\chapter{Exponentials and logarithms}
\section{Logarithmic and exponential models}
There are many real life applications that can be studied by using logarithmic functions.
A typical example is the measurement of sound.
The human ear is extremely
sensitive, because it can detect very faint noises such as the rustling of leaves,
that in comparison to the a very loud noise
such as a jet engine is one trillion times less powerful.
If we disregard the human ear, the physical intensity of sound (that is produced by
pressure waves traveling through the air at 767 miles per hour)
is measured in watts per square meter.
But since the sound intensity encountered in typical natural settings
has such a large range of values (from close to zero to about 1 trillion),
it is more convenient to use a logarithmic function in order to describe how loud a sound is as perceived
by a human ear.
The \textit{decibel} (db for short) is the unit commonly
used to measure how loud a sound is.
See the picture below for the measurement of common sources of sound in decibels.
\[\img{Decibel_Level_Infographic.jpeg}{}{40em}{}\]
If the smallest possible audible sound (near total silence) is 0 db, then a sound 100 times
more powerful is 20 db,
and a sound 1000 times more powerful is 30 db.
The fact that the number of decibels changes so slowly when the sound intensity
changes reminds us the logarithmic functions, that grow very slowly.
The precise relationship is described in the next example.
\begin{example}
The relationship between the number of decibels $\mbox{Db}$
and the intensity of a sound $I$ (in watts per square meter) is
\[ \mbox{Db}(I) = 10 \log I +120. \]
So the number of decibels of a sound with an intensity of 1 watt per square meter will be
\[\mbox{Db}(1)=10\log(1)+120= 10(0)+120=120.\]
If the sound has intensity \(0.01\) watts per square meter, then the number of decibels is
\[\mbox{Db}(0.01)=10\log (10^{-2})+120=-20+120=100.\]
Notice that even though the intensity decreased 100 times (from 1 to 0.01), the number of decibels only decreased from 120 to 100.
This is because the logarithm function changes very slowly.
\end{example}
\subsection{Exponential growth and decay}
When an exponential function is used to study a problem arising from an application, it
is called an \textit{exponential model}.
If an increasing exponential function (with base greater than \(1\)) is used in an application
problem, then it is called an \textit{exponential growth model}.
If a decreasing exponential function (with base less than \(1\)) is used, then it is called
an \textit{exponential decay model}.
In this course, we will always use \(e\) as the base for exponential models. If the exponent is
positive, then it is a growth model. If the exponent is negative, then it is a decay model.
Since in all the application problems we will study the independent variable is time, we will use
\(t\) instead of \(x\).
We will also include two constants, \(a\) and \(k\), that account for
vertical and horizontal stretching or shrinking. Choosing the right values for these constants will allow us
to find the right function for a given application problem.
The definition of exponential growth and decay models is summarized below:
| Exponential models |
|---|
| \begin{itemize}
\item
Exponential growth: \( \hspace{1ex} f(t)=a e^{kt}\)
\item
Exponential decay: \( \hspace{1ex} f(t)=a e^{-kt}\)
\end{itemize}
|
Examples of quantities that can be modeled with exponential growth include:
\begin{itemize}
\item Biology:
\begin{enumerate}
\item The population of a city.
\item The number of bacteria in a Petri dish.
\end{enumerate}
\item Finance:
The interest on a savings account
\item Economics:
The GDP growth of the US economy since World War II.
\end{itemize}
So for example to model the population of a city, we will use the function
\[f(t) = ae^{kt},\]
and then \(f(t)\) will represent the number of people in the city at time \(t\).
Examples of quantities that can be modeled with exponential decay include:
\begin{itemize}
\item Physics:
\begin{enumerate}
\item The mass of a radioactive element.
\item The amount of fluid emptying from a tube open at the bottom.
\end{enumerate}
\item Pharmacy:
The concentration of an administered drug in the bloodstream.
\end{itemize}
So for example to model the mass of a radioactive element, we will use the function
\[f(t)=ae^{-kt},\]
and then \(f(t)\) will represent the amount of the element at time \(t\).
\subsection{The meaning of \(a\) and \(k\)}
We now discuss the practical meaning of the two constants \(a\) and \(k\).
Since the variable \(t\) represents time, \(t=0\) is the starting or \textit{initial} time.
If we substitute \(t=0\) in the function \(f(t) = e^{kt}\), we find:
\[f(0) = a e^{k(0)} = a e^0 = a(1) = a.\]
This gives us the meaning of the constant \(a\): it is the initial amount of the quantity
we are studying.
\begin{example}
In a population growth model \(f(t)=a e^{kt}\), we are given that \(t=0\) means
the year 1990. So \(t=1\) means 1991, \(t=5\) means
1995, \(t=10\) means 2000, and so on.
We are also given that the population in the year 1990 was
25,000, and the value of \(k\) is \(k=0.0125\). We then know that \(a=25,000\), and the model becomes
\[f(t)=25,000e^{0.0125t}.\]
We can then find the the population in any year by evaluating \(f(t)\) at the number of
years since 1990. So for example in the
year 2000 \(t=10\), and so we find:
\[f(10)=25,\!000 e^{(0.0125)(10)}= 25,\!000e^{0.125}=25,000(1.133148)\approx 28,\!329.\]
This means that the population in the year 2000 was 28,329.
Note that we have rounded the
number to the nearest unit, since we are counting people, and we have used many decimal places in
our calculations, because we are working with fairly large numbers (the initial population
was a five digit number). As a general rule, if we want to have an accurate count, we need to keep
at least a couple of decimal places more than the number of digits in \(a\) or in \(k\).
\end{example}
\begin{example}
In a radioactive decay model \(f(t)=ae^{-kt}\), we know that the initial quantity of the element is
\(a=450\) grams, \(t\) is measured in years, and \(k=0.0025\). So the model is
\[ f(t) = 450 e^{-0.0025t}.\]
Suppose we want to know how
many grams of the element are left after \(1000\) years, rounded to the nearest gram.
We then calculate \(f(1000)\):
\[f(1000)=450e^{-0.0025(1000)}=450e^{-2.5}=450(0.082) \approx 37 \mbox{ grams}.\]
\end{example}
We now discuss the practical meaning of the constant \(k\). If \(k\) is a very small number,
then it will take a large value of \(t\) for the exponent \(kt\) to be a sizable number, and so for
the quantity to move away from its initial value. So \(k\) affects the length of time
it takes for the quantity to change substantially.
A useful way to determine change in a given quantity is to consider the length of time it takes
for the quantity to double (in case of exponential growth) or to become half its original value
(in case of exponential decay).
In case of exponential growth, we call this length of time the \textit{ doubling time}, and in case of
exponential decay we call it the \textit{half life}. In both cases, we denote it by \(T\).
It is easy to see how \(T\) is related to \(k\). First we look at the exponential growth:
\[f(t)=a e^{kt}.\]
We want the quantity to become twice \(a\) when \(t=T\). This means that we need to solve the equation
\[e^{kT}=2.\]
Using the translation formula, this means
\[\ln 2 =kT \]
and so we find \[k=\frac{\ln 2}{T}.\]
The calculation for exponential decay is very similar: we want
\[f(t)=a e^{-kt}\]
to become half of \(a\). This means that we need to solve the equation:
\[e^{-kT} = \frac{1}{2}.\]
Using the translation formula,
\[\ln\left(\frac{1}{2}\right) = - kT, \]
and using property (B) of the logarithms,
\[\ln 1 - \ln 2 = -kT.\]
Since \(\ln 1 =0\), we solve for \(k\) and find again
\[k=\frac{\ln 2}{T}.\]
So in both growth and decay model, the formula to find \(k\) from the doubling time
(or half life) \(T\) is the same. We summarize what we have found below.
| Doubling time and half life |
|---|
| \begin{itemize}
\item In a growth model, the time it takes for the quantity to double is called
the \textit{doubling time}, and denoted by \(T\).
\item
In a decay model, the time it takes for the quantity to become of half its initial
value is called the \textit{half life}, and is also denoted by \(T\).
\end{itemize} |
| Meaning of \(a\) and \(k\) |
|---|
| \begin{itemize}
\item
\(a\) is the initial value of the quantity (at time \(t=0\)).
\item
\(k\) affects the time it takes for the quantity to double (for growth) or to become
\(1/2\) (for decay).
\item The relationship beween \(T\) and \(k\) is given by the formulas
\[k=\frac{\ln 2}{T}, \hspace{5ex} T=\frac{\ln 2}{k}.\]
\end{itemize} |
\subsection{Applications}
We now discuss some examples of how exponential growth or decay models are used to
solve application problems.
\begin{example}
The population of a certain town is growing according to an exponential growth model, and we
take the initial time \(t=0\) to be the year 1980. Suppose that in 1980 the population was
\(47,\!000\), and then it grew to \(56,\!850\) in the year 2000.
We want to find the growth model function for this problem. We already know that \(a=47,\!000\),
because that is the given initial value at \(t=0\). So we can write
\[f(t)=47,\!000 e^{kt}.\]
We now need to find \(k\). We use the given information: in the year 2000 (that is, \(t=20\),
the population was \(56,\!850\). This means that \(f(20)=56,\!850\), so substituting in our model
we find
\[56,850=47,\!000e^{k(20)}.\]
We now solve this exponential equation for \(k\). First divide both sides by \(47,\!000\):
\[\frac{56,\!850}{47,\!000}=e^{20k},\]
and using a calculator we find
\[1.20957447\approx e^{20k}.\]
Note the large number of decimals: since the initial population is a five digit number, we keep
a few decimals more than that.
Now take \(\ln\) of both sides:
\[\ln(1.20957447)=20k,\]
and dividing by \(20\) we find \(k\):
\[k=\frac{\ln(1.20957447)}{20}\approx 0.00951343.\]
So we have found our model to be:
\[f(t)=47,\!000 e^{0.00951343 t}.\]
Using our newly found model, we can now answer all sorts of questions about the population of the
town. For example, suppose we want to know what the population will be in the year 2020.
All we need to do is figure out how many years that is from 1980: \(2020 -1980=40\), so we use
\(t=40\) in our equation and find:
\[f(40)=47,\!000e^{0.00951343(40)}=47,\!000 e^{0.3805372}=47,\!000(1.46307033)\approx 68,\!764.\]
Another common question we may want to answer is this: how long will it take for the population
to reach a certain number of people? For example, suppose we want to know when the population
of that city will reach \(100,\!000\). This means that we want to know what \(t\) will give us
\(f(t)=100,\!000\), or in other words we need to solve the equation:
\[100,\!000 = 47,\!000 e^{0.00951343t}.\]
Dividing by \(47,\!000\) as before and then taking \(\ln\) we find:
\[\frac{100,\!000}{47,\!000}= e^{0.00951343t},\]
\[2.12765957\approx e^{0.00951343t},\]
\[\ln(2.12765957) = 0.00951343t.\]
Solving for \(t\):
\[t=\frac{\ln(2.12765957)}{0.00951343}=\frac{0.755022584}{0.00951343}\approx 79 \mbox{ years}.\]
So the population will reach \(100,\!000\) \(79\) years after 1980, or in the year 2059.
\end{example}
In the next example the half life for an exponential decay model is given.
\begin{example}
Suppose we know that a radioactive element is decaying according to an exponential decay model,
and its half life is \(T=1,\!800 \) years. We also know that the initial quantity of the element
is \(680\) grams. We want to find the exponential decay model, \(f(t)=ae^{-kt}\).
We know that
\(a=680\). To find \(k\), we use the formula
\[k=\frac{\ln 2 }{T}.\] Substituting \(T=1,\!800\), we find
\[k=\frac{\ln 2}{1,\!800} \approx 0.00038508.\]
So the exponential decay model is
\[f(t)=680 e^{-0.00038508 t}.\]
Suppose we now want to find when the mass of this element will be reduced to \(1/10\) of its
original value. That means we want to find \(t\) for which \(f(t)=680/10\), or \(f(t)=68\).
Substituting in our model we get:
\[68 = 680 e^{-0.00038508t},\]
and solving as before we find:
\[
\begin{array}{rcl}
\displaystyle \frac{68}{680} & = & \displaystyle \frac{680e^{-0.00038508t}}{680}\\[2ex]
0.1 & = & e^{-0.00038508t}\\[2ex]
\ln(0.1) & = & -0.00038508t\\[2ex]
\displaystyle \frac{\ln 0.1}{-0.00038508} & = & \displaystyle \frac{-0.00038508t}{-0.00038508}\\[2ex]
t&\approx & 5,\!979
\end{array}
\]
So the element will be reduced to \(1/10\) of its initial mass in \(5,\!979\) years.
\end{example}
Problems
\problem
The population of a city is increasing according to an exponential growth model, and it was
\(135,\!000\) in 1999. It then grew to be \(187,\!500\) in the year 2015.
\begin{enumerate}
\item
Use \(t=0\) to be 1999, and find the exponential growth model for the population.
\item
Find the doubling time.
\item
Find the population in the year 2025.
\item
When will the population reach \(500,\!000\)?
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
The exponential growth model is
\[f(t)=a e^{kt},\]
so we need to find \(a\) and \(k\). We take \(t=0\) to be 1999, and so \(a=135,\!000\).
The year 2015 corresponds to \(t=16\), so we know that \(f(16)=187,\!500\).
Substituting this in the model, we find
\[187,\!500= 135,\!000 e^{k(16)}.\]
We now solve this equation for \(k\):
\[
\begin{array}{rcl}
\displaystyle \frac{187,\!500}{135,\!000} & = & \displaystyle \frac{135,\!000 e^{16k}}{135,\!000}\\[2ex]
1.38888888 & \approx & e^{16k}\\[2ex]
\ln(1.38888888) & = & 16k\\[2ex]
\displaystyle \frac{\ln(1.38888888)}{16} & = & \displaystyle \frac{16k}{16}\\[2ex]
k &\approx & 0.020531504
\end{array}
\]
So the exponential growth model is
\[f(t)=135,\!000e^{0.020531504t}.\]
\item
The doubling time \(T\) is related to \(k\) by the formula
\[k=\frac{\ln 2}{T}.\]
So we can use the value of \(k\) we found and solve for \(T\):
\[
\begin{array}{rcl}
0.020531504&=& \displaystyle \frac{\ln2}{T}\\[2ex]
T&=& \displaystyle \frac{\ln 2}{0.020531504}\\[2ex]
T & \approx & 34.
\end{array}
\]
So the doubling time is about \(T=34\) years.
\item
The year 2025 corresponds to \(t=26\). So we need to find \(f(26)\):
\[
\begin{array}{rcl}
f(26)&=& 135,\!000 e^{0.020531504(26)}\\[2ex]
& = & 135,\!000 e^{0.533819104}\\[2ex]
& \approx & 230,\!233
\end{array}.
\]
So the population will be \(230,\!233\) in 2025.
\item
Wa need to find \(t\) for which \(f(t)=500,\!000\). Substituting in the model:
\[500,\!000= 135,\!000 e^{0.020531504 t}.\]
We solve for \(t\):
\[
\begin{array}{rcl}
\displaystyle \frac{500,\!000}{135,\!000} & = & \displaystyle
\frac{135,\!000 e^{0.020531504t}}{135,\!000}\\[2ex]
3.703703703& \approx & e^{0.020531504t}\\[2ex]
\ln(3.703703703) & = & 0.020531504t \\[2ex]
\displaystyle \frac{\ln(3.703703703)}{0.020531504} & = & \displaystyle
\frac{0.020531504t}{0.020531504}\\[2ex]
t&\approx & 64.
\end{array}.
\]
So the population will reach \(500,\!000\) about \(64\) years after 1999, or in the year 2063.
\end{enumerate}
\end{sol} \mproblem
The population of a city is increasing according to an exponential growth model, and it was
\(110,\!000\) in 1991. It then grew to be \(123,\!800\) in the year 2005.
\begin{enumerate}
\item
Use \(t=0\) to be 1991, and find the exponential growth model for the population.
\item
Find the doubling time.
\item
Find the population in the year 2020.
\item
When will the population reach \(400,\!000\)?
\end{enumerate}
\problem
A radioactive substance is decaying according to an exponential decay model, and its initial
mass is \(1,\!235\) grams. Suppose the half life is \(2,\!900\) years.
\begin{enumerate}
\item
Find the exponential decay model.
\item
Find how much of the substance is left after \(5,\!000\) years.
\item
How long will it take until only \(5\%\) of the initial amount is left?
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
We are given that \(a=1,\!235\). To find \(k\), we use the equation
\[T=\frac{\ln 2}{k}:\]
\[
\begin{array}{rcl}
2,\!900 & = & \displaystyle \frac{\ln 2}{k}\\[2ex]
k & = & \displaystyle \frac{\ln 2}{2,\!900}\\[2ex]
k & \approx & 0.0002390162\\[2ex]
\end{array}
\]
So the model is
\[f(t)=1,\!235 e^{-0.0002390162t}.\]
\item
We compute \(f(5,\!000)\):
\[
\begin{array}{rcl}
f(5,\!000)&=& 1,235 e^{-0.0002390162(5,\!000)}\\[2ex]
& = & 1,235 e^{-1.1948}\\[2ex]
& \approx & 373.9
\end{array}
\]
So after \(5,\!000\) years, \(373.9\) grams of the substance are left.
\item
We want the ratio \(f(t)/1,\!235\) to be \(5\%\) or \(0.05\). So we need to solve the equation
\(e^{-0.0002390162t} = 0.05\) for \(t\):
\[
\begin{array}{rcl}
e^{-0.0002390162t}& = & 0.05\\[2ex]
-0.0002390162t & = & \ln(0.05)\\[2ex]
\displaystyle \frac{-0.0002390162t}{-0.0002390162} & = &
\displaystyle \frac{\ln(0.05)}{-0.0002390162}\\[2ex]
t & \approx & 12,\!534
\end{array}.
\]
So it will take \(12,\!534\) years for the substance to be reduced to \(5\%\) of its initial value.
\end{enumerate}
\end{sol} \mproblem
A radioactive substance is decaying according to an exponential decay model, and its initial
mass is \(2,\!358\) grams. Suppose the half life is \(3,\!500\) years.
\begin{enumerate}
\item
Find the exponential decay model.
\item
Find how much of the substance is left after \(2,\!000\) years.
\item
How long will it take until only \(10\%\) of the initial amount is left?
\end{enumerate}