%% Math 131, Fall, 2003

\documentclass[12pt]{article}
\usepackage{latexsym,amsmath}

\pagestyle{empty}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
%% \setlength{\topmargin}{-.5in}
\setlength{\textheight}{8.25in}

\def\ds{\displaystyle}
\def\R{\textbf{R}}
\def\Item{\item{$\Box$}\quad}
\newcommand{\Neg}{\phantom{-}}

\newcommand{\clog}{\textrm{log}_{10}\,}
\newcommand{\soln}{\paragraph{Solution}}

\def\be{\begin{enumerate}}
\def\ee{\end{enumerate}}

\def\bd{\begin{description}}
\def\ed{\end{description}}

\def\bi{\begin{itemize}}
\def\ei{\end{itemize}}

\begin{document}

\begin{center}
\textbf{Holy Cross College} \\
\textbf{Math 131 (Professor Hwang), Fall Semester, 2003} \\
Solutions to Worksheet for Thursday, September~11
\end{center}

Recall that $\clog 2 \simeq 0.30103$ and $\clog 3\simeq0.47712$.

\be

\item Without using a calculator, find $\clog 5$, $\clog 500$, and 
  $\clog 0.5=\clog(1/2)$.
  
  \soln Since $5=10/2$, we have 
  \begin{align*}
  \clog 5 &= \clog 10 - \clog 2 \simeq 1-0.30103\simeq0.69897 \\
  \intertext{Similarly,}
  \clog 500 &= 3-\clog 2\simeq\Neg2.69897, \\
  \clog 0.5 &= \phantom{3}-\clog 2\simeq -0.30103
  \end{align*}

\vfil
\item Without using a calculator, find $\clog 6$, $\clog 12$, and
  $\clog(3/2)$. Verify your answers with a calculator. \\ Hint:
  $6=2\times3$ and $12=2^2\times3$.

\soln Since ``the log of a product is the sum of the logs'',
\begin{align*}
  \clog 6  &= \phantom{2} \clog 2+\clog 3\simeq0.77815, \\
  \clog 12 &= 2\clog 2 + \clog 3\simeq1.07918, \\
  \intertext{and} 
  \clog(3/2) &= \clog 3-\clog 2\simeq0.176091
  \end{align*}

\vfill\eject
\item The population of Borduria is given by the formula
  $P=(22.6)(1.035)^t$ million, with $t$~measured in years
  since~1990. In what year will the population reach 40~million?

\soln The population is 40~million when $40=(22.6)(1.035)^t$. To solve
for~$t$, divide both sides by~$22.6$ and take the natural logarithm:
\[
\ln\left(\frac{40}{22.6}\right)=t\ln(1.035), \qquad
\text{so}\quad t=\frac{1}{\ln(1.035)}\ln\left(\frac{40}{22.6}\right)
\simeq 16.6~\text{years}.
\]
The population will reach 40~million around 2006~or 2007.

\vfil
\item Referring to the previous question, find the doubling time of
  the population of Borduria, both exactly (as a symbolic answer
  involving logarithms) and numerically.

\soln The doubling time~$t$ satisfies $P=2P_0$, or
$(1.035)^t=2$. Taking logs and solving for~$t$ gives
\[
t\ln(1.035)=\ln 2,\qquad
t=\frac{\ln 2}{\ln(1.035)} \simeq 20.15~\text{years}.
\]

\vfil
\item The population of a certain species of bacteria triples in one
  hour. How long does it take the population to double in size?

\soln The number of bacteria present at time~$t$ (measured in hours)
is $P=P_0\,e^{kt}$ for some constants $P_0$~and $k$. We are given that
the triping time is~$1$: $P_0\,e^k=3P_0$, or $k=\ln 3$. The doubling
time~$t$ satisfies $2=e^{kt}=3^t$, so $\ln 2=t\,\ln 3$, or
\[
t=\frac{\ln 2}{\ln 3}\simeq 0.631~\text{hours}
\simeq 38~\text{min}.
\]

\vfil
\item In this question, you will show that $\clog$~and $\ln$ are
  proportional functions. Start with the equation $y=10^x$, which is
  another way of saying $x=\clog y$. Take the natural logarithm, and
  solve for~$x$. What do you conclude? 

\soln Taking~$\ln$, we find that $\ln y=\ln(10^x) = x\ln 10$. Since
$x=\clog y$,
\[
\ln y = (\clog y)(\ln 10)\simeq(2.3025851)\clog y,\qquad
\text{or}\quad \clog y=\frac{\ln y}{\ln 10}.
\]
In words, the two logarithm functions are proportional.

\ee
\vfill\eject

\end{document}
