Exponential and Logarithmic Models

We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling.

Modeling Exponential Growth and Decay

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

where $A_0$ is equal to the value at time zero, $e$ is Euler’s constant, and $k$ is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.

On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form $y=A_0e^{kt}$ where $A_0$ is the starting value, and $e$ is Euler’s constant. Now $k$ is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see below. It is important to remember that, although parts of each of the two graphs seem to lie on the $x$-axis, they are really a tiny distance above the $x$-axis.

Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is $40{,}113{,}497{,}200{,}000$ kilometers. Expressed in scientific notation, this is $4.01134972\times10^{13}$. So, we could describe this number as having order of magnitude $10^{13}$.

Example. A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

Solution. When an amount grows at a fixed percent per unit time, the growth is exponential. To find $A_0$ we use the fact that $A_0$ is the amount at time zero, so $A_0=10$. To find $k$, use the fact that after one hour $(t=1)$ the population doubles from $10$ to $20$. The formula is derived as follows

so $k=\ln(2)$. Thus the equation we want to graph is $y=10e^{(\ln2)t}=10\left(e^{\ln2}\right)^t=10\cdot2^t$. The graph is shown below.

Analysis. The population of bacteria after ten hours is $10{,}240$. We could describe this amount is being of the order of magnitude $10^4$. The population of bacteria after twenty hours is $10{,}485{,}760$ which is of the order of magnitude $10^7$, so we could say that the population has increased by three orders of magnitude in ten hours.

Half-Life

We now turn to exponential decay. One of the common terms associated with exponential decay, as stated above, is half-life, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

We find that the half-life depends only on the constant $k$ and not on the starting quantity $A_0$.

The formula is derived as follows

Since $t$, the time, is positive, $k$ must, as expected, be negative. This gives us the half-life formula

Example. The half-life of carbon-14 is $5{,}730$ years. Express the amount of carbon-14 remaining as a function of time, $t$.

Solution. This formula is derived as follows.

The function that describes this continuous decay is $f(t)=A_0e^{\left(\tfrac{\ln(0.5)}{5{,}730}\right)t}$. We observe that the coefficient of $t$, $\tfrac{\ln(0.5)}{5{,}730}\approx-1.2097\times10^{-4}$, is negative, as expected in the case of exponential decay.

Radiocarbon Dating

The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last $60{,}000$ years.

Carbon-14 is a radioactive isotope of carbon that has a half-life of $5{,}730$ years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last $60{,}000$ years, using tree rings and other organic samples of known dates — although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is $5{,}730$ years, the formula for the amount of carbon-14 remaining after $t$ years is

where

• $A$ is the amount of carbon-14 remaining
• $A_0$ is the amount of carbon-14 when the plant or animal began decaying

This formula is derived as follows:

To find the age of an object, we solve this equation for $t$:

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately $0.0000000001\%$. Let $r$ be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation $A\approx A_0e^{-0.000121t}$ we know the ratio of the percentage of carbon-14 in the object we are dating to the initial amount of carbon-14 in the object when it was formed is $r=\tfrac{A}{A_0}\approx e^{-0.000121t}$. We solve this equation for $t$, to get

Example. A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?

Solution. We substitute $20\%=0.20$ for $r$ in the equation and solve for $t$:

The bone fragment is about $13{,}301$ years old.

Analysis. The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as $13{,}301$ years $\pm$ 1%, or $13{,}301$ years $\pm$ 133 years.

Calculating Doubling Time

For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time.

Given the basic exponential growth equation $A=A_0e^{kt}$, doubling time can be found by solving for when the original quantity has doubled, that is, by solving $2A_0=A_0e^{kt}$.

The formula is derived as follows:

Thus the doubling time is

Example. According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.

Solution. The formula is derived as follows:

The function is $A_0e^{\tfrac{\ln2}{2}t}$.

Using Newton’s Law of Cooling

Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature

This formula is derived as follows:

Example. A cheesecake is taken out of the oven with an ideal internal temperature of $165\degree$F, and is placed into a $35\degree$F refrigerator. After 10 minutes, the cheesecake has cooled to $150\degree$F. If we must wait until the cheesecake has cooled to $70\degree$F before we eat it, how long will we have to wait?

Solution. Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation

We know the initial temperature was 165, so $T(0)=165$.

We were given another data point, $T(10)=150$, which we can use to solve for $k$.

This gives us the equation for the cooling of the cheesecake: $T(t)=130e^{-0.0123t}+35$.

Now we can solve for the time it will take for the temperature to cool to 70 degrees.

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to $70\degree$F.

Using Logistic Growth Models

Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.

The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants $a$, $b$, and $c$, the logistic growth of a population over time $t$ is represented by the model

The graph below shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.

Example. An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.

For example, at time $t=0$ there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is $b=0.6030$. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.

Solution. We substitute the given data into the logistic growth model

Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is $c=1000$. To find $a$, we use the formula that the number of cases at time $t=0$ is $\tfrac{c}{1+a}=1$, from which it follows that $a=999$. This model predicts that, after ten days, the number of people who have had the flu is $f(t)=\tfrac{1000}{1+999e^{-0.6030x}}\approx293.8$. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, $c=1000$.

Analysis. Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.

The graph below gives a good picture of how this model fits the data.

Choosing an Appropriate Model for Data

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.

Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.

In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.

A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.

After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.

Example. Does a linear, exponential, logarithmic, or logistic model best fit the values listed below? Find the model, and use a graph to check your choice.

Solution. First, plot the data on a graph as below. For the purpose of graphing, round the data to two decimal places.

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try $y=a\ln(bx)$. Plugging in the first point, $(1,0)$, gives $0=a\ln b$. We reject the case that $a=0$ (if it were, all outputs would be 0), so we know $\ln(b)=0$. Thus $b=1$ and $y=a\ln(x)$. Next we can use the point $(9,4.394)$ to solve for $a$:

Because $a=\tfrac{4.394}{\ln(9)}\approx2$, an appropriate model for the data is $y=2\ln(x)$.

To check the accuracy of the model, we graph the function together with the given points below.

We can conclude that the model is a good fit to the data.

Compare to the graph of $y=\ln\left(x^2\right)$ shown below.

The graphs appear to be identical when $x>0$. A quick check confirms this conclusion: $y=\ln\left(x^2\right)=2\ln(x)$ for $x>0$.

However, if $x<0$, the graph of $y=\ln\left(x^2\right)$ includes an “extra” branch, as shown below. This occurs because, while $y=2\ln(x)$ cannot have negative values in the domain (as such values would force the argument to be negative), the function $y=\ln\left(x^2\right)$ can have negative domain values.

Expressing an Exponential Model in Base $e$

While powers and logarithms of any base can be used in modeling, the two most common bases are $10$ and $e$. In science and mathematics, the base $e$ is often preferred. We can use laws of exponents and laws of logarithms to change any base to base $e$.

Example. Change the function $y=2.5(3.1)^x$ so that this same function is written in the form $y=A_0e^{kx}$.

Solution. The formula is derived as follows

Key equations

Key concepts

• The basic exponential function is $f(x)=ab^x$. If $b>1$, we have exponential growth; if $0<b<1$, we have exponential decay.
• We can also write this formula in terms of continuous growth as $A=A_0e^{kx}$, where $A_0$ is the starting value. If $A_0$ is positive, then we have exponential growth when $k>0$ and exponential decay when $k<0$.
• In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay.
• We can find the age, $t$, of an organic artifact by measuring the amount, $k$, of carbon-14 remaining in the artifact and using the formula $t=\tfrac{\ln(k)}{-0.000121}$ to solve for $t$.
• Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay.
• We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time.
• We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.
• We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data.
• Any exponential function with the form $y=ab^x$ can be rewritten as an equivalent exponential function with the form $y=A_0e^{kx}$ where $k=\ln b$.

Practice

Model exponential growth and decay

Use Newton’s Law of Cooling

Use logistic-growth models

A different logistic growth model is given by $f(x)=\tfrac{150}{1+8e^{-2x}}$.

Choose an appropriate model for data

Express an exponential model in base $e$

This section is adapted from Precalculus 2e, Section 4.7: Exponential and Logarithmic Models by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: excluded a “coreq-skills” block present in the pinned CNXML module (a corequisite-course skills review covering compound interest, exponential growth/decay applications, decibel levels, and the Richter scale, with its own short exercise set) that does not appear in the printed Precalculus 2e text — pages 494–495 of the source PDF confirm the printed section runs directly from the chapter introduction into “Modeling Exponential Growth and Decay,” with no corequisite-skills material between them; omitted the decorative nuclear-reactor photograph, which carries no mathematics; omitted the section’s five Media links to external graphing-calculator resources; recreated every graph as an accessible inline SVG — the labeled-points growth curve $y=2e^{3x}$ and decay curve $y=3e^{-2x}$; the generic growth/decay panels of $y=A_0e^{kt}$ (drawn with concrete $A_0=2$, $k=\pm1$ so the curves are plottable, keeping the source’s own symbolic point labels, which name the high point and low point of each panel rather than literal coordinates); the bacteria-growth curve $y=10e^{(\ln2)t}$, labeled by its equivalent simplified form $y=10\cdot2^t$; the generic logistic S-curve with its carrying-capacity asymptote, initial-value point, and point-of-maximum-growth point; the flu-epidemic logistic curve $f(t)=\tfrac{1000}{1+999e^{-0.6030t}}$ with its four data-point annotations; the nine-point scatter plot of Table 1; the logarithmic curve $y=2\ln(x)$ and the identical-looking $y=\ln(x^2)$ over $x>0$, each overlaid with the same nine data points; and the full two-branch graph of $y=\ln(x^2)$ showing the extra branch for $x<0$. The generic logistic S-curve, the flu-epidemic logistic curve, and the left-hand ($x<0$) branch of $y=\ln(x^2)$ are each drawn from their exact closed-form equation by an analytic primitive: this authoring pass added a logistic curve kind and a reflect option on the log curve to graph-core.mjs, which previously had neither, so all three had to be hand-sampled as dense point lists. Presented the compound-model comparison table (Try It data) and Table 1/Table 2 data as Markdown tables; renamed the rumor model’s population function from the source’s $N(t)$ to $p(t)$, since $N$ is reserved by this project’s compute engine for numeric evaluation; converted the “Try It” practice problems into interactive exercises with instant feedback — a fill-in for the plutonium-244 half-life function, a multiple-choice plus fill-in for the cesium-137 comparison (splitting the source’s combined “which, and what is the exact value” answer into a categorical choice and a decimal value), a fill-in for the three-year Moore’s-Law doubling function, a fill-in for the pitcher-of-water cooling time, a fill-in for the day-15 flu estimate, a fill-in for the data-table exponential model, and a fill-in for the base-$e$ conversion of $y=3(0.5)^x$; and adapted eighteen selected end-of-section exercises with an answer present in the CNXML solution key — a half-life verbal identification, the Erbium-165 hourly decay rate (split into its decimal and percentage parts), the twenty-minute bacteria-doubling model and its three-hour population, the Iodine-125 half-life, the turkey and soup Newton’s-Law-of-Cooling problems, the fish-farm logistic doubling time and time-to-900, a separate logistic carrying-capacity identification, the town-rumor logistic evaluation, two Technology-section model-classification tables, and the drug-decay base-$e$ conversion together with its three-hour evaluation and the general $b^x=e^{x\ln(b)}$ identity — into nineteen interactive components in a closing Practice block, one group per objective.