8.4: The Logarithm Functions
- Page ID
- 22688
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 \(x>1 .\)
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.
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.
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.
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.
\(\lim _{x \rightarrow+\infty} \log (x)=+\infty\) 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 \((-\infty,+\infty)\).
Show that for any rational number \(\alpha>0\),
\[\lim _{x \rightarrow+\infty} x^{\alpha}=+\infty .\]
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.
Show that
\[\lim _{x \rightarrow 0^{+}} x^{\alpha} \log (x)=0\]
for any rational number \(\alpha>0\).