
8.4: The Logarithm Functions


Definition

Given a positive real number $$x,$$ we call

$\log (x)=\int_{1}^{x} \frac{1}{t} d t$

the logarithm of $$x .$$

Note that $$\log (1)=0, \log (x)<0$$ when $$0<x<1,$$ and $$\log (x)>0$$ when $$x>1 .$$

Proposition $$\PageIndex{1}$$

The function $$f(x)=\log (x)$$ is an increasing, differentiable function with

$f^{\prime}(x)=\frac{1}{x}$

for all $$x>0 .$$

Proof

Using the Fundamental Theorem of Calculus, we have

$f^{\prime}(x)=\frac{1}{x}>0$

for all $$x>0,$$ from which the result follows. $$\quad$$ Q.E.D.

Proposition $$\PageIndex{2}$$

For any $$x>0$$,

$\log \left(\frac{1}{x}\right)=-\log (x).$

Proof

Using the substitution $$t=\frac{1}{u},$$ we have

$\log \left(\frac{1}{x}\right)=\int_{1}^{\frac{1}{x}} \frac{1}{t} d t=\int_{1}^{x} u\left(-\frac{1}{u^{2}}\right) d u=-\int_{1}^{x} \frac{1}{u} d u=-\log (x).$

Q.E.D.

Proposition $$\PageIndex{3}$$

For any positive real numbers $$x$$ and $$y,$$

$\log (x y)=\log (x)+\log (y).$

Proof

Using the substitution $$t=x u,$$ we have

\begin{aligned} \log (x y) &=\int_{1}^{x y} \frac{1}{t} d t \\ &=\int_{\frac{1}{x}}^{y} \frac{x}{x u} d u \\ &=\int_{\frac{1}{x}}^{1} \frac{1}{u} d u+\int_{1}^{y} \frac{1}{u} d u \\ &=-\int_{1}^{\frac{1}{x}} \frac{1}{u} d u+\log (y) \\ &=-\log \left(\frac{1}{x}\right)+\log (y) \\ &=\log (x)+\log (y). \end{aligned}

Q.E.D.

Proposition $$\PageIndex{4}$$

If $$r \in \mathbb{Q}$$ and $$x$$ is a positive real number, then

$\log \left(x^{r}\right)=r \log (x).$

Proof

Using the substitution $$t=u^{r},$$ we have

$\log \left(x^{r}\right)=\int_{1}^{x^{r}} \frac{1}{t} d t=\int_{1}^{x} \frac{r u^{r-1}}{u^{r}} d u=r \int_{1}^{x} \frac{1}{u} d u=r \log (x).$

Q.E.D.

Proposition $$\PageIndex{5}$$

$$\lim _{x \rightarrow+\infty} \log (x)=+\infty$$ and $$\lim _{x \rightarrow 0+} \log (x)=-\infty .$$

Proof

Given a real number $$M,$$ choose an integer $$n$$ for which $$n \log (2)>M$$ (there exists such an $$n$$ since $$\log (2)>0$$ ). Then for any $$x>2^{n}$$, we have

$\log (x)>\log \left(2^{n}\right)=n \log (2)>M.$

Hence $$\lim _{x \rightarrow+\infty} \log (x)=+\infty$$.

Similarly, given any real number $$M,$$ we may choose an integer $$n$$ for which $$-n \log (2)<M .$$ Then for any $$0<x<\frac{1}{2^{n}},$$ we have

$\log (x)<\log \left(\frac{1}{2^{n}}\right)=-n \log (2)<M.$

Hence $$\lim _{x \rightarrow 0+} \log (x)=-\infty . \quad$$ Q.E.D.

Note that the logarithm function has domain $$(0,+\infty)$$ and range $$(-\infty,+\infty)$$.

Exercise $$\PageIndex{1}$$

Show that for any rational number $$\alpha>0$$,

$\lim _{x \rightarrow+\infty} x^{\alpha}=+\infty .$

Proposition $$\PageIndex{6}$$

For any rational number $$\alpha>0$$,

$\lim _{x \rightarrow+\infty} \frac{\log (x)}{x^{\alpha}}=0.$

Proof

Choose a rational number $$\beta$$ such that $$0<\beta<\alpha .$$ Now for any $$t>1$$,

$\frac{1}{t}<\frac{1}{t} t^{\beta}=\frac{1}{t^{1-\beta}}.$

Hence

$\log (x)=\int_{1}^{x} \frac{1}{t} d t<\int_{1}^{x} \frac{1}{t^{1-\beta}} d t=\frac{x^{\beta}-1}{\beta}<\frac{x^{\beta}}{\beta}$

whenever $$x>1 .$$ Thus

$0<\frac{\log (x)}{x^{\alpha}}<\frac{1}{\beta x^{\alpha-\beta}}$

for $$x>1 .$$ But

$\lim _{x \rightarrow+\infty} \frac{1}{\beta x^{\alpha-\beta}}=0,$

so

$\lim _{x \rightarrow+\infty} \frac{\log (x)}{x^{\alpha}}=0.$

Q.E.D.

Exercise $$\PageIndex{2}$$

Show that

$\lim _{x \rightarrow 0^{+}} x^{\alpha} \log (x)=0$

for any rational number $$\alpha>0$$.