maths.free › Precalculus › 4. Exponential and Logarithmic Functions › Exponential and Logarithmic Models
Exponential and Logarithmic Models
Model exponential growth and decay.
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:
\[y={A}_{0}{e}^{kt}\]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 \(y={A}_{0}{e}^{kt}\) 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 in and . 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 \(4.01134972\ \times \ {10}^{13}.\) So, we could describe this number as having order of magnitude \({10}^{13}.\)
Condensed — the full section is in OpenStax Precalculus 2e.
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
\[T(t)=A{e}^{kt}+{T}_{s}\]This formula is derived as follows:
\[\begin{array}{lllll}T(t)=A{b}^{ct}+{T}_{s} & \\ T(t)=A{e}^{\ln ({b}^{ct})}+{T}_{s}\begin{array}{llll} & & & \end{array} & \text{Laws of logarithms}. \\ T(t)=A{e}^{ct\ln b}+{T}_{s} & \text{Laws of logarithms}. \\ T(t)=A{e}^{kt}+{T}_{s} & \text{Rename the constant }c\ln b,\ \text{calling it }k.\end{array}\]Condensed — the full section is in OpenStax Precalculus 2e.
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 \(\text{a, b,}\) and \(\text{c,}\) the logistic growth of a population over time \(t\) is represented by the model
\[f(t)=\frac{c}{1+a{e}^{-bt}}\]The graph in 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.
Condensed — the full section is in OpenStax Precalculus 2e.
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.
Condensed — the full section is in OpenStax Precalculus 2e.
Expressing an Exponential Model in Base
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
Try it.
Change the function \(y=2.5{(3.1)}^{x}\) so that this same function is written in the form \(y={A}_{0}{e}^{kx}.\)
Solution
The formula is derived as follows
\[\begin{array}{ll}y=2.5{(3.1)}^{x} & \\ =2.5{e}^{\ln ({3.1}^{x})} & \text{Insert exponential and its inverse}\text{.} \\ =2.5{e}^{x\ln 3.1} & \text{Laws of logs}\text{.} \\ =2.5{e}^{(\ln 3.1)}{}^{x} & \text{Commutative law of multiplication}\end{array}\]Key Equations
| Half-life formula | If \(\ A={A}_{0}{e}^{kt},\) \(k<0,\) the half-life is \(\ t=-\frac{\ln (2)}{k}.\) |
| Carbon-14 dating | \(t=\frac{\ln (\frac{A}{{A}_{0}})}{-0.000121}.\)
\({A}_{0}\) is the amount of carbon-14 when the plant or animal died \(A\) is the amount of carbon-14 remaining today \(t\) is the age of the fossil in years |
| Doubling time formula | If \(\ A={A}_{0}{e}^{kt},\) \(k>0,\) the doubling time is \(\ t=\frac{\ln 2}{k}\) |
| Newton’s Law of Cooling | \(T(t)=A{e}^{kt}+{T}_{s},\) where \(\ {T}_{s}\) is the ambient temperature, \(\ A=T(0)-{T}_{s},\) and \(\ k\) is the continuous rate of cooling. |
Key Concepts
- The basic exponential function is \(f(x)=a{b}^{x}.\) If \(b>1,\) we have exponential growth; if \(0
- We can also write this formula in terms of continuous growth as \(A={A}_{0}{e}^{kx},\) where \({A}_{0}\) is the starting value. If \({A}_{0}\) is positive, then we have exponential growth when \(k>0\) and exponential decay when \(k<0.\) See .
- 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. See .
- 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 \(t=\frac{\ln (k)}{-0.000121}\) to solve for \(t.\) See .
- Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay. See .
- 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. See .
- 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. See .
- 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. See .
- Any exponential function with the form \(y=a{b}^{x}\) can be rewritten as an equivalent exponential function with the form \(y={A}_{0}{e}^{kx}\) where \(k=\ln b.\) See .
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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A total of \(\text{\$}10,000\) was invested in a college fund for a new grandchild.
ⓐ If the interest rate is \(5\text{\%},\) how much will be in the account in 18 years by each method of compounding?
ⓑ compound quarterly
ⓒ compound monthly
ⓓ compound continuously
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
ⓐ
\(\ A=?\) Identify the values of each variable in the formulas. \(\ P=\text{\$}10,000\) Remember to express the percent as a decimal. \(\ r=0.05\) \(\ t=18\ \text{years}\) ⓑ
For quarterly compounding, \(n=4\). There are 4 quarters in a year. \(\ A=P{(1+\frac{r}{n})}^{nt}\) Substitute the values in the formula. \(\ A=10,000{(1+\frac{0.05}{4})}^{4\cdot 18}\) Compute the amount. Be careful to consider the order of operations as you enter the expression into your calculator. \(\ A=\text{\$}24,459.20\) ⓒ
For monthly compounding, \(n=12\) . There are 12 months in a year. \(\ A=P{(1+\frac{r}{n})}^{nt}\) Substitute the values in the formula. \(\ A=10,000{(1+\frac{0.05}{12})}^{12\cdot 18}\) Compute the amount. \(\ A=\text{\$}24,550.08\) ⓓ
For compounding continuously, \(\ A=P{e}^{rt}\) Substitute the values in the formula. \(\ A=10,000{e}^{0.05\cdot 18}\) Compute the amount. \(\ A=\text{\$}24,596.03\) -
Chris is a researcher at the Center for Disease Control and Prevention and he is trying to understand the behavior of a new and dangerous virus. He starts his experiment with 100 of the virus that grows at a rate of 25% per hour. He will check on the virus in 24 hours. How many viruses will he find?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Identify the values of each variable in the formulas. \(\ A=?\) Be sure to put the percent in decimal form. \({A}_{0}=100\) Be sure the units match—the rate is per hour and the time is in hours. \(r=0.25\text{/hour}\) \(t=24\ \text{hours}\) Substitute the values in the formula: \(A={A}_{0}{e}^{rt}\) . \(A=100{e}^{0.25\cdot 24}\) Compute the amount. \(A=40,342.88\) Round to the nearest whole virus. \(A=40,343\) The researcher will find 40,343 viruses. -
Angela invested \(\text{\$}15,000\) in a savings account. If the interest rate is \(4\text{\%},\) how much will be in the account in 10 years by each method of compounding?
ⓐ compound quarterly
ⓑ compound monthly
ⓒ compound continuously
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Another researcher at the Center for Disease Control and Prevention, Lisa, is studying the growth of a bacteria. She starts her experiment with 50 of the bacteria that grows at a rate of \(15\text{\%}\) per hour. She will check on the bacteria every 8 hours. How many bacteria will she find in 8 hours?
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Extended exposure to noise that measures 85 dB can cause permanent damage to the inner ear which will result in hearing loss. What is the decibel level of music coming through earphones with intensity \({10}^{-2}\) watts per square inch?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Substitute in the intensity level, I. \(D=10\text{log}(\frac{{10}^{-2}}{{10}^{-12}})\) Simplify. \(D=10\log ({10}^{10})\) Since \(\text{log}{10}^{10}=10\) \(D=10\times 10\) Multiply. \(D=100\) -
In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. Over 80% of the city was destroyed by the resulting fires. In 2014, Los Angeles experienced a moderate earthquake that measured 5.1 on the Richter scale and caused $108 million dollars of damage. Compare the intensities of the two earthquakes.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
To compare the intensities, we first need to convert the magnitudes to intensities using the log formula. Then we will set up a ratio to compare the intensities.
Convert the magnitudes to intensities. \(R=\text{log}\ I\) 1906 earthquake \(7.8=\text{log}\ I\) Convert to exponential form. \(I={10}^{7.8}\) 2014 earthquake \(5.1=\text{log}\ I\) Convert to exponential form. \(I={10}^{5.1}\) Form a ratio of the intensities. \(\frac{\text{Intensity}\ \text{for}\ 1906}{\text{Intensity}\ \text{for}\ 2014}\) Substitute in the values. \(\frac{{10}^{7.8}}{{10}^{5.1}}\) Divide by subtracting the exponents. \({10}^{2.7}\) Evaluate. \(501\) Answer: The intensity of the 1906 earthquake was about 501 times the intensity of the 2014 earthquake. -
What is the decibel level of one of the new quiet dishwashers with intensity \({10}^{-7}\) watts per square inch?
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In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. In 1989, the Loma Prieta earthquake also affected the San Francisco area, and measured 6.9 on the Richter scale. Compare the intensities of the two earthquakes.
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A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
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
\[\begin{array}{ll}20=10{e}^{k⋅1} & \\ 2={e}^{k} & \text{Divide by 10} \\ \ln 2=k & \text{Take the natural logarithm}\end{array}\]so \(k=\ln (2).\) Thus the equation we want to graph is \(y=10{e}^{(\ln 2)t}=10{({e}^{\ln 2})}^{t}=10\cdot {2}^{t}.\) The graph is shown in .
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The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, \(t.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
This formula is derived as follows. \[\begin{array}{ll}A={A}_{0}{e}^{kt} & \text{The continuous growth formula}. \\ 0.5{A}_{0}={A}_{0}{e}^{k⋅5730} & \text{Substitute the half-life for }t\ \text{and }0.5{A}_{0}\ \text{for }f(t). \\ 0.5={e}^{5730k} & \text{Divide by }{A}_{0}. \\ \ln (0.5)=5730k & \text{Take the natural log of both sides}. \\ k=\frac{\ln (0.5)}{5730} & \text{Divide by the coefficient of }k. \\ A={A}_{0}{e}^{(\frac{\ln (0.5)}{5730})t} & \text{Substitute for }r\ \text{in the continuous growth formula}.\end{array}\]
The function that describes this continuous decay is \(f(t)={A}_{0}{e}^{(\frac{\ln (0.5)}{5730})t}.\) We observe that the coefficient of \(t,\) \(\frac{\ln (0.5)}{5730}\approx -1.2097\times {10}^{-4}\) is negative, as expected in the case of exponential decay.
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The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of plutonium-244 remaining as a function of time, measured in years.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(f(t)={A}_{0}{e}^{-0.0000000087t}\)
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A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
We substitute \(20\%=0.20\) for \(r\) in the equation and solve for \(t:\)
\[\begin{array}{lllll}t=\frac{\ln (r)}{-0.000121} & \text{Use the general form of the equation}. \\ =\frac{\ln (0.20)}{-0.000121}\begin{array}{llll} & & & \end{array} & \text{Substitute for }r. \\ \approx 13301 & \text{Round to the nearest year}.\end{array}\]The bone fragment is about 13,301 years old.
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Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
less than 230 years, 229.3157 to be exact
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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.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The formula is derived as follows:
\[\begin{array}{lllll}t=\frac{\ln 2}{k} & \text{The doubling time formula}. \\ 2=\frac{\ln 2}{k} & \text{Use a doubling time of two years}. \\ k=\frac{\ln 2}{2} & \text{Multiply by}\ k\ \text{and divide by 2}. \\ A\ ={A}_{0}{e}^{\frac{\ln 2}{2}t}\begin{array}{llll} & & & \end{array} & \text{Substitute}\ k\ \text{into the continuous growth formula}.\end{array}\]The function is \({A}_{0}{e}^{\frac{\ln 2}{2}t}.\)
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Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(f(t)={A}_{0}{e}^{\frac{\ln 2}{3}t}\)
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A cheesecake is taken out of the oven with an ideal internal temperature of \(\text{165^{\circ}F,}\) and is placed into a \(35^{\circ}F\) refrigerator. After 10 minutes, the cheesecake has cooled to \(\text{150^{\circ}F}\text{.}\) If we must wait until the cheesecake has cooled to \(\text{70^{\circ}F}\) before we eat it, how long will we have to wait?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation
\[T(t)=A{e}^{kt}+35\]We know the initial temperature was 165, so \(T(0)=165.\)
\[\begin{array}{ll}165=A{e}^{k0}+35 & \text{Substitute }(0,165). \\ \ \ A=130 & \text{Solve for }A.\end{array}\]We were given another data point, \(T(10)=150,\) which we can use to solve for \(k.\)
\[\begin{array}{ll}150=130{e}^{k10}+35 & \text{Substitute (10, 150)}. \\ 115=130{e}^{k10} & \text{Subtract 35}. \\ \frac{115}{130}={e}^{10k} & \text{Divide by 130}. \\ \ln (\frac{115}{130})=10k & \text{Take the natural log of both sides}. \\ k=\frac{\ln (\frac{115}{130})}{10}\approx -0.0123 & \text{Divide by the coefficient of }k.\end{array}\]This gives us the equation for the cooling of the cheesecake: \(T(t)=130{e}^{-0.0123t}+35.\)
Now we can solve for the time it will take for the temperature to cool to 70 degrees.
\[\begin{array}{ll}70=130{e}^{-0.0123t}+35 & \text{Substitute in 70 for }T(t). \\ 35=130{e}^{-0.0123t} & \text{Subtract 35}. \\ \frac{35}{130}={e}^{-0.0123t} & \text{Divide by 130}. \\ \ln (\frac{35}{130})=-0.0123t & \text{Take the natural log of both sides} \\ t=\frac{\ln (\frac{35}{130})}{-0.0123}\approx 106.68 & \text{Divide by the coefficient of }t.\end{array}\]It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to \(\text{70^{\circ}F}\text{.}\)
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A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
6.026 hours
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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.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
We substitute the given data into the logistic growth model
\[f(t)=\frac{c}{1+a{e}^{-bt}}\]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 \(\frac{c}{1+a}=1,\) from which it follows that \(a=999.\) This model predicts that, after ten days, the number of people who have had the flu is \(f(t)=\frac{1000}{1+999{e}^{-0.6030x}}\approx 293.8.\) 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, \(c=1000.\)
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Using the model in , estimate the number of cases of flu on day 15.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
895 cases on day 15
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Does a linear, exponential, logarithmic, or logistic model best fit the values listed in ? Find the model, and use a graph to check your choice.
\(x\) 1 2 3 4 5 6 7 8 9 \(y\) 0 1.386 2.197 2.773 3.219 3.584 3.892 4.159 4.394 ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First, plot the data on a graph as in . 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, \((\text{1,0})\text{,}\) 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 \(y=a\ln (\text{x}).\) Next we can use the point \((\text{9,4}\text{.394})\) to solve for \(a:\)
\[\begin{array}{l}y=a\ln (x) \\ 4.394=a\ln (9) \\ a=\frac{4.394}{\ln (9)}\end{array}\]Because \(a=\frac{4.394}{\ln (9)}\approx 2,\) 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 as in .
We can conclude that the model is a good fit to the data.
Compare to the graph of \(y=\ln ({x}^{2})\) shown in .
The graphs appear to be identical when \(x>0.\) A quick check confirms this conclusion: \(y=\ln ({x}^{2})=2\ln (x)\) for \(x>0.\)
However, if \(x<0,\) the graph of \(y=\ln ({x}^{2})\) includes a “extra” branch, as shown in . 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 \(y=\ln ({x}^{2})\) can have negative domain values.
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Does a linear, exponential, or logarithmic model best fit the data in ? Find the model.
\(x\) 1 2 3 4 5 6 7 8 9 \(y\) 3.297 5.437 8.963 14.778 24.365 40.172 66.231 109.196 180.034 ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Exponential. \(y=2{e}^{0.5x}.\)
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Change the function \(y=2.5{(3.1)}^{x}\) so that this same function is written in the form \(y={A}_{0}{e}^{kx}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The formula is derived as follows
\[\begin{array}{ll}y=2.5{(3.1)}^{x} & \\ =2.5{e}^{\ln ({3.1}^{x})} & \text{Insert exponential and its inverse}\text{.} \\ =2.5{e}^{x\ln 3.1} & \text{Laws of logs}\text{.} \\ =2.5{e}^{(\ln 3.1)}{}^{x} & \text{Commutative law of multiplication}\end{array}\] -
Change the function \(y=3{(0.5)}^{x}\) to one having \(e\) as the base.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(y=3{e}^{(\ln 0.5)x}\)
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With what kind of exponential model would half-life be associated? What role does half-life play in these models?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.
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What is carbon dating? Why does it work? Give an example in which carbon dating would be useful.
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With what kind of exponential model would doubling time be associated? What role does doubling time play in these models?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.
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Define Newton’s Law of Cooling. Then name at least three real-world situations where Newton’s Law of Cooling would be applied.
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What is an order of magnitude? Why are orders of magnitude useful? Give an example to explain.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about \({10}^{\text{2}}\) times, or 2 orders of magnitude greater, than the mass of Earth.
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The temperature of an object in degrees Fahrenheit after t minutes is represented by the equation \(T(t)=68{e}^{-0.0174t}+72.\) To the nearest degree, what is the temperature of the object after one and a half hours?
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Find and interpret \(f(0).\) Round to the nearest tenth.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(f(0)\approx 16.7;\) The amount initially present is about 16.7 units.
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Find and interpret \(f(4).\) Round to the nearest tenth.
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Find the carrying capacity.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
150
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Graph the model.
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Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data.
\(x\) \(f(x)\) –2 0.694 –1 0.833 0 1 1 1.2 2 1.44 3 1.728 4 2.074 5 2.488 ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
exponential; \(f(x)={1.2}^{x}\)
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Rewrite \(f(x)=1.68{(0.65)}^{x}\) as an exponential equation with base \(e\) to five decimal places.
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\(x\) \(f(x)\) 1 2 2 4.079 3 5.296 4 6.159 5 6.828 6 7.375 7 7.838 8 8.238 9 8.592 10 8.908 ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
logarithmic
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\(x\) \(f(x)\) 1 2.4 2 2.88 3 3.456 4 4.147 5 4.977 6 5.972 7 7.166 8 8.6 9 10.32 10 12.383 -
\(x\) \(f(x)\) 4 9.429 5 9.972 6 10.415 7 10.79 8 11.115 9 11.401 10 11.657 11 11.889 12 12.101 13 12.295 ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
logarithmic
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\(x\) \(f(x)\) 1.25 5.75 2.25 8.75 3.56 12.68 4.2 14.6 5.65 18.95 6.75 22.25 7.25 23.75 8.6 27.8 9.25 29.75 10.5 33.5 -
Graph the function.
Symbols used here
The exponent b must be raised to for x; ln uses base e.
1/360 of a full turn. 180° = π radians.
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
Number of k-element subsets of n things: n!/(k!(n−k)!).
Ratio of a circle's circumference to its diameter, 3.14159…
2.71828…, the base whose exponential is its own derivative.
i² = −1.
The usual name for an angle.
A quantity with magnitude and direction; a column of numbers.
How to: Exponential and Logarithmic Models
- Model exponential growth and decay.
- Use Newton’s Law of Cooling.
- Use logistic-growth models.
- Choose an appropriate model for data.
- Express an exponential model in base
- Use exponential models in applications. (IA 10.2.3)
- Use logarithmic models in applications. (IA 10.3.5)
- one-to-one function
Questions people ask
What is a function, really?
A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.
Why do we need complex numbers?
Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.
ನಿಮ್ಮದೇ ಆದದ್ದನ್ನು ಪ್ರಯತ್ನಿಸಿ
Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
ಇನ್ನಷ್ಟು Precalculus
Complex numbersPolynomial functionsRational functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates