The Function log x

Let . Then the function $f(t)=\frac{1}{t}$ is defined and continuous on the interval between $t=1$ and $t=x$. We define a function for , which we shall call $\log x$, by

$$\log x={\int }_1^x\frac{1}{t}dt$$

Obviously this function has one property which we have attributed to that rather indefinite function which we have previously written as $\log x$: namely, (\log x)' = \frac{1}{x}. The next basic property we prove is

$$\log (ab)=\log a+\log b$$

But for any fixed $a$, consider $\log (ax)$ as a function of $x$. Then

(\log (ax))' = \frac{1}{ax} \cdot a = \frac{1}{x} = (\log x)'

Therefore $\log (ax)=\log x+c$, $c$ some constant. For $x=1$, we have

$$\log a=\log 1+c$$

Since

$$\log 1=0(={\int }_1^1\frac{1}{t}dt),$$

$c=\log a$, and $\log (ax)=\log x+\log a$. This is what we wished to prove.

Now let ; then $\log a\leqq a-1$, for this is the area of the larger rectangle in the diagram below.

The smaller rectangle has area $(a-1)\cdot \frac{1}{a}\leqq \log a$. Thus we have for ,

$$a-1\geqq \log a\geqq \frac{a-1}{a}=1-\frac{1}{a}.$$

This is an equality when $a=1$, so is true whenever $a\geqq 1$. Now if , then , and we have

$$\frac{1}{a}-1\geqq \log \frac{1}{a}\geqq 1-\frac{1}{1/a}$$

by the above. Now

$$\log a+\log \frac{1}{a}=\log 1=0,$$

so

$$\log \frac{1}{a}=-\log a.$$

This gives

$$\frac{1}{a}-1\geqq -\log a\geqq 1-a,$$

or, reversing signs

\tag{*} a - 1 \geqq \log a \geqq 1 - \frac{1}{a}

Thus we have proved the inequality (*) for all .

Let us use this inequality to derive some properties of $\log x$.
Using $a=1+\frac{1}{n}$, $n$ a positive integer, we find

$$\frac{1}{n}\geqq \log (1+\frac{1}{n})\geqq 1-\frac{1}{1+\frac{1}{n}}=1-\frac{n}{n+1}=\frac{1}{n+1}$$

Since $n(\log a)=\log (a^n)$ (prove by induction), we have

$$1=\frac{n}{n}\geqq n\log (1+\frac{1}{n})=\log (1+\frac{1}{n}{)}^n\geqq \frac{n}{n+1}\cdot$$

Now as $n$ becomes large, $\frac{n}{n+1}$ approaches $1$. Therefore $\log (1+\frac{1}{n}{)}^n$ is between $1$ and a quantity which approaches $1$, and hence must itself approach $1$. We write this as

$$\underset{n\to \mathrm{\infty }}{lim}\log {(1+\frac{1}{n})}^n=1.$$

Next consider the expression $\log (2^n)=n\log 2\geqq n(1-\frac{1}{2})=\frac{n}{2}$. As $n$ becomes large, $\frac{n}{2}$ also becomes large, and consequently $\log (2^n)$ becomes as large as we wish. Similarly, $\log (2^{-n})\leqq -\frac{n}{2}$, and as $n$ becomes arbitrarily large, $\log (2^{-n})$ becomes arbitrarily small (i.e., far negative). Thus we see that $\log x$ takes on arbitrarily large and arbitrarily small values.

$\log x$ is continuous at every , since it is differentiable. Moreover, it is strictly increasing. If , it is clear that . If , then the area under $y=1/x$ from $1$ to $x_1$ is evidently less than that from $1$ to $x_2$, so again . Finally, if , the area from $x_1$ to $1$ is greater than that from $x_2$ to $1$; but the logarithms are these areas taken with a negative sign, giving once more . Thus we have proved the assertion.