6.7 Integrals, exponential functions, and logarithms  (Page 2/4)

Note that if we use the absolute value function and create a new function $\text{ln}\left|x\right|,$ we can extend the domain of the natural logarithm to include $x<0.$ Then $\left(d\text{/}\left(dx\right)\right)\text{ln}\left|x\right|=1\text{/}x.$ This gives rise to the familiar integration formula.

Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.

Proof

1. By definition, $\text{ln}1={\displaystyle {\int }_1^1\frac{1}{t}}dt=0.$
2. We have
   
   $\text{ln}\left(ab\right)={\displaystyle {\int }_1^{ab}\frac{1}{t}}dt={\displaystyle {\int }_1^a\frac{1}{t}}dt+{\displaystyle {\int }_a^{ab}\frac{1}{t}}dt.$
   
   Use $u\text{-substitution}$ on the last integral in this expression. Let $u=t\text{/}a.$ Then $du=\left(1\text{/}a\right)dt.$ Furthermore, when $t=a,u=1,$ and when $t=ab,u=b.$ So we get
   
   $\text{ln}\left(ab\right)={\displaystyle {\int }_1^a\frac{1}{t}}dt+{\displaystyle {\int }_a^{ab}\frac{1}{t}}dt={\displaystyle {\int }_1^a\frac{1}{t}}dt+{\displaystyle {\int }_1^{ab}\frac{a}{t}·\frac{1}{a}}dt={\displaystyle {\int }_1^a\frac{1}{t}dt+}{\displaystyle {\int }_1^b\frac{1}{u}du=\text{ln}a+\text{ln}b.}$
3. Note that
   
   $\frac{d}{dx}\text{ln}\left(x^r\right)=\frac{r{x}^{r-1}}{x^r}=\frac{r}{x}.$
   
   Furthermore,
   
   $\frac{d}{dx}\left(r\text{ln}x\right)=\frac{r}{x}.$
   
   Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have
   
   $\text{ln}\left(x^r\right)=r\text{ln}x+C$
   
   for some constant $C.$ Taking $x=1,$ we get
   
   $\begin{array}{ccc}\hfill \text{ln}\left(1^r\right)& =\hfill & r\text{ln}\left(1\right)+C\hfill \\ \hfill 0& =\hfill & r\left(0\right)+C\hfill \\ \hfill C& =\hfill & 0.\hfill \end{array}$
   
   Thus $\text{ln}\left(x^r\right)=r\text{ln}x$ and the proof is complete. Note that we can extend this property to irrational values of $r$ later in this section.
   Part iii. follows from parts ii. and iv. and the proof is left to you.

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Defining the number e

Now that we have the natural logarithm defined, we can use that function to define the number $e.$

To put it another way, the area under the curve $y=1\text{/}t$ between $t=1$ and $t=e$ is $1$ ( [link] ). The proof that such a number exists and is unique is left to you. ( Hint : Use the Intermediate Value Theorem to prove existence and the fact that $\text{ln}x$ is increasing to prove uniqueness.)

The number $e$ can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and Maclaurin Series ). Its approximate value is given by

The exponential function

We now turn our attention to the function $e^x.$ Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by $\text{exp}x.$ Then,