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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Exponential and Logarithmic functions}

\begin{prop}
For all \( b \), \( c \), and \( d \), \( c \) and \( d>0 \),  
\begin{enumerate}
\item \( \log _{b}(cd)=\log _{b}(c)+\log _{b}(d) \), 
\item \( \log _{b}(c^{d})=d\log _{b}(c) \), Here \( d \) can be negative. 
\item \[
\frac{\log _{b}(c)}{\log _{b}(d)}=\log _{d}(c).\]
 This last fact means that you can find the logarithm of \( c \) to any base,
if you know all logarithms of one base.
\end{enumerate}
\end{prop}
\begin{proof}
The proofs consist of turning around the rules for exponents. I'll show the
first one as an example, then you will do the other ones as (written) exercises.

To show the first of these rules, start by labelling all the relevant expressions:\begin{eqnarray*}
\log _{b}(c) & := & C\\
\log _{b}(d) & := & D,\: \textrm{and}\\
\log _{b}(cd) & := & E.
\end{eqnarray*}
Then, unwind what these expressions mean, by the definition of the logarithms:\begin{eqnarray*}
\log _{b}(c)=C & \iff  & c=b^{C}\\
\log _{b}(d)=D & \iff  & d=b^{D},\, \textrm{and}\\
\log _{b}(cd)=E & \iff  & cd=b^{E}.
\end{eqnarray*}
But if we look harder at that last equation, and substitute for \( c \) and
\( d \) according to the earlier equations, \begin{eqnarray*}
cd & = & b^{E}\\
b^{C}b^{D} & = & b^{E}.
\end{eqnarray*}
Finally, use the rules for exponents: \begin{eqnarray*}
b^{E} & = & b^{C}b^{D}\\
 & = & b^{C+D}
\end{eqnarray*}
which implies that the two expoenents have to be the same,\[
E=C+D.\]
This is, however, the end, since we recall what these letters meant:\[
\log _{b}(cd)=\log _{b}(c)+\log _{b}(d).\]
\end{proof}

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