/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Identify the doubling time or ha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 2

Identify the doubling time or half-life of each of the following exponential functions. Assume \(t\) is in years. [Hint: What value of \(t\) would give you a growth (or decay) factor of 2 (or \(1 / 2\) )?] a. \(Q=70(2)^{t}\) b. \(Q=1000(2)^{t / 50}\) c. \(Q=300\left(\frac{1}{2}\right)^{t}\) d. \(Q=100\left(\frac{1}{2}\right)^{t / 250}\) e. \(N=550(2)^{t / 10}\) f. \(N=50\left(\frac{1}{2}\right)^{t / 20}\)

Short Answer

Expert verified
a. 1 yearb. 50 yearsc. 1 yeard. 250 yearse. 10 yearsf. 20 years

Step by step solution

01

Understanding Doubling Time and Half-Life

Doubling time is the time it takes for a quantity to double, while half-life is the time it takes for a quantity to reduce to half. In exponential functions, doubling time is found by setting the base of the exponential function to 2, and half-life is found by setting the base to 1/2.
02

Identify the Doubling Time for Each Function

For functions with a form similar to \(Q = a(2)^{bt}\) where we need to find the time when the factor is 2. For functions like \(Q = a(2)^{t}\), doubling occurs in 1 unit of time if t is in years.
03

Identify the Half-Life for Each Function

For functions with a form similar to \(Q = a\left(\frac{1}{2}\right)^{bt}\), we need to find the time when the factor is 1/2. For functions like \(Q = a\left(\frac{1}{2}\right)^{t}\), halving occurs in 1 unit of time if t is in years.
04

Solve for Each Given Function

a. \(Q = 70(2)^{t}\): The doubling time is 1 year since the base is 2 raised to t.b. \(Q = 1000(2)^{t/50}\): Set \(2^{t/50}=2\). Therefore, t/50 = 1, resulting in doubling time = 50 years.c. \(Q = 300\left(\frac{1}{2}\right)^{t}\): The half-life is 1 year since the base is \(\frac{1}{2}\) raised to t.d. \(Q = 100\left(\frac{1}{2}\right)^{t/250}\): Set \(\left(\frac{1}{2}\right)^{t/250}=\frac{1}{2}\). Therefore, t/250 = 1, resulting in half-life = 250 years.e. \(N = 550(2)^{t/10}\): Set \(2^{t/10}=2\). Therefore, t/10 = 1, resulting in doubling time = 10 years.f. \(N = 50\left(\frac{1}{2}\right)^{t/20}\): Set \(\left(\frac{1}{2}\right)^{t/20}=\frac{1}{2}\). Therefore, t/20 = 1, resulting in half-life = 20 years.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Doubling Time
Doubling time is a crucial concept in understanding exponential growth. It tells us how long it takes for a quantity to double from its initial value. In an exponential function, this is represented by a base of 2, like in the function \(Q=70(2)^t\). Here, the quantity \(Q\) doubles whenever \(t\) increases by one unit. This is due to the base, which is 2, showing that the quantity multiplies by 2 for every increment of \(t\).

This means:
  • If \(t\) is in years, the doubling time in the equation \(Q=70(2)^t\) is 1 year.
  • For an equation like \(Q=1000(2)^{t/50}\), you can find the doubling time by solving \(2^{t/50}=2\). Simplifying, you get \(t/50=1\), so \(t=50\) years.
Half-Life
Half-life is another fundamental concept in exponential functions, closely related to exponential decay. Half-life refers to the time required for a quantity to reduce to half its initial value. This is represented by a base of \(1/2\) in the function, such as \(Q=300\big(\frac{1}{2}\big)^t\). Here, the quantity \(Q\) halves whenever \(t\) increases by one unit. This happens because the base, \(1/2\), indicates that the quantity multiplies by \(1/2\) (or gets halved) for every unit increase in \(t\).

This concept can be better understood with examples:
  • If \(t\) is in years, the half-life in the equation \(Q=300\big(\frac{1}{2}\big)^t\) is 1 year.
  • For an equation like \(Q=100\big(\frac{1}{2}\big)^{t/250}\), find the half-life by solving \(\big(\frac{1}{2}\big)^{t/250}=\big(\frac{1}{2}\big)\). Simplifying, you get \(t/250=1\), so \(t=250\) years.
Growth Factor
The growth factor in an exponential function indicates how much the quantity grows for each unit increase in time. This factor is the base of the exponential function. If the base is greater than 1, it implies growth. For example, in the function \(N=550(2)^{t/10}\), the base 2 is the growth factor, indicating that the quantity doubles every 10 units of time. Thus, the growth factor tells us the rate at which the quantity increases over time.

So:
  • In \(N=550(2)^{t/10}\), the growth factor is 2, indicating doubling every 10 units of time.
  • Growth factors can vary, but any base more than 1 indicates exponential growth.
Decay Factor
The decay factor in an exponential function represents the rate at which a quantity decreases over time whenever the base is between 0 and 1. An example of this can be seen in \(N=50\big(\frac{1}{2}\big)^{t/20}\). Here, the base \(1/2\) is the decay factor, meaning that the quantity halves every 20 units of time. Thus, the decay factor helps us understand how quickly a quantity diminishes.

Breaking it down:
  • In \(N=50\big(\frac{1}{2}\big)^{t/20}\), the decay factor is \(1/2\), indicating halving every 20 units of time.
  • Decay factors are always between 0 and 1, indicating exponential decay.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(Graphing program recommended.) a. As \(x \rightarrow+\infty\), which function will dominate, $$f(x)=100+500 x \text { or } g(x)=2(1.005)^{x} ?$$ b. Determine over which \(x\) interval(s) \(g(x)>f(x)\)

MCI, a phone company that provides long-distance service, introduced a marketing strategy called "Friends and Family." Each person who signed up received a discounted calling rate to ten specified individuals. The catch was that the ten people also had to join the "Friends and Family" program. a. Assume that one individual agrees to join the "Friends and Family" program and that this individual recruits ten new members, who in turn each recruit ten new members, and so on. Write a function to describe the number of new people who have signed up for "Friends and Family" at the \(n\) th round of recruiting. b. Now write a function that would describe the total number of people (including the originator) signed up after \(n\) rounds of recruiting. c. How many "Friends and Family" members. stemming from this one person, will there be after five rounds of recruiting? After ten rounds? d. Write a 60 -second summary of the pros and cons of this recruiting strategy. Why will this strategy eventually collapse?

On November \(25,2003,\) National Public Radio did a report on Under Armour, a sports clothing company, stating that their "profits have increased by \(1200 \%\) in the last 5 years." a. Let \(P(t)\) represent the profit of the company during every 5-year period, with \(A_{0}\) the initial amount. Write the exponential model for the company's profit. b. Assuming an initial profit of \(\$ 100,000,\) what would be the profit in year 5 ? Year \(10 ?\) c. Determine the \(a n n u a l\) growth rate for Under Armour.

Each table has values representing either linear or exponential functions. Find the equation for each function. $$ \begin{array}{cccccc} \hline x & -2 & -1 & 0 & 1 & 2 \\ h(x) & 160 & 180 & 200 & 220 & 240 \\ \hline \end{array} $$ $$ \begin{aligned} &\text { b. }\\\ &\begin{array}{cccccc} \hline x & 0 & 10 & 20 & 30 & 40 \\ j(x) & 200 & 230 & 264.5 & 304.17 & 349.8 \\ \hline \end{array} \end{aligned} $$

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?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.