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Chapter 4: Problem 1

Population Growth A population is increasing according to the exponential function \(y=2 e^{0.02 x},\) where \(y\) is in millions and \(x\) is the number of years. Match each question in Column I with the correct procedure in Column II to answer the question. A. Evaluate \(y=2 e^{0.02(1 / 3)}\) B. Solve \(2 e^{0.02 x}=6\) C. Evaluate \(y=2 e^{0.02(3)}\) D. Solve \(2 e^{0.02 x}=3\) How long will it take for the population to triple?

Short Answer

Expert verified
It will take approximately 54.93 years for the population to triple.

Step by step solution

01

Identify the corresponding equation

The question is asking how long it will take for the population to triple. The population at time 0 is given by the equation: \(y=2 e^{0.02 x}\)
02

Set up the equation for the tripled population

If the population triples, the population will be 3 times its initial value. Therefore, the equation becomes: \(2 e^{0.02 x} = 6\)
03

Solve for \(x\)

To isolate \(x\), divide both sides of the equation by 2: \(e^{0.02 x} = 3\). Then, take the natural logarithm of both sides: \(\ln(e^{0.02 x}) = \ln(3)\). Since \(\ln(e^a) = a\), it simplifies to \(0.02 x = \ln(3)\).
04

Solve for \(x\)

Now, solve for \(x\): \(x = \frac{\ln(3)}{0.02}\)
05

Calculate the value of \(x\)

Finally, compute the value using a calculator: \(x \approx \frac{1.0986}{0.02} \approx 54.93\)

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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. They are often used to model real-life situations involving growth or decay, such as population increase, radioactive decay, and compound interest. In the given exercise, the population growth is represented by the exponential function \(y=2 e^{0.02 x}\), where 'y' is the population size in millions, and 'x' is the number of years.

Key characteristics of exponential functions include:
  • They have a constant relative growth rate.
  • The base of the exponent in our problem is 'e', which is approximately 2.71828.
  • The function \(y=2e^{0.02x}\) indicates that the population starts at 2 million (when \(x=0\)) and grows as time passes due to the factor \(e^{0.02x}\).

Understanding how exponential functions work is crucial for solving problems related to growth phenomena, including population growth.
natural logarithm
The natural logarithm, denoted as \(\ln\), is the logarithm to the base 'e'. It is an important concept in mathematics, particularly in calculus and its applications including solving exponential growth problems.

Here are some helpful points on natural logarithms:
  • \(\ln(e^a) = a\) for any real number 'a'.
  • It is the inverse function of the exponential function with base 'e'.
  • In our problem, taking the natural logarithm helps in solving equations where the variable is in the exponent.

When we solved the equation for the tripling population, we used the natural logarithm to isolate the exponent 'x':
\[ \ln(e^{0.02x}) = \ln(3) \]
This simplifies to \(0.02x = \ln(3)\) because \(\ln(e^a)=a\).
solving equations
Solving equations involves finding the value(s) of the variable(s) that satisfy the given equation. In the context of exponential growth, it often requires using the properties of logarithms. Let's break down the general steps taken in this specific example to solve for 'x'.

1. **Identifying the equation**: We start by recognizing that the population triples, leading to:
\[ 2e^{0.02x} = 6 \]
2. **Isolating the exponential term**: To simplify, divide both sides by 2:
\[ e^{0.02x} = 3 \]
3. **Using the natural logarithm**: Apply \(\ln\) to both sides to handle the exponent:
\[ \ln(e^{0.02x}) = \ln(3) \]
4. **Simplifying the expression**: Use the property of the natural logarithm \(\ln(e^a) = a\)
\[ 0.02x = \ln(3) \]
5. **Solving for 'x'**: Divide both sides by 0.02:
\[ x = \frac{\ln(3)}{0.02} \]
By calculating the above expression with a calculator, we find that:
\[ x \approx 54.93 \] years. Therefore, it will take approximately 54.93 years for the population to triple from its initial size.

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Most popular questions from this chapter

Growth of Bacteria The growth of bacteria makes it necessary to time-date some food products so that they will be sold and consumed before the bacteria count is too high. Suppose for a certain product the number of bacteria present is given by $$ f(t)=500 e^{0.1 t} $$ where \(t\) is time in days and the value of \(f(t)\) is in millions. Find the number of bacteria present at each time. (a) 2 days (b) 4 days (c) 1 week

Product Sales Sales of a product, under relatively stable market conditions but in the absence of promotional activities such as advertising, tend to decline at a constant yearly rate. This rate of sales decline varies considerably from product to product, but it seems to remain the same for any particular product. The sales decline can be expressed by the function $$ S(t)=S_{0} e^{-a t} $$ where \(S(t)\) is the rate of sales at time \(t\) measured in years, \(S_{0}\) is the rate of sales at time \(t=0,\) and \(a\) is the sales decay constant. (a) Suppose the sales decay constant for a particular product is \(a=0.10 .\) Let \(S_{0}=50,000\) and find \(S(1)\) and \(S(3)\) (b) Find \(S(2)\) and \(S(10)\) if \(S_{0}=80,000\) and \(a=0.05\)

(Modeling) Solve each problem. See Example 11 . Employee Training A person learning certain skills involving repetition tends to learn quickly at first. Then learning tapers off and skill acquisition approaches some upper limit. Suppose the number of symbols per minute that a person using a keyboard can type is given by $$ f(t)=250-120(2.8)^{-0.5 t} $$ where \(t\) is the number of months the operator has been in training. Find each value. (a) \(f(2)\) (b) \(f(4)\) (c) \(f(10)\) (d) What happens to the number of symbols per minute after several months of training?

Use the definition of inverses to determine whether \(f\) and \(g\) are inverses. $$f(x)=\frac{-1}{x+1}, \quad g(x)=\frac{1-x}{x}$$

Deer Population The exponential growth of the deer population in Massachusetts can be calculated using the model $$ f(x)=50,000(1+0.06)^{x} $$ where \(50,000\) is the initial deer population and 0.06 is the rate of growth. \(f(x)\) is the total population after \(x\) years have passed. (a) Predict the total population after 4 yr. (b) If the initial population was \(30,000\) and the growth rate was \(0.12,\) approximately how many deer would be present after 3 yr? (c) How many additional deer can we expect in 5 yr if the initial population is \(45,000\) and the current growth rate is \(0.08 ?\) (IMAGE CANT COPY)
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