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Show that the doubling time of \(y=A 2^{k t}\) is \(1 / k\).

Short Answer

Expert verified
The doubling time of \( y = A \cdot 2^{kt} \) is \( t_d = \frac{1}{k} \).

Step by step solution

01

Set Up the Equation for Doubling Time

To find the doubling time, we need the time it takes for the function value to double. Start with the function: \( y = A \cdot 2^{kt} \). To find the doubling time, let the new value be \( 2y \) and set \( y_f = 2y_i \), where \( y_f \) is the future value and \( y_i \) is the initial value.
02

Substitute Into the Equation

Since \( y_i = A \cdot 2^{kt_0} \) and \( y_f = A \cdot 2^{kt} \), set \( 2 \cdot y_i = A \cdot 2^{kt_d} \). This implies that: \( 2 \cdot (A \cdot 2^{kt_0}) = A \cdot 2^{k(t_0 + t_d)} \).
03

Simplify Equation

Divide both sides by \( A \cdot 2^{kt_0} \) to isolate the growth factor: \( 2 = 2^{k(t_0 + t_d) - kt_0} \). This further simplifies to \( 2 = 2^{kt_d} \).
04

Apply Logarithms

To solve for \( t_d \), take the base-2 logarithm of both sides: \( \log_2(2) = kt_d \). This simplifies to \( 1 = kt_d \) since \( \log_2(2) = 1 \).
05

Solve for Doubling Time

Divide both sides by \( k \) to find the doubling time: \( t_d = \frac{1}{k} \).

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Key Concepts

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

Doubling Time
Doubling time is a concept used in exponential growth to determine how quickly something grows. It's the time taken for a quantity to double in size or value. In the specific case of the equation \(y = A \cdot 2^{kt}\), which represents exponential growth, doubling time helps us to understand how fast the function is increasing.

Here's how it works: to find the doubling time \(t_d\), you need to determine the length of time until the initial value, \(y_i = A\), becomes twice as large. By setting the future value \(y_f\) to \(2y_i\), you can use the equation:\[2 = 2^{kt_d}\]
This simplification comes from dividing both sides of the equation by \(A \cdot 2^{kt_0}\), the initial growth condition. Solving the equation for \(t_d\), using logarithms as demonstrated, shows that the doubling time is \(t_d = \frac{1}{k}\).

Understanding doubling time is crucial because it provides an easy and clear way to measure exponential growth. This concept is used in diverse fields ranging from finance to population studies.
Logarithms
Logarithms transform multiplicative processes, such as exponential growth, into additive ones, making them easier to work with. In the context of finding the doubling time for exponential growth, logarithms are a key tool.

When we have the equation \(2 = 2^{kt_d}\), and we want to solve for \(t_d\), a base-2 logarithm becomes very handy. The property of logarithms tells us that \(\log_b(b^x) = x\). So, applying the base-2 logarithm to both sides gives us \(\log_2(2) = kt_d\). This reduces our problem to \(1 = kt_d\), because \(\log_2(2) = 1\).

Logarithms are incredibly useful in solving equations involving exponentials because they allow us to simplify and solve for variables that are exponents, such as \(t_d\) here. Understanding how to use logarithms can immensely simplify working with exponential functions.
Growth Factor
The growth factor is the rate at which a quantity increases during exponential growth. In the function \(y = A \cdot 2^{kt}\), the growth factor is defined by the base of the exponential term, which is \(2^k\).

The value of \(k\) determines how quickly or slowly the quantity grows. A larger \(k\) results in a faster growth rate, leading to a shorter doubling time, and vice versa. This is evident from the derived formula for doubling time: \(t_d = \frac{1}{k}\).

In practical terms, understanding the growth factor helps us predict how fast a process will increase over time, whether it's cells doubling in biology, interest accruing in finance, or information multiplying in a network. The growth factor provides a measure for exponential growth, indicating how each time step leads to exponential change.

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Most popular questions from this chapter

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

Suppose a population is initially of size 1,000,000 and grows at the rate of \(2 \%\) per year. What will be the size of the population after 50 years?

There is a suggestion that the world human population is growing exponentially. Shown below are the human population numbers in billions of people for the decades \(1940-2010\). \(\begin{array}{lrrrrrrrr}\text { Year } & 1940 & 1950 & 1960 & 1970 & 1980 & 1990 & 2000 & 2010 \\ \text { Index, } t & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text { Human Population } \times 10^{6} & 2.30 & 2.52 & 3.02 & 3.70 & 4.45 & 5.30 & 6.06 & 6.80\end{array}\) 1\. Test the equation $$ P_{t}=2.21 .19^{t} $$ against the data where \(t\) is the time index in decades after 1940 and \(P_{t}\) is the human population in billions. 2\. What percentage increase in human population each decade does the model for the equation assume? 3\. What world human population does the equation predict for the year \(2050 ?\)

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Determine the doubling times of the following exponential equations. (a) \(y=2^{t}\) (b) \(y=2^{3 t}\) (c) \(y=2^{0.1 t}\) (d) \(y=10^{t}\) (e) \(y=10^{3 t}\) (f) \(y=10^{0.1 t}\)

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