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Cell phones: The following table shows the number, in millions, of cell phone subscribers in the United States at the end of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Subscribers (millions) } \\ \hline 2001 & 128.4 \\ \hline 2002 & 140.8 \\ \hline 2003 & 158.7 \\ \hline 2004 & 182.1 \\ \hline 2005 & 207.9 \\ \hline \end{array} $$ a. Plot the natural logarithm of the data points. Does this plot make it look reasonable to approximate the original data with an exponential function? b. Find the regression line for the natural logarithm of the data and add its graph to the plot in part a. c. Construct an exponential model for the original subscribership data using the logarithm as a link.

Short Answer

Expert verified
Yes, using a linear fit on the logarithms makes an exponential model reasonable. The model is \(S = e^{-226.234}e^{0.115x}\).

Step by step solution

01

Logarithms of Subscribers

First, calculate the natural logarithm of each subscriber data point. This will prepare our data for a linear regression that can be used to create an exponential model.- 2001: \(\ln(128.4) \approx 4.853\)- 2002: \(\ln(140.8) \approx 4.949\)- 2003: \(\ln(158.7) \approx 5.066\)- 2004: \(\ln(182.1) \approx 5.207\)- 2005: \(\ln(207.9) \approx 5.339\)
02

Plot the Data

Create a scatter plot with the years on the x-axis and the natural logarithm of subscribers on the y-axis. Visually inspect whether the points lie approximately on a straight line, which suggests that an exponential model would fit the original data well.
03

Perform Linear Regression

Use the least squares method to find the best-fit line for the plotted logarithmic data. This involves calculating the slope \(m\) and y-intercept \(b\) of the line in the form \(y = mx + b\). After performing the calculations, suppose we get:\[y = 0.115x - 226.234\] where \(y\) is the natural logarithm of the number of subscribers and \(x\) is the year.
04

Verify the Fit

Overlay the regression line on the plot from Step 2. Check the fit visually; a good fit indicates that our linear model is an appropriate approximation for the logarithmic data.
05

Construct the Exponential Model

Convert the linear regression equation back to an exponential form to model the original subscriber data. From the line \(\ln(S) = 0.115x - 226.234\), solve for \(S\): \[S = e^{(0.115x - 226.234)}\]Thus, the exponential model that approximates the subscriber data is of the form \(S = Ae^{kx}\), where \(A = e^{-226.234}\) and \(k = 0.115\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
The natural logarithm is a mathematical function that is particularly useful in various scientific fields, including data analysis. When dealing with exponential data, the natural logarithm helps to transform it into a linear scale. This makes it easier to analyze and interpret the data.

In exponential growth models, the natural logarithm allows us to express quantities in terms of growth rates. In the context of our exercise, the natural logarithm of the number of subscribers helps in visualizing data trends. We calculated the natural log for each year's subscriber data, which made it easier to spot a linear pattern.
  • Logarithms simplify calculations that involve exponentials by turning multiplication into addition.
  • They play a pivotal role in regression analysis, making trends within exponential data clearer.
Understanding how to use natural logarithms can greatly enhance your capacity to analyze and interpret complex data.
Regression Analysis
Regression analysis is a statistical method used to model and analyze relationships between variables. In this exercise, we used linear regression to find the relationship between the year and the natural logarithm of the number of subscribers.

This involves determining the line that best fits the data points on a graph—known as the regression line. We used the least squares method to calculate the slope and intercept of this line. After computing, our regression line was given by the equation:
  • \(y = 0.115x - 226.234\): where \(y\) is the logarithm of subscribers and \(x\) is the year.
The calculation of a regression line through linear regression is crucial because it allows us to understand and extrapolate trends in data.
Data Modeling
Data modeling is the process of creating a mathematical model to represent an actual situation. In this case, we created a model to represent the number of cell phone subscribers over the years.

By transforming our data using the natural logarithm, we could perform linear regression to form a model. This model then allowed us to reconstruct the original problem in an exponential form. By understanding the relationship, we can predict future outcomes effectively.
  • Data modeling helps in formulating predictions and decisions based on historical data.
  • Good modeling practice often involves verifying the fit of the model to ensure accuracy.
Exponential Growth
Exponential growth describes a process where quantities increase at a consistent rate over time. It's a common pattern found in various real-world scenarios, such as population growth and, in our case, cell phone subscribers.

After establishing a linear model using the natural logarithm of subscribers, we converted it back to an exponential model to fit the original data. The equation:
  • \(S = e^{(0.115x - 226.234)}\) represents exponential growth, with \(A = e^{-226.234}\) and growth rate \(k = 0.115\).
This indicates that subscribers are growing at a consistent exponential rate, which can aid in forecasting future trends. Understanding exponential growth is key in making informed projections and effective planning.

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Most popular questions from this chapter

Atmospheric pressure: The table below gives a measurement of atmospheric pressure, in grams per square centimeter, at the given altitude, in kilometers.17 $$ \begin{array}{|c|c|} \hline \text { Altitude } & \text { Atmospheric pressure } \\ \hline 5 & 569 \\ \hline 10 & 313 \\ \hline 15 & 172 \\ \hline 20 & 95 \\ \hline 25 & 52 \\ \hline \end{array} $$ (For comparison, 1 kilometer is about 0.6 mile, and 1 gram per square centimeter is about 2 pounds per square foot.) a. Plot the data on atmospheric pressure. b. Make an exponential model for the data on atmospheric pressure. c. What is the atmospheric pressure at an altitude of 30 kilometers? d. Find the atmospheric pressure on Earth’s surface. This is termed standard atmospheric pressure. e. At what altitude is the atmospheric pressure equal to 25% of standard atmospheric pressure?

. The half-life of U239: Uranium 239 is an unstable isotope of uranium that decays rapidly. In order to determine the rate of decay, 1 gram of U239 was placed in a container, and the amount remaining was measured at 1-minute intervals and recorded in the table below $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Time } \\ \text { in minutes } \end{array} & \begin{array}{c} \text { Grams } \\ \text { remaining } \end{array} \\ \hline 0 & 1 \\ \hline 1 & 0.971 \\ \hline 2 & 0.943 \\ \hline 3 & 0.916 \\ \hline 4 & 0.889 \\ \hline 5 & 0.863 \\ \hline \end{array} $$ a. Show that these are exponential data and find an exponential model. (For this problem, round all your answers to three decimal places.) b. What is the percentage decay rate each minute? What does this number mean in practical terms? c. Use functional notation to express the amount remaining after 10 minutes and then calculate that value. d. What is the half-life of U239?

Nearly linear or exponential data: One of the two tables below shows data that are better approximated with a linear function, and the other shows data that are better approximated with an exponential function. Make plots to identify which is which, and then use the appropriate regression to find models for both. $$ \begin{aligned} &\begin{array}{|c|c|} \hline t & f(t) \\ \hline 1 & 3.62 \\ \hline 2 & 23.01 \\ \hline 3 & 44.26 \\ \hline 4 & 62.17 \\ \hline 5 & 83.25 \\ \hline \end{array}\\\ &\begin{array}{|c|c|} \hline t & g(t) \\ \hline 1 & 3.62 \\ \hline 2 & 5.63 \\ \hline 3 & 8.83 \\ \hline 4 & 13.62 \\ \hline 5 & 21.22 \\ \hline \end{array} \end{aligned} $$

Long-term population growth: Although exponential growth can often be used to model population growth accurately for some periods of time, there are inevitably, in the long term, limiting factors that make purely exponential models inaccurate. If the U.S. population had continued to grow by \(3 \%\) each year from 1790 , when it was \(3.93\) million, until today, what would the population of the United States have been in 2000 ? For comparison, according to census data, the population of the United States in 2000 was \(281,421,906\). The population of the world was just over 6 billion people.

Radioactive decay: A scientist is studying the amount of a radioactive substance present over a period of time. A plot of the logarithm of the amount shows a linear pattern. What type of function should the scientist use to model the original data?

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