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\chapter{Scales of Measurement}

\specialsection{Introduction}
Dealing with real-world data in a quantitative way means we must
select one of the scientific systems of \emph{units} for our 
measurements.  We will begin by reviewing the \emph{metric system}, 
different forms of which are used most commonly in science.  We 
will also recall the techniques needed to convert measurements given
in metric units to the perphaps more familiar English units used
in the U.S.  To prepare for material to come in later chapters, we will 
recall some of the mathematics of \emph{logarithms} and discuss measurements
on logarithmic scales.\footnote{Depending on how comfortable the class is with this material, 
we might omit covering some or all of this in class.}    Each chapter of these notes will end with an
extended Chapter Project.  The first of these will apply the material
developed to estimate the total amount of water contained in the
Greenland ice sheet and the expected increase in average sea level
if all the freshwater ice sheets melt and that water is introduced into the oceans.  

\specialsection{Metric Units}
The metric system
\index{metric system} 
\index{units!metric}
commonly used in science today was originally
standardized by the First French Republic in 1799 as part of the 
``new start'' following the revolution that deposed the king Louis XVI and 
abolished the French monarchy. In its basic form, the basic units were: 
\begin{itemize}
\item The \emph{meter}, abbreviated as $m$, a idealized measure of length equal to one ten millionth
of the distance from the North Pole to the Equator along the line of 
longitude through Paris.  
\item Areas can be measured in square length units (e.g. square kilometers). 
Another unit, the \emph{hectare}, abbreviation $ha$, is often used for plots of land and
other everyday uses.  One hectare is the area of a square $100$ meters
on a side, or $10,000$ square meters.
\item The \emph{kilogram}, abbreviation $kg$, an idealized measure of the mass of 
a cubical volume of pure water $1/10$ of a meter on a side, at the melting point
of ice.
\item The \emph{liter}, abbreviation $l$, is the volume given by the cube $1/10$ of a meter
on a side.
\item The \emph{second}, abbreviation $sec$, is the standard unit of time.
\item The degree Celsius (centigrade) is the standard unit of temperature,  where
the freezing point of water is $0^\circ$ C and the boiling point
of water is $100^\circ$ C.
\end{itemize}
To facilitate calibration, these notional definitions of the meter and the kilogram were 
replaced almost immediately by standard metal prototypes kept in the 
French national archives.  But such physical models change with temperature,
air pressure, and other variables.  Later, in 1960, the definition of the 
meter was changed to make it even more precise and reproducible -- a meter
is now taken to be a certain number of wavelengths of a precisely
specified color of light (electromagnetic radiation) that can be produced
by lasers of a certain type.  

For scientific work, the great advantage of the metric system is the 
way it follows the standard way we represent numbers -- the Hindu-Arabic
numerals with base-10 arithmetic.  A standard series of prefixes 
represent positive and negative powers of 10:

\begin{table}[ht]
\caption{Metric Prefixes}\label{MetricTable}
\renewcommand\arraystretch{1.5}
\noindent\[
\begin{array}{|c|c|c|}
\hline
{\rm prefix} & {\rm abbrev.} & {\rm power} \\
\hline
{\rm tera}& {\rm T}&10^{12}\\
\hline
{\rm giga}& {\rm G}&10^9\\
\hline
{\rm mega}& {\rm M}&10^6\\
\hline
{\rm kilo}& {\rm k}&10^3\\
\hline
{\rm hecto}& {\rm h}&10^2\\
\hline
{\rm deca}& {\rm da}&10^1\\
\hline
{\rm deci}& {\rm d}&10^{-1}\\
\hline
{\rm centi}& {\rm c}&10^{-2}\\
\hline
{\rm milli}& {\rm m}&10^{-3}\\
\hline
{\rm micro}& \mu &10^{-6}\\
\hline
{\rm nano}& {\rm n}&10^{-9}\\
\hline
{\rm pico}& {\rm p}&10^{-12}\\
\hline
\end{array}
\]
\end{table}
As you might guess, the ``hect-'' in the name \emph{hectare} comes from the multiple
for 100.  A hectare is 100 \emph{ares}; the \emph{are} is rarely used, though.  
Note how this system makes it possible to talk about lengths, masses, volumes, etc.
over a huge range of sizes.  This means that the same basic system of units applies
to study both extremely small things, such as the cells in our bodies, atoms within our
cells, etc. and extremely large things, such as our solar system, the Milky Way galaxy, 
etc.  In fact, there are standard names for even larger and
even smaller powers of $10$ as well, but we will not have any need for them.

\begin{example}
\label{Ex1}
Here are some examples of working with various metric units.
\begin{enumerate}
\item a \emph{kilogram} equals $10^3 = 1,000$ grams (a gram is roughly the mass of the plastic cap of a ballpoint pen;
body masses of humans typically run in the range 50 - 100kg.)
\item a \emph{microgram} equals $10^{-6} = \frac{1}{1,000,000} = .000001$ of a gram (one one-millionth of a 
gram -- roughly the mass of a large human cell like the egg cells produced in the female reproductive
system) 
\item a \emph{milliliter} equals $10^{-3} = .001$ of a liter (note or recall that a milliliter is 
also the same as a \emph{cubic centimeter}, since the cube $1/10$ of a meter (a decimeter!) on a side
that gives the liter has a side equal to $10$ centimeters, so $1$ liter equals $10^3 = 1000$
cubic centimeters)
\item a \emph{terameter} is $10^{12} = 1,000,000,000,000$ meters (roughly the distance light travels in one hour
in vacuum; the distance
from the Sun to the planet Saturn is about $1.4$ terameters)
\item a \emph{megasecond} is $10^6 = 1,000,000$ seconds (a bit over $11.5$ days) 
\end{enumerate}
Measurements of time are rarely expressed in purely metric terms when they are communicated
to humans(!)
$\triangle$ \end{example}

If you have taken physics, you may recall that in discussions of that subject 
a big deal is often made of the difference between the ``cgs'' ($=$ centimeters, grams, seconds) and 
``mks'' ($=$ meters, kilograms, seconds) versions of the metric system and what are the associated units of 
velocity, acceleration, work, energy, etc. depending on whether the cgs or mks system is 
adopted.  We will not need to stress this distinction, though, and 
we will generally use units chosen for convenience according to the sizes of the quantities being measured.  

\specialsection{Conversions}
Because of the ``powers of 10'' prefix system from Table~\ref{MetricTable},
conversions \emph{within} the metric system essentially just involve shifting decimal points
in numbers.  This is the main reason why the metric system is so easy and why it has been 
adopted virtually universally both for scientific use and for everyday measurements!
\index{conversion factors}
\index{English units}
\index{units!English}

\begin{example}
Here are some examples of conversions within the metric system.
\begin{enumerate}
\item A distance of $17.3$ kilometers can also be expressed in terms of meters like this:
$$17.3\ {\rm km} \times 1000\ {\rm m/km} = 17,300\ {\rm m}.$$
As usual in ``dimensional analysis,'' the units of km in the first number cancels the km
in the denominator of the conversion factor $1000$ meters per kilometer.  
\item Similarly, to convert a volume of $343.2$ milliliters to the equivalent
number of liters, we just need to remember that a milliliter is $10^{-3} = .001$ liters,
and we have 
$$343.2\ {\rm ml} \times .001\ {\rm l/ml} =  .3432\ {\rm l}.\qquad \triangle$$
\end{enumerate}
\end{example}

As of 2017, there are \emph{only three} countries in the world that do not
use the metric system for everyday measurements -- the African country of Liberia, the Asian 
country of Myanmar (formerly known as Burma), and the United States.\footnote{Editorial comment:  this is 
an example of the dark side of the idea of ``American exceptionalism.''  It can be argued that 
we do some things differently and better than many other nations; using the English system of units 
is not one of those things!}  This means that the metric units above are probably less intuitively 
familiar than the English units:
\begin{itemize}
\item lengths in inches, feet, yards, miles
\item masses or weights in ounces, pounds, tons
\item volumes in fluid ounces, quarts, gallons 
\item temperatures in degrees Fahrenheit
\end{itemize}
There are others too, of course! These are just the most basic and common ones.
Using these units, we first need to remember conversions within the system. 
For instance, rounding to three significant digits in all cases we have the following.  

\begin{definition}
\label{EngConvs}
The following constants give conversions \emph{within the English system}.
\begin{enumerate}
\item $1$ foot $= 12$ inches, so $1$ inch $= 1/12 \doteq
.0833$ foot.\footnote{We will use the symbol 
$\doteq$ consistently to mean that the two quantities are \emph{approximately equal}.  The
exact value of the rational number $\frac{1}{12}$ is the infinite repeating decimal
$.08\overline{3}$.  The approximate value $.0833$ is thus slightly smaller than
$\frac{1}{12}$.}     $1$ yard $= 3$ feet $= 36$ inches.  
Then $1$ mile = $5280$ feet, so $1$ foot $ = 1/5280 \doteq .000189$ mile.
\item $1$ pound $= 16$ oz., so $1$ oz. $=1/16 = .0625$ pound.  
Then $1$ ton $= 2000$ pounds, so $1$ pound $= 1/2000 = .0005$ ton.  
\item $1$ quart $ = 32$ fl. oz., so $1$ fl. oz. $=1/32 \doteq .0313$ quart.  
Then $1$ gallon $= 4$ quarts, so $1$ quart $= .25$ gallon.
\item $1$ acre $= 1/640 \doteq .0015625$ square mile
\end{enumerate}
\end{definition}

Then, if we want to convert between the English system and the metric system, 
we need to remember another set of conversion factors.

\begin{definition}
\label{MetricEngConvs}
The following constants give metric to English and English to metric conversions.
\begin{enumerate}
\item $1$ inch $\doteq 2.54$ centimeters, and $1$ cm $=1/2.54 \doteq .394$ inch.
\item $1$ foot $\doteq 12 \times 2.54$ cm $\doteq 30.5$ cm $=.305$ m.
\item $1$ mile $\doteq 5280 \times .305$ m $\doteq 1610$ m $\doteq 1.61$ km.
Hence $1$ km $\doteq 1/1.61 \doteq .621$ mile. 
\item $1$ ha $\doteq$ 2.471 acres $\doteq .00386$ square mile. 
\item $1$ pound $\doteq .454$ kilogram, and $1$ kg $\doteq 1/.454 \doteq 2.20$ pounds.
\item $1$ quart $\doteq .946$ liter, and $1$ liter $\doteq 1/.946 \doteq 1.06$ quarts.
\item To convert back and forth between a temperature $F$ in degrees Fahrenheit 
and the equivalent Celsius temperature $C$, use
$$C = \frac{5}{9}(F - 32) \qquad {\rm and} \qquad F = \frac{9}{5}C + 32.$$
\end{enumerate}
\end{definition}

Let's practice using the conversion factors from Definitions~\ref{EngConvs} and 
\ref{MetricEngConvs} on several examples.
\begin{example}
\label{MetricExamples}
Be sure you understand the thinking behind these and check the arithmetic
as you are reading.  Reading a mathematics book should be an active endeavor!
\begin{enumerate}
\item To begin, let's ask how long a distance a distance of 3 kilometers is in miles, yards, and feet.  
From the above, we see
$$3\ {\rm km} \times \frac{1}{1.61}\ {\rm miles/ km} \doteq 1.86\ {\rm miles}$$
Then to get the equivalent length in feet, we multiply by the conversion factor
from miles to feet:
$$1.86\ {\rm miles} \times 5280\ {\rm feet/mile} \doteq 9820\ {\rm feet}.$$
Finally, to get the equivalent distance in yards, we multiply by the conversion factor from feet to yards:
$$9280\ {\rm feet} \times \frac{1}{3}\ {\rm yards/foot} \doteq 3090\ {\rm yards}.$$
\item Next, we ask:  How much does a 1 meter by 1 meter by 1 meter cube of water weigh in 
tons and what is its volume in fluid ounces?  There are many ways to answer the first part of the
question.  Probably the most direct, 
though, is to find the mass of the water in kilograms first, and then convert this to the equivalent weight in pounds and then tons.
The reason for this approach is the fact that a cube with side $1/10$ meter has a mass of 1 kilogram
by the definition of the metric units.  Since our cube has a side that is 10 times as long as this, 
the mass of the water is $10 \times 10 \times 10 = 1000$ kilograms.   Then
$$1000\ {\rm kilograms} \times 2.20\ {\rm pounds/kilogram} \times \frac{1}{2000}\ {\rm tons/pound} \doteq 1.1\ {\rm tons}.$$
This says, for instance, that just the water contained in a 3 meter by 2 meter by 1/6 meter waterbed would also weigh about 1.1 tons!\footnote{Think about
that before setting one up in your 4th-floor apartment bedroom!} 
For the second part, we note that the volume is $10 \times 10 \times 10 = 1000$ liters, and 
$$1000\ {\rm liters} \times 1.06\ {\rm quarts/liter} \times 32\ {\rm fl. oz./quart} \doteq 33920\ {\rm fl. oz.}$$
\item Now, suppose we have a flat 1 square mile field that is flooded with water to a
depth of one inch.  We ask:  what is the total volume of the water in cubic meters?
Thinking of the water as (approximately!) a rectangular solid, the volume is length $\times$ width $\times$ height.
We want the volume in cubic meters, so it makes sense to convert all three dimensions to 
meters first, then compute the volume in cubic meters.  So we have
$1$ mile $\doteq 1610$ meters, and 
$$1\ {\rm inch} \times .083\ {\rm foot/inch} \times .305\ {\rm meter/foot} \doteq .0235\ {\rm meter}.$$
Hence the total volume of the water is approximately
$$1610\ {\rm meters} \times 1610\ {\rm meters} \times .0235\ {\rm meter} \doteq 65580\ {\rm cubic\ meters}.$$
We treated the shape of the field as though it was a perfectly flat square one mile on a side.   The given
information says we may ignore
the curvature of the Earth here and assume the field is perfectly flat (that is, contained in one plane).  This
is reasonable because one mile is so much smaller than the Earth's radius (approximately 4000 miles).
The answer would also be the same no matter what the actual shape of the field was because a solid of uniform 
height over a fixed perfectly planar base
always has volume equal to the area of the base times the height.
\item Finally, a temperature of $35^\circ$ Celsius is equivalent to a Fahrenheit temperature of
$$\frac{9}{5} \cdot 35 + 32 = 95^\circ\ {\rm F}.$$ 
The (much more comfortable) Fahrenheit temperature $72^\circ$ F is equivalent to a Celsius
temperature
$$\frac{5}{9}(72 - 32) \doteq 22.2^\circ\ {\rm C}.\qquad \triangle$$
\end{enumerate}
\end{example}

\specialsection{Estimation}
In many circumstances, exact measurements of quantities in real-world processes may be 
unavailable due to the difficulty or the cost of carrying them out.  In these
situations, a good \emph{estimate} or ``educated guess'' (with emphasis on the ``educated,'' of course!)
may be the most we can hope for.  Making
good estimates requires both a solid understanding of the process and an intuitive grasp 
of the units involved.  Here is a simple example.

\begin{example} 
Suppose we wish to estimate the average per capita daily usage of water for a U.S. resident.  
First we must think of all the ways that we use water in a typical day.  Most of us probably
use
\begin{itemize}
\item about 16 gallons to take a shower lasting 8 minutes (at about 2 gallons per minute)
\item about 24 gallons in bathrooms (halve that if you use efficient 1 gallon per flush toilets, but
most people \emph{do not} have them)
\item about 2 gallons for drinking and cooking purposes
\end{itemize}
That's about 42 gallons a day, but are those the only uses of water most of us have?  If you 
think about it, you'll see that there are large water uses in addition to these:
\begin{itemize}
\item if you water a lawn or use a swimming pool, add 25 gallons a day
\item add 4 gallons per day for use in washing laundry
\item add 4 gallons per day if you use a dishwasher
\end{itemize}
This adds up to an estimate of about 75 gallons per day (and it doesn't even include
uses of water to wash the car or clean the house, etc.)

In fact, this is pretty close to the mark.  The U.S. Environmental Protection Agency
estimates that the average American household uses about 300 gallons of water
per day\footnote{According to {\tt https://www.epa.gov/watersense/how-we-use-water}}.  

By way of contrast, per capita water usage in most other parts of the world (even 
in economically advanced areas such as Western Europe) is significantly lower.  
Even though most parts of the U.S. have adequate water resources to support our
lifestyles at the present time, the same is not true in drier parts of the world.
The cost of purifying water for human consumption is also increasing.   
$\triangle$ \end{example}

\specialsection{A Feature of Human Perception}
To introduce our next topic, we will discuss an interesting feature of the way our senses
(vision, hearing, taste, touch, smell) deal with stimuli from the physical world.  We
will illustrate our point with two series of images of circles.  
In Figure~\ref{fig:ArithCircs} above, the areas of the circles (in suitable square units!) would be 
numbers in the \emph{arithmetic progression} $1,2,3,4,5,6,7,8$.  What this
means is that the change in area from each circle to the next is always the
same: 1 square unit.  
\begin{figure}
\begin{center}
\includegraphics[width=1.5in,width=1.5in]{CircSeqOne.png}
\caption{Circles with Areas in Arithmetic Progression}
\label{fig:ArithCircs}
\end{center}
\end{figure}

Now compare this with the sequence of circles in Figure~\ref{fig:GeomCircs} on the following page.
Here, if the smallest circle at the left had area 1, then areas would be 
$1,2,4,8,16,32,64,128$.  (The scale is not the same as in the first figure because
the range of areas is much greater.)  In other words, each circle is twice as large
in area as the one before it.  We call such a sequence of number values
a \emph{geometric progression}.  

Compare the two sequences of circles carefully.  In the first case (Figure~\ref{fig:ArithCircs}), 
you should notice that the \emph{rate of growth} in the areas as you sweep your vision from the left
to the right seems to get less and less.  In fact, there is apparently very little 
difference between the last two circles: if presented with those two circles in isolation 
you might be hard pressed to see the difference at all (even though their areas differ
by the same amount as the areas of the first two circles).  On the other hand, the growth 
of the areas in Figure~\ref{fig:GeomCircs} is seemingly steadier and we have no difficulty in perceiving
that the areas are increasing. 

\index{Weber-Fechner Law}
Psychologists call the phenomenon we are seeing here the \emph{Weber-Fechner law} of perception.
In rough terms, the Weber-Fechner law says that human sense perception works on a \emph{logarithmic}
scale. We will explain what this means in detail in the next sections.

\specialsection{Logarithms}
\index{logarithms|(}
You have probably seen logarithms in your high school algebra or precalculus classes. Recall that
the idea is the following.  Given a positive number $a \ne 1$, called the base of the logarithms,
and any positive number $x$ we say that 
\begin{equation}
\label{eq:LogDef}
\framebox{$y = \log_a(x)\quad {\rm if\ (and\ only\ if)}\quad a^y = x$.}
\end{equation}
In other words, the  base $a$ logarithm of $x$ is the \emph{exponent} to which $a$ must be
raised to yield the number $x$.\footnote{The reason
for the restriction $a \ne 1$ can be seen if we think of trying to find $y$ with $1^y = x$
if $x \ne 1$.  In practice, values $a > 1$ are used much more commonly than values $0 < a < 1$.}  
\begin{figure}
\begin{center}
\includegraphics[width=1.5in,height=1.5in]{CircSeqTwo.png}
\caption{Circles with Areas in Geometric Progression}
\label{fig:GeomCircs}
\end{center}
\end{figure}

\begin{example}
For instance
$$\log_3(81) = 4\quad {\rm since}\quad 3^4 = 81.$$
Similarly,
$$\log_{10}\left(\frac{1}{1000}\right) = \log_{10}(10^{-3}) = -3.$$
For numbers $x$ that are not exactly equal to whole number powers of the base $a$, a calculator, 
mathematical computer software, 
or a table is used to find values of the logarithm $\log_a(x)$.  For example by any of these
methods, we find
$$\log_{10}(2) \doteq .30103.$$
This is true since
$$10^{.30103} \doteq 2.$$
Similarly, using a calculator
$$\log_{10}(.58) \doteq -0.23657.$$
Because $.58 < 1$, a negative exponent $y$ is needed to produce an equation $10^y = .58$.  
$\triangle$ \end{example}

From \eqref{eq:LogDef}, if we know $y=\log_a(x)$, then recovering $x$ is simply a matter 
of computing $x = a^y$.  For instance, if we know $\log_{10}(x) = 4$, then $x = 10^4 = 10,000$.  

Properties of exponents carry over into corresponding properties for logarithms.  

\begin{proposition} 
\label{LogProps}
For all $a > 0$ different from 1 and all positive
$x,x_1,x_2,b$, the base-$a$ logarithms satisfy 
\begin{enumerate}
\item $\log_a(x_1 x_2) = \log_a(x_1) + \log_a(x_2)$ -- in words, ``the log of a product
is the sum of the logs.''
\item $\log_a\left(\frac{x_1}{x_2}\right) = \log_a(x_1) - \log_a(x_2)$ -- in words, ``the log
of a quotient is the difference of the logs.''
\item $\log_a(1) = 0$.
\item $\log_a(x^b) = b\cdot \log_a(x)$ -- in words, ``the log of a base to a power is the power
times the log of the base.''
\end{enumerate}
\end{proposition}

\begin{proof}
We are not going to be doing a lot of proofs, but we will look at these because
understanding how logarithms work is important and the proofs show that.  
In fact, all of these statements follow from properties of exponents.
\begin{enumerate}
\item For instance, if $y_1 = \log_a(x_1)$ and $y_2 = \log_a(x_2)$, then we have equations
$$a^{y_1} = x_1 \quad {\rm and} \quad a^{y_2} = x_2.$$
Hence multiplying and using the fact that the exponents add when we multiply two
powers of the same base $a$, we have 
$$x_1x_2 = a^{y_1}a^{y_2} = a^{y_1 + y_2}.$$
Therefore 
$$\log_a(x_1x_2) = y_1 + y_2 = \log_a(x_1) + \log_a(x_2),$$
since that is the exponent in the equation giving $x_1x_2$ as a power of $a$.  
\item This follows in the same way since
$$\frac{x_1}{x_2} = \frac{a^{y_1}}{a^{y_2}} = a^{y_1 - y_2}.$$
\item The equation $\log_a(1) = 0$ follows since $a^0 = 1$.
\item If $y = \log_a(x)$, then $a^y = x$.  Hence, raising both sides to the $b$
power, we get $x^b = (a^y)^b$.  But $(a^y)^b = a^{b\cdot y}$ since the exponents multiply.
This gives $\log_a(x^b) = b\cdot y = b\cdot \log_a(x)$.  
\end{enumerate}
\end{proof}

The idea of logarithms is usually ascribed to the Scottish mathematician John Napier (1550 - 1617).
\index{Napier, John}
Napier was interested mostly in the ways parts (1), (2) and (4) of Proposition~\ref{LogProps}
can be used to simplify numerical calculations.  The idea is that you have a table
of logarithms computed for a suitably ``dense'' set of numbers between $1$ and $10$, say
including $\log_{10}(2.31) \doteq 0.3636120$.  
Then you can deal with numbers of any magnitude by using scientific notation like this:
$$\log_{10}(2.31\times 10^4) = \log_{10}(2.31) + \log_{10}(10^4) = \log_{10}(2.31) + \log_{10}(10^4) \doteq 4.3636120.$$
Most importantly, using this approach, complicated jobs of multiplying, dividing, or raising numbers to powers 
can be replaced by the simpler computations of addition and subtraction of the logarithms,
or multiplication of the logarithm by the power.  As recently as the early 1970's, when 
I was in high school,\footnote{I hesitate to admit this, but that was before the advent of electronic calculators!}
it was still possible to find whole semester courses devoted to these
calculations  in some school curricula.   The textbooks usually contained 7-place log tables
making up most of the book(!) Of course the availability of
calculators has made the very idea of such courses hopelessly old-fashioned today(!)

Most scientific calculators have keys for computing both $\log_{10}$, the base-$10$ or 
\emph{common logarithm}, and $\ln$, the so-called \emph{natural logarithm}, where the 
base is a number called $e \doteq 2.71828$.  The reasons for calling this odd-looking
choice of $a$ ``natural'' are studied in calculus classes and we will not go into the 
details here.  


Much scientific work uses the base-10 or common logarithms pretty exclusively and we will 
follow that practice in these notes as well.  You should be aware, however, that converting 
back and forth between these two (and actually \emph{any} two) systems of logarithms is 
very easy if you ever need to do it.  
The reason is a consequence of the reasoning in part 4 of Proposition~\ref{LogProps}.  

For instance, if $e^y = x$, so $y = \ln(x)$, then 
using the equation $e = 10^{\log_{10}(e)}$ and substituting in for the $e$ in
$e^y = x$, we also have
$$(10^{\log_{10}(e)})^y = 10^{y\log_{10}(e)} = x,$$
so 
$$\log_{10}(x) = y\log_{10}(e) = \ln(x)\log_{10}(e).$$
In other words, to convert from the natural log of $x$ to the common log of $x$, 
you just multiply by the constant $\log_{10}(e) \doteq .43429.$
To convert the other way, you divide the common log by $.43429$, which is the 
same as multiplying by $2.3026 \doteq \ln(10)$. 

The most general statement along these lines is given in the following general conversion formulas.
For any positive numbers $a,b,x$:
\begin{equation}
\label{LogConv}
\framebox{$\displaystyle{\log_b(x) = \log_a(x) \cdot \log_b(a)}\quad {\rm and} \quad 
\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$.}
\end{equation}
\noindent
Here is an example.

\begin{example}
Say $b = e$, so $\log_b = \ln$ is the natural logarithm.  Also take $a = 10$.
Using a calculator, we have 
$$\ln(4.33) \doteq 1.4656.$$
Hence rounding to 5 decimal digits,
$$\log_{10}(4.33) \doteq \frac{1.4656}{\ln(10)} \doteq \frac{1.4656}{2.3026} \doteq .63650.$$
All the values are rounded here; if you compute $\log_{10}(4.33)$ directly you 
will get a value closer to $.63649$. These small differences generally do not
make much difference when dealing with real-world data, where a measurement
may only be known to within two or three significant digits anyway due to possible
experimental or observational errors.  But if more precise values are
required, it is always possible to use all of the decimal digits returned
in the calculator values of logarithms. 
$\triangle$ \end{example}

\specialsection{Logarithmic Scales}
We say a quantity or a magnitude is defined using a \emph{logarithmic scale}
\index{logarithmic scale}
if
it is computed using a logarithm of some other quantity.  For instance, if instead
of plotting values of a distance along the $x$-axis in a plot we used the logarithms of 
the distances, then we would be using a logarithmic scale.  

\begin{example}
The Weber-Fechner law
\index{Weber-Fechner Law}
we discussed before
can be stated in a more precise way by saying that the human visual system deals 
with information about the areas of plane regions using a logarithmic scale.  That is, it is 
effectively the logarithm of the area of the region that ``registers'' 
more directly in our minds than the area itself.  Recall that in Figure~\ref{fig:ArithCircs},
we saw circles of areas $1,2,3,4,5,6,7,8$.  If we take the (common) logarithms of these
values, we get the numbers in Table~\ref{ArithLogTable}.  
\begin{table}[ht]
\caption{Logarithms of the Areas in Figure~\ref{fig:ArithCircs}}\label{ArithLogTable}
\renewcommand\arraystretch{1.5}
\noindent\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
{\rm Area} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
\log_{10}({\rm Area}) & 0 & .30103&  .47712& .60206& .69897& .77815& .845100& .90309 \\
\hline
\end{array}
\]
\end{table}
The successive differences are decreasing as we go to the right.  For instance, 
the first two logs differ by $.30103$, while the last two differ by much less, namely 
approximately $.06$.  Thus, the logarithmic scale seems to be reproducing the intuitive idea
that the last two areas differed much less in visual terms than the first two. 

On the other hand, in Figure~\ref{fig:GeomCircs}, we saw circles of areas
$1,2,4,8,16,32,64,128$.  The common logarithms of these numbers are (approximately) shown
in Table~\ref{GeomLogTable}.
\begin{table}[ht]
\caption{Logarithms of the Areas in Figure~\ref{fig:GeomCircs}}\label{GeomLogTable}
\renewcommand\arraystretch{1.5}
\noindent\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
{\rm Area} & 1 & 2 & 4 & 8 & 16 & 32 & 64 & 128 \\
\hline
\log_{10}({\rm Area}) & 0 & .30103&  .60206& .90309& 1.20412& 1.50515& 1.80618& 2.10720\\
\hline
\end{array}
\]
\end{table}
Here the successive differences are all $\log_{10}(2)\doteq .30103$ and this
seems to match the perception that those circles are growing steadily in size.  
$\triangle$ \end{example}

There are many other important examples of logarithmic scales used in science.
We will discuss one of these next, and you will see others in the exercises and in future chapters.

\begin{example}
In chemistry, the $pH$ (from ``potential of hydrogen'')  of an aqueous (water-based)
solution is a measure of its acidity or alkalinity.
Acids and bases are distinguished by the level  of $H^+$ ions present.  Formally,
\begin{equation}
\label{pHDef}
\framebox{$\displaystyle{pH = -\log_{10}(a_{H^+}) = \log_{10}\left(\frac{1}{a_{H^+}}\right)}$,}
\end{equation}
where $a_{H^+}$ is the \emph{hydrogen ion activity} (essentially the concentration
in units of moles per liter of solution).  For instance pure water has 
$a_{H^+} \doteq 1.0 \times 10^{-7}$, and hence 
$$pH = -\log_{10}(1.0 \times 10^{-7}) = 7.$$
Acids are aqueous solutions with $pH < 7$ and the smaller the pH, the 
``stronger'' the acid.  For instance, lemon juice has $pH \doteq 2$, while
concentrated hydrochloric acid can have $pH \doteq 0$.  At the other
end of the scale, on average, human blood is slightly alkaline or basic ($pH \doteq 7.4$)
and strongly basic solutions like extra-strength Drano (a liquid 
drain cleaner) can have $pH \doteq 14$.  Because of the $\log_{10}$
in the definition (and the negative sign), if solution A has $pH = 4$ and solution B has
$pH = 5$, then the hydrogen ion activity for solution A is 
$10$ times as large as that for solution B.  The $a_{H^+}$ for solution A is $1.0 \times 10^{-4}$
and the $a_{H^+}$ for solution B is $1.0 \times 10^{-5}$.  
$\triangle$ \end{example}
\index{logarithms|)}

\specialsection{Chapter Project}
\index{Greenland ice sheet|(}
The ice sheet covering the island of Greenland is currently one of the major
concentrations of frozen fresh water situated over a land mass on the Earth.  It is 
estimated to contain about $\frac{1}{9}$ of the total freshwater ice on the planet.
Because
this water is presently on land, if it should melt, that water would
add to the depth of the oceans.  This is different from the situation 
for the Arctic sea ice, or much of the ice in the shelves surrounding 
Antarctica, which rest on sea water.  Melting in those cases would not
appreciably change sea levels because that ice is floating on and supported
by water already.\footnote{This is good since as of July 12, 2017 the Larsen C ice shelf, with an area
of about $44,200$ square kilometers has broken off from the Antarctic coast and become the largest
iceberg ever observed by humans.}   There are important freshwater ice sheets covering the 
land mass of Antarctica as well as these sea ice formations.   
\begin{figure}
\centering
\includegraphics[height=7in,width=4.0in]{GreenlandIce.png}
\caption{The Greenland Ice Sheet, as of about 2013, source 
{\tt https://en.wikipedia.org/wiki/Greenland$\_$ice$\_$sheet}}
\label{fig:Greenland}
\end{figure}

As we will see, there is a large volume of water contained in the 
Greenland ice sheet.  For this reason, its fate is of more
than passing interest for all humans--many of our major cities
and settled coastal areas occur in regions that are close enough to the current 
sea level that any significant increase will cause major disruptions.  
For instance, the \emph{highest} (natural) elevation of any point 
on the island of Manhattan in New York City is only 81 meters ($\doteq 265$ feet)
above sea level and most of the island lies much lower than that. 
The effects of the relatively small increase in average sea level
that have already occurred were evident, for instance, in the 
flooding of Manhattan that happened during ``super-storm Sandy''
in October, 2012.   

The news on this front is not encouraging because
major melting has been observed on Greenland in recent summers.  This is 
thought to be a 
result of a combination of higher air temperatures 
and decreases in the reflectivity of the ice due to 
deposition of soot particles from sources on other
continents (this increases the amount of energy absorbed from solar
radiation during the summer months and increases the rate of melting).  

Refer to the map of Greenland showing the depth of the ice sheet covering
most of its land area in Figure~\ref{fig:Greenland}.
We want to use the information here to estimate the volume of the 
water contained in this ice sheet and understand the possible (now
maybe even probable!) effects if it melts completely.  

Our methods will yield rough estimates or approximations of 
the ice volume.  They will be based on the following simplifying 
assumptions:
\begin{itemize}
\item Even though it appears as a large area on the familiar Mercator
projection maps you may have seen, that map projection distorts areas
of regions near the poles and makes them look much larger than
they actually are.  Greenland does not make up a very large
a portion of the surface area of the (roughly spherical) Earth.  
As a result we will not lose too much if we simply estimate areas as though 
they corresponded to areas on the flat map (i.e. without trying to take the 
curvature of the Earth into account).  The distance scale marked in
the legend of the map can be used, together with a ruler, to approximate linear 
dimensions of regions.  
\item The ice depths (in meters) of the various portions of the ice sheet are encoded
in the map by the 6 different colors.  For the purposes of this 
exercise, let's take the ice depth in a region to be \emph{constant}
at the \emph{lower limit} of the range of depths shown in the legend
of the map.  For example, this will mean that all points colored
with the darkest shade of blue will have ice depths equal to 3000 m
(even though the actual depths can be larger).  Note that this will
produce estimates on the small side of the actual depth.
\end{itemize}  

\subsection*{Questions}
The chapter project will involve investigating the following questions
and writing up your results as directed below.
\begin{enumerate}
\item[(A)] Estimate the areas of each of the six different ice sheet
regions identified by the colors in the map.  Explain how you are doing
this in a clearly-written paragraph.   Note:  There are \emph{many ways} 
to do this in a reasonable fashion and there is not a single correct answer!  
One suggestion would be to print out paper copies of the page with the map 
and draw in collections of rectangles covering, or nearly covering, each of 
the regions by hand.  You don't need to get super-detailed or picky, but
be as accurate as is reasonably possible.  
Then use the distance scale to estimate the areas of those rectangles and
add up a total for each region.
\item[(B)] Multiply each of your area estimates by the depth estimate
to get a volume estimate.  Add the ice volume estimates to get a 
total volume and express in units of \emph{cubic kilometers}.  (As a 
``reality check'' for your method, the total volume of the Greenland
ice sheet is often estimated to be about $3,000,000$ cubic kilometers.  How
close did you come to that?)  Recall that the Greenland ice sheet accounts
for about $\frac{1}{9}$ of the total freshwater ice on Earth.  What is your
estimate of the total volume of freshwater ice on Earth?
\item[(C)]  Now imagine that an amount of water equal to your total volume estimate 
is added to the oceans all at once.  How much would sea levels rise as a result?  One
way to estimate that is to use the same idea as in part (3) of 
Example~\ref{MetricExamples} above.  What is the total surface area of the oceans
on the Earth? (You should look this up online. If you find different estimates, 
how will you choose which one to use?  Explain your thinking.)  
If the water from the melted Greenland ice sheet was spread evenly over that
area, how deep would it be, in meters?  Would the actual change in sea level 
be less than or greater than this estimated height?  Explain.\footnote{Technical note:  
In case you are worried about the fact that we
are ignoring the spherical shape of the Earth, this method is actually
sufficient (i.e. accurate enough) for our purposes because of the fact 
that the depth of the water would be much smaller than the radius of the Earth.  
Here's one way to think about it:  If there is no land area, then 
adding the water from the melted ice sheet is 
equivalent in mathematical terms to changing the radius of the spherical Earth from $r$
to $r + \Delta r$ (with $\Delta r$ representing the depth of the new water, much smaller than $r$ itself). 
The volume of the added water is the difference between the volume of the larger sphere of 
radius $r + \Delta r$ and the volume of the original sphere of radius $r$:
$$\frac{4\pi (r + \Delta r)^3}{3} - \frac{4\pi r^3}{3} = 4\pi r^2\times \Delta r + 4\pi r\times (\Delta r)^2 + 
\frac{4\pi (\Delta r)^3}{3}.$$
Since we assume $\Delta r$ is much smaller than $r$, then the last two terms on the right are 
negligible in size compared to the first term
and we obtain an estimate
$${\rm volume\ of\ added\ water} \doteq 4\pi r^2 \times \Delta r = {\rm surface\ area\ of\ sphere} \times \Delta r.$$
The change in sea level is then approximated by 
$$\Delta r \doteq \frac{\rm volume\ of\ added\ water}{\rm surface\ area\ of\ sphere}.$$
The same idea works even if the water covers only a portion of the surface area of the sphere.}
\item[(D)] Find an elevation contour map of Manhattan.  Use that information 
to estimate what portions of that island would be under water if all the 
freshwater ice sheets melted.
\end{enumerate}
Write up your solutions for these questions as a project report. Include
all of your calculations, the version of the map you used to estimate
the areas of the different regions of the ice sheet, and your answers
to all the ``explain'' portions of the questions above.  
\index{Greenland ice sheet|)}


\begin{xcb}{Chapter Exercises}
\vskip 5pt
\begin{enumerate}
\vskip 5pt
\item Express a volume of $5343$ cubic centimeters in terms of liters and then cubic meters.

\vskip 5pt
\item Express a weight of $4.3$ tons in pounds and then in ounces.  

\vskip 5pt
\item Express the volume $8$ fl. oz. in terms of liters and then milliliters.

\vskip 5pt
\item Express an area of $130$ square kilometers in terms of square miles, square yards, and finally square feet.

\vskip 5pt
\item On July 12, 2017, the Larsen C ice shelf broke free of the rest of the Antarctic ice shelves
and became a free-floating iceberg.  Its total area at the time was estimated to be 44,200 square kilometers (roughly the size
of the state of Delaware!)\vskip 5pt
\begin{enumerate}
\item What was the area of the new iceberg in square miles?  \vskip 5pt
\item The average thickness of the Larsen C ice shelf was estimated at $350$ meters.
Estimate the total volume of ice contained in the new iceberg at the time it separated (``calved''), in cubic kilometers, then in cubic feet.
\end{enumerate}

\vskip 5pt
\item How many minutes have passed since the inauguration of Donald Trump as U.S. President
\index{Trump, Donald J.}
at 12:00noon on January 20, 2017?  (Don't estimate -- find a value as close to exact as possible.
Your answer will depend on when you are doing this problem, of course.  State what that time is!)
\vskip 5pt
\item Light travels at a velocity $c \doteq 299, 792, 458$ m/sec in vacuum.
\vskip 5pt
\begin{enumerate}
\item[(a)] What is the equivalent velocity in units of mi/hr?  (Note:  to get from m/sec
to mi/hr, you will need to multiply by the conversion factors for mi/m and then sec/hr.)
\item[(b)] How long is a \emph{light year} (the distance light travels in one year), measured in km, then in mi?   
\item[(c)] At a speed of $.5 c$, how long would it take to get from Earth to Proxima Centauri, the nearest 
star outside our solar system?  (Look up the distance online.)
\end{enumerate}
\vskip 5pt
\item There is exactly one temperature where the Celsius and Fahrenheit scales give the 
same numerical value.  What is this temperature?  Show your work to determine it.
\vskip 5pt
\item Estimate how many times an average human heart beats over the course of the lifetime of 
its ``owner.''  Your number should be amazing if you think about it--very few things we can
make with moving parts are that durable!
(Hints: First estimate how many times the heart beats over a short time span like one minute.
This depends on lots of things--whether the person is at rest or exercising, what the 
general health of the person is, how old the person is, etc.  Don't worry too much about those, though.   Most
people spend more time in a state close to resting than in intense exercise.  You'll also need to 
estimate the average life span of a human being.  Feel free to look up any information 
you need online, but think carefully about what you find and ask whether it is
general enough for your purposes!)
\vskip 5pt
\item Without using a calculator, compute exactly:
\vskip 5pt
\begin{enumerate}
\item[(a)] $\log_5(125)$ \vskip 5pt
\item[(b)] $\log_3\left(\frac{1}{729}\right)$ \vskip 5pt
\item[(c)] $\log_{10}(.001) \times \log_{7}(49)$ \vskip 5pt
\item[(d)] $\log_2\left(\frac{16^3\times 8}{1024}\right)$.  Do this using Proposition~\ref{LogProps}
first, then check your work by simplifying the fraction $\frac{16^3\times 8}{1024}$ first before
taking the logarithm.
\end{enumerate}
\vskip 5pt
\item Using a calculator (and converting as necessary using \eqref{LogConv}) find
\vskip 5pt
\begin{enumerate}
\item $\log_{10}(5.34689)$, then $\log_{10}(53.4689)$, then $\log_{10}(534.689)$.  Explain 
the pattern you are seeing using Proposition~\ref{LogProps}.  Note that the decimal
point is shifting one space to the right each time. \vskip 5pt
\item $\log_7(34.333)$ \vskip 5pt
\item $\ln(100.3)$ 
\end{enumerate}
\vskip 5pt
\item (Refer to \eqref{pHDef}.)  What is the hydrogen ion activity $a_{H^+}$
for a solution with $pH = 8$?  Same question if $pH = 3$?
\vskip 5pt
\item 
\label{ex:SoundPressure}
\index{sound pressure}
\index{logarithmic scale}
\index{decibels}
The intensities of sounds are often measured in units called \emph{decibels}, or dB. 
The method involves comparing the sound pressure level measured as a result of the sound
with a standard reference level corresponding to ``silence,'' 
where only the resting air pressure is experienced by the measuring device.  In acoustics, the following definition
is used for the sound pressure level, $L$, measured in decibels:
$$L = 20 \times \log_{10}\left(\frac{p_m}{p_r}\right).$$
Here, $p_r$ is the reference sound pressure $p_r = 20$ micropascals\footnote{The pascal is
a metric system unit with dimensions of force per unit area.  A pressure of 1 pascal is 1 
newton per square meter.  The pascal has dimensions kg/(m $\times\ {\rm sec}^2$).} 
in air and $p_m$ is the measured sound pressure.  From this formula, we can see that sound pressures
are measured along a \emph{logarithmic scale}, in agreement with what we said before
about the Weber-Fechner law and human sense perception.
\index{Weber-Fechner Law}
 \vskip 5pt
\begin{enumerate}
\item What is the sound pressure level in decibels corresponding to a measured sound
pressure $p_m = 5000$ micropascals?  \vskip 5pt
\item What is $p_m$ if $L$ has the value $1$ dB, $2$ dB, $10$ dB? \vskip 5pt
\item A jet engine produces sound pressure at the level of $150$ dB.  What would the measured
sound pressure $p_m$ be then?
\end{enumerate}
\vskip 5pt
\item 
\label{ex:QuakeMagnitude}
Strength of earthquakes is currently measured using the \emph{moment magnitude}
scale. 
\index{moment magnitude scale (earthquakes)}
The moment magnitude $M$ is a quantity defined as
$$M = \frac{2}{3} \log_{10}(S) - 10.7$$
where $S$ is the seismic moment (a quantity measured in units of force times distance
and representing the energy released).  This quantity is also given on a logarithmic scale.
The constants $\frac{2}{3}$ and $-10.7$ are
chosen so that the numbers generated are roughly equal to the older \emph{Richter scale}
magnitude, which used a different method.   \vskip 5pt
\begin{enumerate}
\item If two earthquakes have seismic moments $S_1$ and $S_2$ and
 moment magnitudes $M_1$ and $M_2$, show that the
ratio between the two seismic moments is
$$\frac{S_1}{S_2} = 10^{\frac{3}{2}(M_1 - M_2)}.$$
\item If one earthquake has moment magnitude $M_1 = 6$ and a second one has moment magnitude $M_2 = 5$,
how much stronger is the first one in terms of the ratio between the two seismic moments?  Express
your ratio in decimal form and explain its meaning.
\end{enumerate}
\vskip 5pt
\item (For the more mathematically minded.)  Look at the argument given in footnote 9 on page 16.  
If $\Delta r$ is much smaller than $r$, why can we say that $t_2 = 4\pi r \times (\Delta r)^2$
is negligible compared to $t_1 = 4\pi r^2 \times \Delta r$?    Similarly, why can we say
$\displaystyle{t_3 = \frac{4\pi (\Delta r)^3}{3}}$ is negligible compared to $t_1$?  Hint:  What are the ratios
$\frac{t_2}{t_1}$ and $\frac{t_3}{t_1}$ after you simplify?  
\vskip 5pt

\item
\label{ex:Kudzu} 
\index{invasive species}
\index{kudzu}
\emph{Invasive species} are a problem in many habitats.  If
you have spent any time in the southeastern part of the U.S. (especially
South Carolina, Georgia, Alabama)  you have surely seen one
of the most successful and persistent invasive plant species--\emph{kudzu}
vines ({\it Pueraria montana}), ``the vines that ate the South.''  
See Figure~\ref{fig:Kudzu}.\footnote{Source: 
{\tt https://en.wikipedia.org/wiki/Kudzu$\_$in$\_$the$\_$United$\_$States}, downloaded
June 28, 2017.} 
Like many of the most notorious examples of invasives, kudzu was intentionally 
introduced by humans into its present habitat.
\begin{figure}
\begin{center}
\includegraphics[width=2.5in,width=2.5in]{KudzuAtlanta.jpg}
\caption{Kudzu vines in Atlanta, GA}
\label{fig:Kudzu}
\end{center}
\end{figure}
People imported kudzu into the U.S. from Japan and other parts of Asia in the late 1800's.
The Soil Erosion Service and Civilian Conservation Corps promoted its use
as a ground cover to prevent soil erosion; other people sold the plants as ornamental
ground cover.  Unfortunately, the climate and
environment of the South were a perfect match for kudzu and it has 
proliferated.  The major problems caused by kudzu come from the fact that
it is what has been called a ``structural parasite''--rather than supporting itself, it 
covers and smothers human-built structures (including power
and telephone lines), and other trees and plants.  It has very rapid growth
and a versatile ``double-barrelled'' reproductive strategy.  Kudzu uses
both sexual reproduction
via flowers and seeds, and asexual self-cloning whereby a kudzu plant generates 
vines that root, separate from the parent plant, and grow into new genetically 
identical individuals.  \vskip 5pt
\begin{enumerate}
\item There were no kudzu plants in the U.S. before 1876.  Today it covers
about $3,000,000$ hectares (about $7,400,000$ acres).  How much 
is that in square miles?  \vskip 5pt 
\item (Preview of Chapter 2.)  What is the average growth rate of the area covered
by kudzu since 1876 in hectares per year, and also in acres per year? \vskip 5pt
\item Another advantage of kudzu is that it has a very efficient symbiotic
relationship with soil bacteria that lets the plants fix large quantities of nitrogen 
as ammonia ($NH_3$) in the soil it grows in.  This lets kudzu plants essentially 
make their own fertilizer(!) and grow initially in areas that are too nitrogen poor to support
other plants.  According to data reported by Forseth and 
Innis\footnote{Kudzu (\emph{Pueraria montana}): History, Physiology, and Ecology Combine
to Make a Major Ecosystem Threat, Critical Reviews in Plant Sciences vol. 23 (2004) online, 
consulted June 28, 2017.} a stand of kudzu can fix 235 kg of nitrogen per hectare per year.
How much is that in pounds per square mile per month?  Not all of this ammonia stays in
the soil either; it can leach into streams and lakes and cause changes in the plants and animals
there as well.
\end{enumerate}

\end{enumerate}
\end{xcb} 
\endinput


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