Additive characterisation of the generic logarithm

At this point, let's define the generic logarithm a little more carefully.  The generic logarithm is a function from the positive real numbers onto a set ${\displaystyle L}$,



satisfying certain conditions.  We have some choice as to the conditions we use.

One possibility is to require that, for all positive real numbers ${\displaystyle x}$ and ${\displaystyle y}$,

,

where this defines the addition on the right hand side.  This is the "the log of a product is the sum of the logs of the factors" condition.  (We shall need another condition eventually, but let's not introduce it until the need is clear.)

Then it is equally true that

.

Letting ${\displaystyle w=xy}$ and substituting,

,
,

and we have the rather obvious-looking result that 

,

By repeating this argument, we can extend it to finite products of any length.

For ${\displaystyle n}$ a counting number, we can break the logarithm of ${\displaystyle n}$ factors of positive real ${\displaystyle x}$ into ${\displaystyle n}$ copies of the logarithm of ${\displaystyle x}$ :

,
,
.

The last two lines define multiplication of quantities in ${\displaystyle L}$ by counting numbers.

Substituting ${\displaystyle y=1}$ in ,  we get

,

i.e.

.

Subtracting the quantity from each side and rearranging, we have that

.

This is useful in its own right.

Now ${\displaystyle 0}$ is the number (or quantity) that does nothing when you add it.  ${\displaystyle 1}$ is the number that does nothing when you multiply it.  For any positive real number ${\displaystyle x}$, multiplying by no factors of ${\displaystyle x}$ is the same as multiplying by ${\displaystyle 1}$, i.e.

${\displaystyle x^0=1}$.

So if , then

,

since ${\displaystyle 0}$ times any quantity is still ${\displaystyle 0}$.

We have therefore established that



is true not only for ${\displaystyle n}$ a counting number, but also when ${\displaystyle n=0}$.

Now let ${\displaystyle y=1/x}$ in . We can do this, because if ${\displaystyle x}$ is positive real, so is ${\displaystyle 1/x}$.  Then

.

Since ${\displaystyle x\cdot 1/x=1}$ and

.

The astute reader will notice that this defines a suitably behaved ${\displaystyle 0}$ in ${\displaystyle L}$.

Subtracting from each side and rearranging, we get that

,

 which can also be written

.

This, too, is a definition---it tells us what negation means in ${\displaystyle L}$.

So now we have established that



also holds when ${\displaystyle n=-1}$.

We can find the logarithm of a product of a counting number ${\displaystyle m}$ copies of ${\displaystyle x^{-1}}$ as we did for multiple copies of ${\displaystyle x}$ :

,
,
.

Then, since "lg"(x^{-1})= -1\cdot"lg"(x)${\displaystyle ,wehavet\hat{}}$

,
.

So now, setting ${\displaystyle n=-m}$, we have established that 



for any integer ${\displaystyle n}$.  i.e., we can always move an integer power outside the logarithm as a multiplier.

Now let ${\displaystyle p}$ be any rational number, which we write as ${\displaystyle p=\frac{r}{s}}$, where ${\displaystyle s}$ is a counting number (and therefore not zero), and ${\displaystyle r}$ is an integer.   Let ${\displaystyle y=x^p=x^{\frac{r}{s}}}$.  Then, raising this to the power of ${\displaystyle s}$, we find that

${\displaystyle y^s=x^r}$,

and so

,

Since ${\displaystyle r}$ and ${\displaystyle s}$ are both integers, we know we can move the powers outside the logarithms to get

.

We now divide both sides by ${\displaystyle s}$---we know ${\displaystyle s}$ is nonzero---and write

,

or

.

This, again, is a definition.  It tells us what it means to multiply a quantity in ${\displaystyle L}$ by a rational number.  But it also allows us to assert that



is true for any rational number ${\displaystyle n}$, not just integers.

It is reasonable to hope that this would be true for any real ${\displaystyle n}$, not just rationals.  Practical uses of mathematics should be insensitive to whether a number is rational or not.

The various ways to ensure reasonable behaviour amount to various assertions of that ${\displaystyle lg}$ is a continuous function, i.e. that numbers that are close in ${\displaystyle ℝ+}$ have logarithms that are close in ${\displaystyle L}$. 

The simplest way is to ensure that we get what we want is simply to assert that



is true for all real numbers ${\displaystyle n}$, not just rational ${\displaystyle n}$.  One needs to establish that no contradictions arise, but the existence of models (such as ordinary logarithms to a particular base) guarantees contradictionlessness in this less specific setting.

The real question here is how to justify the equation




as a definition for multiplication of quantities in ${\displaystyle L}$ by irrational numbers ${\displaystyle n}$.

One way to do this is to import into ${\displaystyle L}$ the notion of order already found in ${\displaystyle {ℝ}^{+}}$.  One can then transfer the definition of an irrational in ${\displaystyle ℝ}$ using, say, Dedekind cuts over the rationals, into a definition of irrational powers.

To import that order, simply assert that whenever positive real numbers ${\displaystyle x}$ and ${\displaystyle y}$ are such that ${\displaystyle x\le y}$, then .  This defines the meaning of ${\displaystyle \le }$ in ${\displaystyle L}$.  Since  "${\displaystyle \le }$" is, in fact, a total order in ${\displaystyle {ℝ}^{+}}$, the version of  "${\displaystyle \le }$" defined for ${\displaystyle L}$ is also a total order, and the function ${\displaystyle lg}$ is monotonic, and therefore continuous.

Consider, now, the quantities ${\displaystyle x^u}$ and ${\displaystyle x^w}$, with ${\displaystyle u}$ and ${\displaystyle w}$ rational, , and ${\displaystyle x}$ a positive real number.  There are three cases :

• if ${\displaystyle x>1}$, then ,
• if ${\displaystyle x=1}$, then ${\displaystyle x^u=x^w}$, and
• if , then ${\displaystyle x^u>x^w}$.

Now consider ${\displaystyle x^v}$, where ${\displaystyle u\le v\le w}$, where ${\displaystyle v}$ is real, possibly irrational.

In all three cases, ${\displaystyle x^v}$ lies between ${\displaystyle x^u}$ and ${\displaystyle x^w}$, i.e.

${\displaystyle min(x^u,x^w)\le x^v\le max(x^u,x^w).}$

Then, by the monotonicity of ${\displaystyle lg}$, the quantity lies between "lg"(x^u), and so lies between "u\cdotlg"(x).

Let ${\displaystyle v}$ now be irrational, and for definiteness, assume the first case, ${\displaystyle x>1}$.  Consider the Dedekind cut for ${\displaystyle v}$, i.e. the partition of the rational numbers into two subsets, those that are below ${\displaystyle v}$ and those that are above ${\displaystyle v}$.  Then here ${\displaystyle u}$ belongs to the lowside set, and ${\displaystyle w}$ to the highside set.

Since we know how find rational multiples of ${\displaystyle lg\left(x\right)}$, the logarithms of the lowside and highside sets are also well defined in ${\displaystyle L}$.  Since ${\displaystyle lg}$ is monotonic, the order also transfers, so just as every member of the lowside set is lower than every member of the highside set in ${\displaystyle ℝ}$, so also is every member of the image of the lowside set lower---in the sense that is used in ${\displaystyle L}$---than every member of the image of the highside set.  So, for the given ${\displaystyle x}$, we can define a Dedekind cut in ${\displaystyle L}$, and can use this to define multiplication of quantities ${\displaystyle lg\left(x\right)}$ in ${\displaystyle L}$, when ${\displaystyle x>1}$.

The argument for the case switches the ends, but is otherwise the same.

For the case ${\displaystyle x=1}$,



Here, the Dedekind cut argument fails, but it still seems reasonable to define the product of ${\displaystyle v}$ and ${\displaystyle 0}$ by

,

i.e. that ${\displaystyle n\cdot 0=0}$, whether ${\displaystyle n}$ is rational or irrational.