- 7 *Elementary 'Iranscendental Functions 423
Theorem 7.7.3 (Laws of Logarithms) \:/x, y E (0, +oo), and \:/n EN,
(a) ln(xy) = lnx + ln y (b) ln ( ; ) = ln x - ln y( c) ln ( xn) = n ln x (d) ln y'x = ~ lnx(e) ln (~) = -lnx (f) ln (xr) = r lnx, \:/r E Q
Proof of (a): Let y > 0 be fixed. By Theorem 7.7.2, \:/x > 0,
d 1
-lnx = -
dx x'
and by the chain rule,d 1 d 1 1
-ln(xy) = - -(xy) = - · y = -.
dx xy dx xy xSince ln x and ln(xy) have the same derivative, they must differ by a con-
stant. That is, 3 C E IR such that
\:/x > 0, ln(xy) = lnx + C. (36)
Letting x = 1, we find t hat lny = C. Plugging this result into (36), we
have ln(xy) = lnx + lny.
Proof of (f): Let r E Q be fixed. By the chain rule, \:/x > 0,
d 1 d 1 r d
-ln(xr) = - -(xr) = -. rxr-l = - = - r lnx.
dx xr dx xr x dxSince ln(xr) and r ln x have the same derivative, they must differ by a
constant. That is, 3 C E IR such that
ln (xr) =rlnx+C. (37)
Letting x = 1, we find that 0 = C. Plugging this result into (37), we have
ln(xr) = r ln x. •
THE NUMBER e
Remark 7. 7.4 In calculus, we define e to be the number that satisfies the
equation ln x = 1.