# The Unapologetic Mathematician

## The Logarithmic Property

Whoops.. Between preparing my exam, practicing my rumba, and adapting to the new WordPress interface, I forgot to actually post today’s installment

Yesterday we defined the natural logarithm as the function

$\displaystyle\ln(x)=\int\limits_1^x\frac{dt}{t}$

on the interval $\left(0,\infty\right)$. This function is differentiable everywhere in this interval, and its derivative is $\frac{1}{x}$ at each point $x$.

We call this function a logarithm because it satisfies the “logarithmic property”. Simply put, it’s a homomorphism of groups from the group of positive real numbers under multiplication to the group of all real numbers under addition.

That is, since the real numbers are an ordered field they are a fortiori a group if we just throw away the multiplication and order structures. Also, if we get rid of that pesky noninvertible ${0}$ element, they’re a group under multiplication, and the positive elements are a subgroup. The logarithm takes elements of this group and sends them to the additive group, and the homomorphism property reads: $f(xy)=f(x)+f(y)$. In particular, we must have $f(1)=0$.

So is our “natural logarithm” a logarithm? First off, we can easily check that

$\displaystyle\ln(1)=\int\limits_1^1\frac{dt}{t}=0$

As for the other property, let’s write

$\displaystyle\ln(xy)=\int\limits_1^{xy}\frac{dt}{t}=\int\limits_1^x\frac{dt}{t}+\int\limits_x^{xy}\frac{dt}{t}=\ln(x)+\int\limits_x^{xy}\frac{dt}{t}$

Now let’s take the second term on the right here and perform a change of variables, setting $u=\frac{t}{x}$. Then we have $du=\frac{dt}{x}$, and as $t$ runs over $\left[x,xy\right]$ the new variable $u$ runs over $\left[1,y\right]$. That is, we have

$\displaystyle\int\limits_x^{xy}\frac{dt}{t}=\int\limits_1^y\frac{du}{u}=\ln(y)$

and the logarithmic property holds.