Monday, March 13, 2006

Logarithms

Each elementary mathematical operation has an inverse operation that has the ability to cancel it. Subtraction and addition are inverses as are multiplication and division. From one perspective, n-roots are the inverse operation to exponents. If xn = y, then x = ny.

So that the inverse operation becomes:

nxn = x.

But what about the value n? Can we do an operation on n which cancels out x? The answer is yes, with logarithms.

In logarithms, we say that logx y = n. So from this perspective, a logarithm is an inverse of the power operation in the sense that:

xlogx y = y

This concept of logarithms and methods for deriving them were first popularized in the west by John Napier in 1614. Historically, logarithms were handled using logarithm tables. For those interested in the history of logarithms and logarithm tables, Wikipedia has an excellent article here.

One of the most important logarithm functions in loge which is written as ln. This turns out to have very important mathematical properties such as:

∫(1/x)dx = ln x

Here are some very basic properties of logarithms that I use in other proofs:

Lemma 1: logx a + logx b = logx (ab)

Proof:

(1) Let a,b be any two real, nonzero numbers.

(2) Let a' = logx a so that a = x(a')

(3) Let b' = logx b so that b = x(b')

(4) Now x(a')*x(b') = x(a' + b') [See here for review of exponents if needed]

(5) So (ab) = x(a' + b') [From #4]

(6) So logx (ab) = a' + b' = logx a + logx b

QED

Lemma 2: logx a - logx b = logx (a/b)

Proof:

(1) Let a,b be any two real, nonzero numbers.

(2) Let a' = logx a so that a = x(a')

(3) Let b' = logx b so that b = x(b')

(4) Now x(a')/x(b') = x(a' - b') [See here for review of exponents if needed]

(5) So (a/b) = x(a' - b') [From #4]

(6) So logx (a/b) = a' - b' = logx a - logx b

QED

Lemma 3: b*logx a = logx (ab)

Proof:

(1) Let a,b be any two real, nonzero numbers.

(2) b*logx a = logx a + ... + logx a [So that we have b items to add together]

(3) From (#2) and Lemma 1 above, we have
b*logx a = logx a + ... + logx a = logx (a*a*...*a)

(4) Since we have b multiples of a, we get:
b*logx a = logx (ab)

QED

References

1 comment :

adu said...

Thank you. This was actually the first google result which actually proved the identities.

Very useful, thanks.