/*! 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 Show that the doubling time of \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that the doubling time of \(y=A B^{t}\) is \(1 /\left(\log _{2} B\right)\)

Short Answer

Expert verified
The doubling time of \(y=AB^t\) is \(1/\log_2 B\).

Step by step solution

01

Understand Doubling Time

Doubling time is the time it takes for a quantity to double in size or value. For the function \(y = AB^t\), doubling time \(T\) is the time after which \(y\) becomes \(2y\).
02

Set Up the Doubling Equation

We want to find \(T\) such that \(AB^{t + T} = 2AB^t\). This equation represents the point at which the initial function doubles in value.
03

Simplify the Doubling Equation

We simplify \(AB^{t + T} = 2AB^t\) by dividing both sides by \(AB^t\):\[ B^T = 2 \]
04

Solve for T Using Logarithms

To solve \(B^T = 2\) for \(T\), take the logarithm base 2 on both sides:\[ \log_2{(B^T)} = \log_2{2} \]This simplifies using logarithm properties to:\[ T \log_2{B} = 1 \]
05

Express T Using Logarithm Properties

Isolate \(T\) by dividing both sides by \(\log_2{B}\):\[ T = \frac{1}{\log_2{B}} \]This shows that \(T\), the doubling time, is indeed \(\frac{1}{\log_2{B}}\).

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 91Ó°ÊÓ!

Key Concepts

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

Exponential Growth
Exponential growth is a process where a quantity increases by a consistent percentage over equal time periods. This type of growth is characterized by its acceleration—instead of adding the same amount each period, it multiplies by a factor. This is often seen in populations, investments, and more, where they grow larger over time. For the mathematical function described as \(y = AB^t\), \(A\) represents the initial quantity, \(B\) the growth factor, and \(t\) the time period. Whenever \(B > 1\), it indicates growth, and the system is doubling at certain intervals.

When modeling exponential growth, it's important to understand that each cycle of growth builds upon the previous amount, leading to exponentially larger amounts. This is why recognizing and calculating doubling time is crucial for forecasting future growth potential.
Logarithms
Logarithms are the inverse operations of exponentiation. They answer the question: "To what exponent must a base be raised to produce a given number?" For instance, \(\log_2(8) = 3\) because \(2^3 = 8\). Logarithms are especially useful when solving exponential equations since they can simplify the process of isolating the variable of interest.

In the context of finding doubling time, logarithms help us to solve equations where the variable appears in the exponent, like \(B^T = 2\). By using a logarithm with base 2, we apply the property \(\log_b{(x^y)} = y\log_b{x}\) to turn the equation into a simple multiplication form \(T \log_2{B} = 1\). This allows us to isolate and solve for \(T\), offering a methodical approach to understand how long it will take for an exponential function to double.
Mathematical Modeling
Mathematical modeling involves using mathematical structures and expressions to represent real-world situations. It's a crucial tool in science, economics, engineering, and many other fields, allowing us to make predictions and understand complex systems. The function \(y = AB^t\) is a classic example of a model designed to predict exponential growth, which is especially common in biological systems and finance.

A well-constructed model provides insights and understanding, enabling reliable forecasts. By calculating doubling time, we gain specific insights into the behavior of the system we're studying. This particular aspect of modeling, where exponential functions are involved, highlights how small changes in growth rates (represented by \(B\)) can lead to significant changes over time. It exemplifies the potential and power of mathematical modeling in analyzing growth dynamics.

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

Consider a chemical reaction $$ A+B \longrightarrow A B $$ in which a chemical, \(A,\) combines with a chemical, \(B,\) to form the compound, \(A B\). Assume that the amount of \(B\) greatly exceeds the amount of \(A,\) and that in any second, the amount of \(A B\) that is formed is proportional to the amount of \(A\) present at the beginning of the second. Write a dynamic equation for this reaction, and write a solution equation to the dynamic equation.

Determine the doubling times or half-lives of the following exponential equations. (a) \(y=0.5^{t}\) (b) \(y=2^{3 t}\) (c) \(y=0.1^{0.1 t}\) (d) \(y=1000.8^{t}\) (e) \(y=45^{3 t}\) (f) \(y=0.00015^{0.1 t}\) (g) \(y=100.8^{2 t}\) (h) \(y=0.01^{3 t}\) (i) \(y=0.01^{0.1 t}\)

The equation, \(B_{t}-B_{t-1}=r B_{t-1},\) carries the same information as \(B_{t+1}-B_{t}=r B_{t}\) a. Write the first four instances of \(B_{t}-B_{t-1}=r B_{t-1}\) using \(t=1, t=2, t=3,\) and \(t=4\). b. Cascade these four equations to get an expression for \(B_{4}\) in terms of \(r\) and \(B_{0}\). c. Write solutions to and compute \(B_{40}\) for (a.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.2 B_{t-1}\) (b.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.1 B_{t-1}\) (c.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=0.05 B_{t-1}\) (d.) \(\quad B_{0}=50 \quad B_{t}-B_{t-1}=-0.1 B_{t-1}\)

The polymerase chain reaction is a means of making multiple copies of a DNA segment from only a minute amounts of original DNA. The procedure consists of a sequence of, say, 30 cycles in which each segment present at the beginning of a cycle is duplicated once; at the end of the cycle that segment and one copy is present. Introduce notation and write a difference equation with initial condition from which the amount of DNA present at the end of each cycle can be computed. Suppose you begin with 1 picogram \(=0.000000000001 \mathrm{~g}\) of DNA. How many grams of DNA would be present after 30 cycles.

Suppose a quail population would grow at \(20 \%\) per year without hunting pressure, and 1000 birds per year are harvested. Describe the progress of the population over 5 years if initially there are a. 5000 birds, b. 6000 birds, and c. 4000 birds.

See all solutions

Recommended explanations on Biology Textbooks

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.