog. (a2—b2)= log. (a+b), or x × log. (a+b)×(a—b)= log. (a+b);

that is, xx log. (a+b)+xx log. (a−b)= log. (a+b). · Hence xxlog. (a—b) =log (a+b)—xXlog. (a+b)= log. (+b)

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(1—x) log. (a+b) ; ··· 1—x log. (a—b)`

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Ex. 4. Given a+b=c, and a-by=d, required the values of x and y.

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Again, by subtraction, we have 26y=c-d, or by

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Ex. 5. Find the value of x in the equation

Ans. x=

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Ex. 6. Find the value of x in the equation a—√(b2—c2)

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¦ log. (b+c)+1 log. (b—c)—3 log. d—i̟ log. e

log, a

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Ex. 7. Find the value of x in the equation +}=}za»

+1.

Ans. x

log. a

Ex. 8. Given log. x+log. y=' to find the values of x and log. X- log. y= and y.

Ans. x 10/10, and y=10. Ex. 9. In the equation 2=10, it is required to find the value of x. Ans. x 3.32198, &c. Ex. 10. Given /729-3, required the value of x.

Ans. x=6.

Ex. 11. Given 57862-8, to find the value of x.

Ans. 5.2734, &c.

Ex. 12. Given (216)=64, to find the value of x.

Ans. x 4.2098, &c.

Ex. 13. Given43=4096, to find the value of x.

Ans. x=

log. 6

log. 3

= 1.6309, &c.

Ex. 14. Given arty=c, and byd, to find the values of

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CHAPTER XV.

ON

THE RESOLUTION OF EQUATIONS

OF THE THIRD AND HIGHER DEGREES.

§ I. THEORY AND TRANSFORMATION OF EQUATIONS.

528. In addition to what has been already said (Art. 192), it may here be observed, that the roots of any equation are the numbers, which, when substituted for the unknown quantity, will make both sides of the equation identically equal. Or, which is the same, the roots of any equation are the numbers, which, substituted for the unknown quantity, reduce the first member to zero, or the proposed equation to the form of 0=0; because every equation may, designating the highest power of the unknown quantity by a", be exhibited

under the form

3

x2+Axm¬¬1+Bxm¬2+Cxm¬3+ . . Tx+V=0.(1),

A, B, C,.... T, V, being known quantities. And the resolution of an equation is the method of finding all the roots, which will answer the required condition.

529. This being premised, it may now be shown, that if a be a root of the equation (1), the left-hand member of that equation will be exactly divisible by x-α.

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agreeably to the above defi

am+Aam-+Bam-2+Cam-3+... Ta+V=0.

And consequentty, by transposition,

Vam-Aam-1-Bam-2-Cam―3

-Ta.

Whence, if this expression be substituted for V in the first equation, we shall have, by uniting the corresponding terms, and placing them all in a line,

(xm—am)+A(xm¬1 —пm▬1)+B(xm-~2—am—2) +T (x — a)=0. Where, since the difference of any two equal powers of

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