/*! 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 51 Escherichia coli (E. coli) is a ... [FREE SOLUTION] | 91Ó°ÊÓ

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Escherichia coli (E. coli) is a bacterium that can reproduce at an exponential rate. The \(E\) coli reproduce by dividing. A small number of E. coli bacteria in the large intestine of a human can trigger a serious infection within a few hours. Consider a particular E. coli infection that starts with \(100 E .\) coli bacteria. Each bacterium splits into two parts every half hour. Assuming none of the bacteria die, the size of the \(E .\) coli population after \(t\) hours is given by \(P(t)=100 \cdot 2^{2 t},\) where \(0 \leq t \leq 16\) a. Find \(P(3)\) and \(P(6)\) b. Use a graphing utility to find the time, to the nearest tenth of an hour, it takes for the \(E .\) coli population to number 1 billion.

Short Answer

Expert verified
a. \(P(3)=6,400\), \(P(6) = 4,096,000\). b. It takes approximately 14.6 hours for the E. coli population to reach 1 billion.

Step by step solution

01

Compute for \(P(3)\) and \(P(6)\)

Substitute \(t = 3\) into the formula to find \(P(3)= 100 \cdot 2^{2 \cdot 3} = 100 \cdot 2^{6} = 6,400\). Similarly, substitute \(t = 6\) into the formula to find \(P(6) = 100 \cdot 2^{2 \cdot 6} = 100 \cdot 2^{12} = 4096000\). Thus, there are 6,400 bacteria after 3 hours, and 4,096,000 bacteria after 6 hours.
02

Compute for the time it takes for the population to reach 1 billion

To solve for time \(t\) when the population reaches 1 billion, set \(P(t) = 1,000,000,000\) and solve for \(t\). Therefore, \(1,000,000,000 = 100 \cdot 2^{2t}\). In order to solve for \(t\) in this case, logarithmic or iterative approximation must be used, or a graphing utility can be taken advantage of, as the problem suggests. Using a graphing utility, we find that \(t\) approximately equals 14.6 hours.

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Key Concepts

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

Exponential Functions
Exponential functions are a type of mathematical function that exhibit rapid growth or decay, represented by an equation of the form \( f(x) = ab^{x} \), where \( a \) is a non-zero constant referred to as the initial value, \( b \) is the base of the exponential (a positive real number not equal to 1), and \( x \) is the exponent. The base, \( b \), determines the rate at which the function grows (when \( b > 1 \)) or decays (when \( 0 < b < 1 \)).

In the context of biological growth such as the reproduction of bacteria, exponential functions model how populations change over time. In our E. coli example, the initial amount of bacteria is represented by \( a = 100 \) and the base of the exponential is \( b = 2^{2} \) due to the bacteria doubling every half hour. The variable \( t \) represents time in hours, making the population at any time \( t \) given by \( P(t) = 100 \times 2^{2t} \).

Understanding the definition and properties of exponential functions is crucial as it allows us to predict and manage real-life situations involving multiplicative growth or decay such as population dynamics, radioactive decay, and even financial investments.
Graphing Exponential Equations
Graphing exponential equations is a visual way to understand the behavior of exponential functions. To graph an exponential equation like \( P(t) = 100 \times 2^{2t} \), we plot points for different time values (\( t \)) and connect them smoothly to exhibit the characteristic 'J-shape' of exponential growth.

The graph passes through the point (0, 100), representing the initial population, and rises rapidly to the right as the value of \( t \) increases due to the population doubling at a constant interval. The steeper the curve, the higher the growth rate, which is noticeable in cases like bacterial multiplication where the base is high, as seen with the base value of 4 in the E. coli growth function.

An important aspect of graphing is to consider the domain and the range of the function. For our E. coli example, the domain is given as \( 0 \leq t \leq 16 \), the time period during which we are observing the bacteria's growth, while the range will be the set of possible population sizes, starting with 100 and potentially growing very large, though practical limits exist such as resource constraints not reflected in the equation.
Solving Exponential Equations
Solving exponential equations, like finding when the E. coli population reaches 1 billion, involves isolating the variable that represents time or the exponent in the function. This process generally uses logarithms to counteract the exponential function and calculate the time needed for the population to reach a certain number.

To solve for a population of 1 billion in our example, we set \( P(t) = 1,000,000,000 \) and then solve for \( t \), which yields \( 1,000,000,000 = 100 \times 2^{2t} \). Dividing both sides by 100 gives us \( 10,000,000 = 2^{2t} \), and applying the logarithm to both sides allows us to write \( \log_{2}(10,000,000) = 2t \). Finally, solving for \( t \) would provide us the time at which the population size is 1 billion. We often rely on a calculator or graphing utility to find this value due to the large numbers involved. In the textbook solution, a graphing utility was used to find that the bacteria population reaches 1 billion in approximately 14.6 hours.

It is notable that, in practice, we use logarithms because they are the inverses of exponential functions. Whenever you have an equation where the variable is in an exponent, logarithms can be utilized to solve for that variable, making them a key tool in the study of exponential functions.

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