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F. E. Smith has reported on population growth of the water flea. \({ }^{24}\) In one experiment, he found that the time \(t\), in days, required to reach a population of \(N\) is given by the relation $$ e^{0.44 t}=\frac{N}{N_{0}}\left(\frac{228-N_{0}}{228-N}\right)^{4.46} $$ Here \(N_{0}\) is the initial population size. If the initial population size is 50 , how long is required for the population to grow to 125 ?

Short Answer

Expert verified
Approximately 7.25 days are required for the population to grow to 125.

Step by step solution

01

Substitute known values

We know that the initial population size, denoted as \(N_0\), is 50, and the final population \(N\) is supposed to be 125. Substitute these values into the given equation:\[e^{0.44 t} = \frac{125}{50}\left(\frac{228 - 50}{228 - 125}\right)^{4.46}\]
02

Simplify the right side of the equation

Calculate each part separately on the right-hand side of the equation.First, \(\frac{125}{50} = 2.5\).Then, calculate \(\frac{228 - 50}{228 - 125} = \frac{178}{103}\). Raise the result to the power of 4.46: \[\left(\frac{178}{103}\right)^{4.46}\].
03

Calculate the expression

Using a calculator, evaluate:\(\left(\frac{178}{103}\right)^{4.46} \approx 9.71\). Multiply this result by 2.5 from Step 2:\[2.5 \times 9.71 = 24.275\].
04

Solve for \(t\) using logarithms

We have:\[e^{0.44 t} = 24.275\] To solve for \(t\), take the natural logarithm on both sides:\[0.44 t = \ln(24.275)\]Calculate \(\ln(24.275)\), which is approximately 3.19.
05

Solve for \(t\)

Solve the equation for \(t\):\[t = \frac{3.19}{0.44} \approx 7.25\].

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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 special type of mathematical function where a constant base is raised to the power of a variable. This kind of function often describes processes that grow or decay at a constant relative rate. The general form of an exponential function is \( y = a \, b^x \), where \( a \) is the initial value, \( b \) is the base, and \( x \) is the exponent.
In the context of population growth, exponential functions can model situations where the rate of growth is proportional to the current population size, leading to faster increases as the population grows.
  • The constant \( e \) is often used in natural exponential functions. Known as the natural exponential base, \( e \approx 2.71828 \).
  • In the given exercise, the function \( e^{0.44 t} \) represents an exponential growth model, where \( e \) is raised to the power of the expression \( 0.44 t \).
Such functions can represent real-world scenarios like bacteria population growth, radioactive decay, or, in this case, the increase in water flea populations.
Natural Logarithm
The natural logarithm is a fundamental mathematical function used to reverse or "undo" an exponential function involving the base \( e \). Denoted as \( \ln(x) \), the natural logarithm of a number \( x \) is the power to which \( e \) must be raised to equal \( x \). For example, when you take \( \ln(e^{0.44 t}) \), you would get \( 0.44 t \) as the natural logarithm reverses the exponentiation.
Understanding the natural logarithm is crucial when you need to solve equations involving exponential functions. In our exercise, to find the time \( t \) required for population growth, we apply the natural logarithm to both sides of the equation \( e^{0.44 t} = 24.275 \).
  • This produces \( \ln(e^{0.44t}) = \ln(24.275) \), simplifying to \( 0.44t = \ln (24.275) \).
  • Taking the natural logarithm allows us to isolate the exponent, which holds the key to solving for the variable \( t \).
The ability to maneuver between exponential and logarithmic forms is valuable, giving us tools to solve and simplify complex equations.
Algebraic Manipulation
Algebraic manipulation involves using operations and properties of equality to rearrange and solve equations. It is an essential skill for solving most mathematical problems, including those involving exponential functions and logarithms.
In the exercise, several algebraic manipulation steps are used to simplify and solve the initial equation.
  • First, known values are substituted into the equation, replacing variables with actual numbers.
  • The right side of the equation is simplified by calculating fractions and exponentiations, which is crucial for managing complex expressions.
  • Lastly, the variable \( t \) is isolated by dividing both sides of the final equation by \( 0.44 \), resulting in solving for \( t \). This step involves precise arithmetic and careful handling of each component of the equation.
These manipulations showcase the importance of careful and systematic steps, ensuring accuracy as we work toward finding the solution. Such steps are applicable not just in population growth problems but in various mathematical contexts such as physics, engineering, and economics.

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Most popular questions from this chapter

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