12.5 Logarithms

12.37   Definition (Logarithm.) Let $t\in\mbox{${\mbox{{\bf R}}}^{+}$}$. The logarithm of $t$ is the unique number $s\in\mbox{{\bf R}}$ such that $e^s=t$. We denote the logarithm of $t$ by $\ln(t)$, Hence

$$e^{\ln (t)}=t\mbox{ for all }t\in\mbox{${\mbox{{\bf R}}}^{+}$}.$$ (12.38)

12.39   Remark. Since $\ln (e^r)$ is the unique number $s$ such that $e^s=e^r$, it follows that

$$\ln(e^r)=r\mbox{ for all }r\in\mbox{{\bf R}}.$$ (12.40)

12.41   Theorem. For all $a,b\in\mbox{${\mbox{{\bf R}}}^{+}$}$,

$$\ln(ab)=\ln a+\ln b.$$

Proof:

$$\begin{eqnarray*} % latex2html id marker 19737\ln(ab)&=&\ln(e^{\ln a}\cdot e^{... ...\ln b \quad\mbox{ (by } (\ref{loge})).\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

12.42   Exercise. A Show that

a) $\ln(a^{-1})=-\ln(a)$ for all $a\in\mbox{${\mbox{{\bf R}}}^{+}$}$.

b) $\ln(a^r)=r\ln(a)$ for all $a\in\mbox{${\mbox{{\bf R}}}^{+}$}$, $r\in\mbox{{\bf Q}}$.

c) $${ \ln \left( {a\over b}\right)=\ln a -\ln b}$$ for all $a,b\in\mbox{${\mbox{{\bf R}}}^{+}$}$.

12.43   Remark. It follows from the fact that $\exp$ is strictly increasing on $\mbox{{\bf R}}$ that $\ln$ is strictly increasing on $\mbox{${\mbox{{\bf R}}}^{+}$}$: if $0<t<s$, then both of the statements $\ln(t)=\ln(s)$ and $\ln(t)>\ln(s)$ lead to contradictions.

12.44   Theorem (Continuity of $\ln$.) $\ln$ is a continuous function on $\mbox{${\mbox{{\bf R}}}^{+}$}$.

Proof: Let $a\in\mbox{${\mbox{{\bf R}}}^{+}$}$, and let $f$ be a sequence in $\mbox{${\mbox{{\bf R}}}^{+}$}$ such that $f\to a$. I want to show that $\ln \circ f \to \ln(a)$. Let $N_{f - \tilde{a}}$ be a precision function for $f - \tilde{a}$. I want to construct a precision function $M$ for $\ln\circ f - \widetilde{\ln(a)}$.

Scratchwork: For all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$, and all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} \vert\ln(f(n)) - \ln(a)\vert < \varepsilon & \mbox{$\Longleftr... ...(a) - \varepsilon} - a < f(n) - a < e^{\ln(a) + \varepsilon} - a \end{eqnarray*}$$

Note that since $\ln$ is strictly increasing, $e^{\ln(a)+\varepsilon} - a$ and $a - e^{\ln(a)- \varepsilon}$ are both positive. This calculation motivates the following definition:

For all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$, let

$$M(\varepsilon) = \max(N_{f - \tilde{a}}(e^{\ln(a) + \varepsilon} - a), N_{f - \tilde{a}}(a - e^{\ln(a) - \varepsilon})).$$

Then for all $n\in\mbox{{\bf N}}$, $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$,

$$\begin{eqnarray*} n \geq M(\varepsilon) & \mbox{$\Longrightarrow$}& \cases{ f(n... ...x{$\Longrightarrow$}& \vert\ln(f(n)) - \ln(a)\vert < \varepsilon \end{eqnarray*}$$

Hence $M$ is a precision function for $\ln\circ f - \widetilde{\ln(a)}$. $\mid\!\mid\!\mid$

12.45   Theorem (Differentiability of $\ln$.)The function $\ln$ is differentiable on $\mbox{${\mbox{{\bf R}}}^{+}$}$ and

$$\ln'(x)={1\over x}\mbox{ for all } x\in\mbox{${\mbox{{\bf R}}}^{+}$}.$$

Proof: Let $a\in\mbox{${\mbox{{\bf R}}}^{+}$}$ and let $\{x_n\}$ be a sequence in $\mbox{${\mbox{{\bf R}}}^{+}$}\backslash\{a\}$. Then

$${{\ln(x_n)-\ln(a)}\over {x_n-a}}={{\ln(x_n)-\ln(a)}\over {e^{... ...( {{e^{\ln(x_n)}-e^{\ln(a)}}\over {\ln (x_n)-\ln(a)}}\right)}}.$$

(Note, I have not divided by $0$.) Since $\ln$ is continuous, I know $\{\ln(x_n)\}\to\ln(a)$, and hence

$$\left\{{{e^{\ln(x_n)}-e^{\ln(a)}}\over {\ln (x_n)-\ln (a)}}\right\}\to\exp'\left(\ln(a)\right)=e^{\ln(a)}=a.$$

Hence,

$$\lim\left\{ {{\ln(x_n)-\ln(a)}\over {x_n-a}}\right\}={1\over a};$$

i.e.,

$$\lim_{x\to a}{{\ln(x)-\ln(a)}\over {x-a}}={1\over a}.$$

This shows that $${\ln'(a) = {1\over a}}$$.