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Chapter 5: Problem 8

a. Complete the following table for the exponential function \(y=20(0.75)^{x}\) b. Choose the correct word in each italicized pair to describe the function: For the exponential function \(y=20(0.75)^{x}\), the differences are constant/decreasing in magnitude and the ratios are constant/decreasing in magnitude.

Short Answer

Expert verified
The differences are decreasing in magnitude, and the ratios are constant.

Step by step solution

01

Understanding the Equation

The function given is an exponential decay function of the form \[ y = 20(0.75)^x \]. The base of the exponent, 0.75, is less than 1, indicating that the function decreases as \( x \) increases.
02

Filling in the Table

To complete the table for \( y = 20(0.75)^x \), input specific values for \( x \) and compute the corresponding \( y \)-values:- For \( x = 0 \): \[ y = 20(0.75)^0 = 20(1) = 20 \]- For \( x = 1 \): \[ y = 20(0.75)^1 = 20(0.75) = 15 \]- For \( x = 2 \): \[ y = 20(0.75)^2 = 20(0.5625) = 11.25 \]- For \( x = 3 \): \[ y = 20(0.75)^3 = 20(0.421875) = 8.4375 \]- For \( x = 4 \): \[ y = 20(0.75)^4 = 20(0.31640625) = 6.3281 \]
03

Determine the Nature of the Differences

To find whether the differences are constant or decreasing, compute the differences between successive \( y \)-values:- Difference for \( x = 0 \) to \( x = 1 \): 20 - 15 = 5- Difference for \( x = 1 \) to \( x = 2 \): 15 - 11.25 = 3.75- Difference for \( x = 2 \) to \( x = 3 \): 11.25 - 8.4375 = 2.8125- Difference for \( x = 3 \) to \( x = 4 \): 8.4375 - 6.3281 = 2.1094These differences are decreasing.
04

Determine the Nature of the Ratios

To determine if the ratios are constant, compute the ratios of successive \( y \)-values:- Ratio for \( x = 0 \) to \( x = 1 \): \( \frac{15}{20} = 0.75 \)- Ratio for \( x = 1 \) to \( x = 2 \): \( \frac{11.25}{15} = 0.75 \)- Ratio for \( x = 2 \) to \( x = 3 \): \( \frac{8.4375}{11.25} = 0.75 \)- Ratio for \( x = 3 \) to \( x = 4 \): \( \frac{6.3281}{8.4375} = 0.75 \)These ratios are constant.
05

Conclusion

Based on the calculations, for the exponential function \( y = 20(0.75)^x \), the differences are decreasing in magnitude and the ratios are constant.

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Key Concepts

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

Exponential Functions

An exponential function is a mathematical expression where a constant base is raised to a variable exponent. It can be written in the form: \( y = ab^x \), where \(a \) is the initial value, \(b \) is the base, and \(x \) is the exponent.


When the base \(b \) is greater than 1, the function exhibits exponential growth, increasing rapidly. Conversely, when the base \(b \) lies between 0 and 1, the function displays exponential decay. This means it decreases as \(x \) increases. For example, the function \( y = 20(0.75)^x \) is an exponential decay function with a base of 0.75. As \( x \) grows, \( y \) gets smaller.

Function Table

A function table helps illustrate how the function behaves as the input values change. For the function \( y = 20(0.75)^x \), you can fill the table by substituting various values of \(x \) and calculating the corresponding \(y \) values:


  • For \( x = 0 \): \( y = 20(0.75)^0 = 20 \)
  • For \( x = 1 \): \( y = 20(0.75)^1 = 15 \)
  • For \( x = 2 \): \( y = 20(0.75)^2 = 11.25 \)
  • For \( x = 3 \): \( y = 20(0.75)^3 = 8.4375 \)
  • For \( x = 4 \): \( y = 20(0.75)^4 = 6.3281 \)

This shows the function decreasing with increasing values of \(x \).

Constant Ratios

In an exponential function, the ratios of successive \(y \) values are constant. For the function \( y = 20(0.75)^x \), you compute the ratios by dividing each \( y \) value by the previous \( y \).


  • Ratio for \( x = 0 \) to \( x = 1 \): \( \frac{15}{20} = 0.75 \)
  • Ratio for \( x = 1 \) to \( x = 2 \): \( \frac{11.25}{15} = 0.75 \)
  • Ratio for \( x = 2 \) to \( x = 3 \): \( \frac{8.4375}{11.25} = 0.75 \)
  • Ratio for \( x = 3 \) to \( x = 4 \): \( \frac{6.3281}{8.4375} = 0.75 \)

These ratios are constant at 0.75, confirming the exponential nature of the function.

Decreasing Differences

While the ratios of successive \(y \) values are constant in an exponential function, the differences between successive \(y \) values decrease. For the function \( y = 20(0.75)^x \), subtract the \( y \) value of each \(x \) from that of the previous \(x \):


  • Difference for \( x = 0 \) to \( x = 1 \): \( 20 - 15 = 5 \)
  • Difference for \( x = 1 \) to \( x = 2 \): \( 15 - 11.25 = 3.75 \)
  • Difference for \( x = 2 \) to \( x = 3 \): \( 11.25 - 8.4375 = 2.8125 \)
  • Difference for \( x = 3 \) to \( x = 4 \): \( 8.4375 - 6.3281 = 2.1094 \)

These differences are decreasing, demonstrating the typical behavior of exponential decay.

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Most popular questions from this chapter

The U.S. Department of Agriculture's data on per capita food commodity consumption for 1980 are listed in the following table. a. Using the data in the following table, construct exponential functions for each food category. Then evaluate each function for the year \(2000 .\) Assume \(t\) is the number of years since \(1980 .\) $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Per Capita } \\ \text { Consumption } \\ \text { (pounds) in } \\ 1980 \end{array} & \begin{array}{c} \text { Yearly } \\ \text { Growth/Decay } \\ \text { Factor } \end{array} & \begin{array}{c} \text { Exponential } \\ \text { Function } \end{array} \\ \hline \text { Beef } & 72.1 & 0.994 & B(t)= \\ \text { Chicken } & 32.7 & 1.024 & C(t)= \\ \text { Pork } & 52.1 & 0.996 & P(t)= \\ \text { Fish } & 12.4 & 1.010 & F(t)= \\ \hline \end{array} $$ b. Which commodities showed exponential growth? Which showed exponential decay? c. Write a 60 -second summary about the consumption of meat, chicken, and fish from 1980 to 2000 .

Which of the following functions have a fixed doubling time A fixed half-life? a. \(y=6(2)^{x}\) b. \(y=5+2 x\) c. \(Q=300\left(\frac{1}{2}\right)^{T}\) d. \(A=10(2)^{t / 5}\) e. \(P=500-\frac{1}{2} T\) f. \(N=50\left(\frac{1}{2}\right)^{t / 20}\)

(Requires technology to find a best-fit function.) In \(1911,\) reindeer were introduced to St. Paul Island, one of the Pribilof Islands, off the coast of Alaska in the Bering Sea. There was plenty of food and no hunting or reindeer predators. The size of the reindeer herd grew rapidly for a number of years, as given in the accompanying table. $$ \begin{array}{cr|rr} \hline & \text { Population } & & \text { Population } \\ \text { Year } & \text { Size } & \text { Year } & \text { Size } \\ \hline 1911 & 17 & 1925 & 246 \\ 1912 & 20 & 1926 & 254 \\ 1913 & 42 & 1927 & 254 \\ 1914 & 76 & 1928 & 314 \\ 1915 & 93 & 1929 & 339 \\ 1916 & 110 & 1930 & 415 \\ 1917 & 136 & 1931 & 466 \\ 1918 & 153 & 1932 & 525 \\ 1919 & 170 & 1933 & 670 \\ 1920 & 203 & 1934 & 831 \\ 1921 & 280 & 1935 & 1186 \\ 1922 & 229 & 1936 & 1415 \\ 1923 & 161 & 1937 & 1737 \\ 1924 & 212 & 1938 & 2034 \\ \hline \end{array} $$ a. Use the reindeer data file (in Excel or graph link form) to plot the data. b. Find a best-fit exponential function. c. How does the predicted population from part (b) differ from the observed ones? d. Does your answer in part (c) give you any insights into why the model does not fit the observed data perfectly? e. Estimate the doubling time of this population.

Lead- 206 is not radioactive, so it does not spontaneously decay into lighter elements. Radioactive elements heavier than lead undergo a series of decays, each time changing from a heavier element into a lighter or more stable one. Eventually, the element decays into lead- 206 and the process stops. So, over billions of years, the amount of lead in the universe has increased because of the decay of numerous radioactive elements produced by supernova explosions. Radioactive uranium- 238 decays sequentially into thirteen other lighter elements until it stabilizes at lead-206. The half-lives of the fifteen different elements in this decay chain vary from 0.000164 seconds (from polonium- 214 to lead- 210 ) all the way up to 4.47 billion years (from uranium- 238 to thorium- 234 ). a. Find the decay rate per billion years for uranium- 238 to decay into thorium- 234 . b. Find the decay rate per second for polonium-214 to decay into lead-2.10.

The price of a home in Medford was \(\$ 100,000\) in 1985 and rose to \(\$ 200,000\) in 2005 . a. Create two models, \(f(t)\) assuming linear growth and \(g(t)\) assuming exponential growth, where \(t=\) number of years after \(1985 .\) b. Fill in the following table representing linear growth and exponential growth for \(t\) years after \(1985 .\) c. Which model do you think is more realistic?
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