The Unapologetic Mathematician

Mathematics for the interested outsider

The Exponential Property

We’ve defined the natural logarithm and shown that it is, in fact, a logarithm. That is, it’s a homomorphism from the multiplicative group of positive real numbers to the additive group of all real numbers. Now I assert that this function is in fact an isomorphism.

First off, the derivative of \ln(x) is \frac{1}{x}, which is always positive for positive x. Thus it’s always strictly increasing. That is, if 0<x<y then \ln(x)<\ln(y). So no two distinct numbers ever have the same natural logarithm, and the function is thus injective.

Flipping this around tells us that we definitely have some nonzero values for the function. For example, we know that 0<\ln(2). Now, since the real numbers are an Archimedean field, no matter how big a number y>0 we pick, there will be some natural number n so that y<n\ln(2)=\ln(2^n), where the latter inequality follows from the logarithmic property.

That is, no matter how large a number we pick, \ln takes values at least that large. But because \ln is continuous on a connected interval there must be some number x with \ln(x)=y. Similarly, if y<0 then there will be some x with \ln(x)=-y, and thus \ln(\frac{1}{x})=y. Thus the natural logarithm is surjective.

So, since our function is one-to-one and onto, it has an inverse function. We will call this function the “exponential” (denoted \exp), and define it to be the unique function satisfying

\exp( \ln(x) )=x
\ln(\exp(y))=y

for all positive real x and all real y.

From here it’s straightforward to see that \exp must be the inverse homomorphism. That is, given two real numbers y_1 and y_2 we know there must be (unique!) positive real numbers x_1 and x_2 with \ln(x_i)=y_i. Then we calculate

\exp(y_1+y_2)=\exp(\ln(x_1)+\ln(x_2))=\exp(\ln(x_1 x_2))=
x_1x_2=\exp(\ln(x_1))\exp(\ln(x_2))=\exp(y_1)\exp(y_2)

And it’s clear from here that \exp(0)=1. A homomorphism from the additive reals to the multiplicative positive reals like this is said to satisfy the “exponential property”, which is just the reverse of the logarithmic property from last time.

April 10, 2008 - Posted by | Analysis, Calculus

8 Comments »

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  2. Another parse error. 🙂

    Comment by Jon | April 16, 2008 | Reply

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