MATH 12002 - CALCULUS I 5.2: Laws of Logarithms Professor Donald L. - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §5.2: Laws of Logarithms

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 4

Laws of Logarithms

The familiar Laws of Logarithms all follow (directly or indirectly) from the fact that d dx ln x = 1 x .

Laws of Logarithms

Let a and b be positive real numbers and let r be any real number.

1 ln(ab) = ln a + ln b 2 ln 1

b = − ln b

3 ln a

b = ln a − ln b

4 ln(ar) = r ln a D.L. White (Kent State University) 2 / 4

Laws of Logarithms

Proof:

1 Let f (x) = ln(ax), for a positive constant a.

Since

d dx ln x = 1 x , we have by the Chain Rule

f ′(x) = 1 ax · a = 1 x . Hence f (x) and ln x have the same derivative, and so f (x) = ln x + C for some constant C, i.e., ln(ax) = ln x + C. But this holds for all x > 0, in particular for x = 1, so ln a = ln(a · 1) = ln 1 + C = 0 + C = C. Hence C = ln a, and so ln(ax) = ln x + ln a = ln a + ln x. Substituting b for x gives ln(ab) = ln a + ln b.

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Laws of Logarithms

[Proof, continued]

2 We have

0 = ln 1 = ln b · 1 b = ln b + ln 1 b by (1), and so ln 1

b = − ln b.

3 We have

ln a b = ln a · 1 b = ln a + ln 1 b = ln a − ln b by (1) and (2).

4 Let f (x) = ln(xr) for x > 0. As before,

f ′(x) = 1 xr · rxr−1 = r · 1 x = d dx (r ln x), and so r ln x = f (x) + C = ln(xr) + C for a constant C. If x = 1 this becomes r ln 1 = ln(1r) + C, or 0 = C. Hence ln(xr) = r ln x, and substituting a for x gives ln(ar) = r ln a.

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