# Use the exponential growth model A=A_0e^{kt} to solve: In 2000, there were 110 million cellphone subscribers in the United States. By 2010, there were 303 million subscribers a. Find the exponential function that models the data. b. According to the model, in which year were there 400 million cellphone subscribers in the United States? Question
Exponential models Use the exponential growth model $$A=A_0e^{kt}$$ to solve: In 2000, there were 110 million cellphone subscribers in the United States. By 2010, there were 303 million subscribers
a. Find the exponential function that models the data.
b. According to the model, in which year were there 400 million cellphone subscribers in the United States? 2021-03-09
Given Data:
The exponential growth model is $$A=A_0e^{kt}$$
In 2000 the number of telephone users was 110 million.
In 2010 the number of telephone users was 303 million.
Taking 2000 as the base year (t=0) and A as the number of cellphone users then, boundary condition can be written as,
$$t=0,A=110$$ million
$$t=10,A=303$$ million
Substituting the boundary condition in the model equation,
101 million $$=A_0e^{k\times0}$$
$$A_0=110$$ million
303 million $$A_0e^{k\times10}$$
303 million = 110 million$$\times e^{k\times10}$$
$$e^{10k}=2.75$$
$$10k=\ln2.75$$
$$k=0.101325$$
Thus, the required model equation is,
$$A=110e^{0.101325t}$$
The time after which the telephone users will be 400 million can be determined as,
$$400=110e^{0.101325t}$$
$$e^{0.101325t}=3.636$$
$$0.101325t=1.290$$
$$t=12.74$$
Thus, the year in which the number of telephone users will be 400 million is
2000+12.74=2012.74
Thus, in the year 2013 the number of telephone users will be 400 million.

### Relevant Questions Use the exponential growth model $$\displaystyle{A}={A}_{{0}}{e}^{{{k}{t}}}$$ to solve: In 2000, there were 110 million cellphone subscribers in the United States. By 2010, there were 303 million subscribers
a. Find the exponential function that models the data.
b. According to the model, in which year were there 400 million cellphone subscribers in the United States? Use the exponential growth model, $$A=A_0e^{kt}$$. In 1975, the population of Europe was 679 million. By 2015, the population had grown to 746 million.
Solve,
a. Find an exponential growth function that models the data for 1975 through 2015.
b. By which year, to the nearest year, will the European population reach 800 million? Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.$$ Solve (8.86) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{e}}}$$ can be estimated from $$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}$$
(b) (Logistic Growth) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}$$ where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{l}}}$$ can be estimated from $$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}$$ for $$\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.$$
(c) Assume that $$\displaystyle{N}{\left({0}\right)}={1}$$ and $$\displaystyle{N}{\left({10}\right)}={1000}.$$ Estimate $$\displaystyle{r}_{{{e}}}$$ and $$\displaystyle{r}_{{{l}}}$$ for both $$\displaystyle{K}={1001}$$ and $$\displaystyle{K}={10000}.$$
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of $$\displaystyle{\left[{r}\right]}$$ to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when $$\displaystyle\frac{{N}}{{K}}$$ is small compared with 1. The following table lists the reported number of cases of infants born in the United States with HIV in recent years because their mother was infected.
Source:
Centers for Disease Control and Prevention.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&\text{amp, Cases}\backslash{h}{l}\in{e}{1995}&{a}\mp,\ {295}\backslash{h}{l}\in{e}{1997}&{a}\mp,\ {166}\backslash{h}{l}\in{e}{1999}&{a}\mp,\ {109}\backslash{h}{l}\in{e}{2001}&{a}\mp,\ {115}\backslash{h}{l}\in{e}{2003}&{a}\mp,\ {94}\backslash{h}{l}\in{e}{2005}&{a}\mp,\ {107}\backslash{h}{l}\in{e}{2007}&{a}\mp,\ {79}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
a) Plot the data on a graphing calculator, letting $$\displaystyle{t}={0}$$ correspond to the year 1995.
b) Using the regression feature on your calculator, find a quadratic, a cubic, and an exponential function that models this data.
c) Plot the three functions with the data on the same coordinate axes. Which function or functions best capture the behavior of the data over the years plotted?
d) Find the number of cases predicted by all three functions for 20152015. Which of these are realistic? Explain. The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially.
(a) Find a function that models the population t years after 1990.
(b) Find the time required for the population to double.
(c) Use the function from part (a) to predict the population of California in the year 2010. Look up California’s actual population in 2010, and compare. The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially.
(a) Find a function that models the population t years after 1990.
(b) Find the time required for the population to double.
(c) Use the function from part (a) to predict the population of California in the year 2010. Look up California’s actual population in 2010, and compare. Solve the given problem related to population growth.
During the first decade of this century the population of a certain city grew exponentially the population of the city was 146210 in 2000 and 217245 in 2010. Find the exponential growth function that models the population growth of the city use t= 0 to represent 2000 and t= 10 to represent 2010 and so on.
N(t)=?
Use your exponential growth function to predict the population of the city in 2016. Round to the nearest thousand. Solve the given problem related to population growth.
During the first decade of this century the population of a certain city grew exponentially the population of the city was 146210 in 2000 and 217245 in 2010. Find the exponential growth function that models the population growth of the city use t= 0 to represent 2000 and t= 10 to represent 2010 and so on.
N(t)=?
Use your exponential growth function to predict the population of the city in 2016. Round to the nearest thousand. The following table shows the approximate average household income in the United States in 1990, 1995, and 2003. ($$\displaystyle{t}={0}$$ represents 1990.)
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{t(Year)}&{0}&{5}&{13}\backslash{h}{l}\in{e}\text{H(Household Income in}\ \{1},{000}{)}&{30}&{35}&{43}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
a) Linear: $$\displaystyle{H}{\left({t}\right)}={m}{b}\ +\ {b}$$
b) Quadratic: $$\displaystyle{H}{\left({t}\right)}={a}{t}^{{{2}}}\ +\ {b}{t}\ +\ {c}$$
c) Exponential: $$\displaystyle{H}{\left({t}\right)}={A}{b}^{{{t}}}$$ The annual sales S (in millions of dollars) for the Perrigo Company from 2004 through 2010 are shown in the table. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&{2004}&{2005}&{2006}&{2007}&{2008}&{2009}&{2010}\backslash{h}{l}\in{e}\text{Sales, S}&{898.2}&{1024.1}&{1366.8}&{1447.4}&{1822.1}&{2006.9}&{2268.9}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ a) Use a graphing utility to create a scatter plot of the data. Let t represent the year, with $$\displaystyle{t}={4}$$ corresponding to 2004. b) Use the regression feature of the graphing utility to find an exponential model for the data. Use the Inverse Property $$\displaystyle{b}={e}^{{{\ln{\ }}{b}}}$$ to rewrite the model as an exponential model in base e. c) Use the regression feature of the graphing utility to find a logarithmic model for the data. d) Use the exponential model in base e and the logarithmic model to predict sales in 2011. It is projected that sales in 2011 will be \$2740 million. Do the predictions from the two models agree with this projection? Explain.