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The number of bacteria in a culture is increasing according to the law of exponential growth. The initial population is \( 250 \) bacteria, and the population after \( 10 \) hours is double the population after \( 1 \) hour. How many bacteria will there be after \( 6 \) hours?

Short Answer

Expert verified
There will be approximately 480 bacteria after 6 hours.

Step by step solution

01

Understand the equation of exponential growth

The standard equation for exponential growth is \( N(t) = N_0e^{rt} \) where \( N(t) \) is the amount at time \( t \), \( N_0 \) is the initial amount, \( r \) is the growth rate, and \( e \) is the base of the natural logarithm.
02

Create equations from the given information

Substitute given values into the standard equation. According to the problem, the initial population \( N_0 \) is 250. So: 1) After 1 hour, we have \( N(1) = 250e^{r} \). 2) After 10 hours, the population is double that of 1 hour, so we get \( N(10) = 2 * N(1) = 2 * 250e^{r} = 500e^{r} \) which leads to \( N(10) = 25e^{10r} \). Notice, we have two unknowns \( r \) and \( N(1) \), that we can solve for.
03

Solve for the growth rate \( r \).

By setting \( N(10) = N(1) \) we can solve for \( r \). This gives \( 500e^{r} = 250e^{10r} \) which simplifies to \( 2 = e^{9r} \). Taking the natural logarithm of both sides we get \( \ln{2} = 9r \). Hence, \( r = \ln{2} / 9 \) which approximates to 0.077.
04

Calculate the population after \( 6 \) hours

Now we know that \( r \approx 0.077 \). We can use this in the original equation to find the number of bacteria after \( 6 \) hours. \( N(6) = 250e^{0.077*6} \). After calculating this, we find that \( N(6) \approx 480 \) bacteria.

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Key Concepts

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

Exponential Growth Equation
Understanding the exponential growth equation is crucial for solving problems involving the rapid increase of quantities, such as populations or investments. Exponential growth is represented by the equation \[ N(t) = N_0e^{rt} \]where
    \t
  • \t\t\( N(t) \) is the amount at time \( t \).\t
  • \t
  • \t\t\( N_0 \) is the initial amount, or starting value.\t
  • \t
  • \t\t\( r \) is the growth rate, which indicates how quickly the amount is increasing over time.\t
  • \t
  • \t\t\( e \) is the base of the natural logarithm, an irrational constant approximately equal to 2.71828.\t
The power to which \( e \) is raised, \( rt \), is the product of the growth rate and time, indicating that growth compounds exponentially. This means that as time progresses, the rate of increase itself becomes larger, leading to a characteristic steep 'curve up' in graphical representations. It's important to note that when the quantity doubles over regular intervals, as in the given exercise, exponential growth is at play.
Solving Exponential Equations
When solving exponential equations, the basic approach is to isolate the term with the exponent and then use logarithms to solve for the variable. As exemplified in the bacteria population problem, we have to create two equations based on the given information and then find the value of the growth rate \( r \). The step-by-step process often involves the following steps:

    \t
  • Setting up the exponential growth equation with the initial conditions.
  • \t
  • Creating additional equations based on given data points or conditions.
  • \t
  • Isolating the exponential term which often includes the unknown variable.
  • \t
  • Taking the natural logarithm of both sides to convert the equation from an exponential form to a linear form, making it easier to solve.
  • \t
  • Finding the numerical value of the variable using algebra.
Using these steps, we can express relationships that initially seem complicated by exponential terms in a more straightforward, solvable manner. This process is fundamental for understanding a wide range of problems involving exponential growth, including those in biology, finance, and physics.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is an operation that is crucial for unraveling equations involving exponential growth. It is the inverse of raising the number \( e \) to a power, answering the question: 'To what power must \( e \) be raised to produce a given number?' For example, if you have \( e^x = y \), then \( \ln(y) = x \).

This concept plays a pivotal role when solving for variables in exponential equations because it allows us to turn the multiplication of the exponent into addition or subtraction, a simpler operation to handle algebraically. As seen in the exercise solution, by taking the natural logarithm of both sides of the equation \( 2 = e^{9r} \), we convert the exponential equation into a linear one, allowing us to solve for the growth rate \( r \) directly. Understanding how to use the natural logarithm is essential for students tackling subjects that involve any sort of exponential changes, and it's a key element of logarithmic properties that extend into many areas of mathematics and science.

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Most popular questions from this chapter

A sport utility vehicle that costs \( \$23,300 \) new has a book value of \( \$12,500 \) after \( 2 \) years. (a) Find the linear model \( V = mt + b \). (b) Find the exponential model \( V = ae^{kt} \). (c) Use a graphing utility to graph the two models in the same viewing window. Which model depreciates faster in the first \( 2 \) years? (d) Find the book values of the vehicle after \( 1 \) year and after \( 3 \) years using each model. (e) Explain the advantages and disadvantages of using each model to a buyer and a seller.

In Exercises 81 - 112, solve the logarithmic equation algebraically. Approximate the result to three decimal places. \( \log x = 6 \)

A cup of water at an initial temperature of \( 78^{\circ}C \) is placed in a room at a constant temperature of \( 21^{\circ}C \). The temperature of the water is measured every 5 minutes during a half-hour period.The results are recorded as ordered pairs of the form \( (t \), \( T) \), where \( t \) is the time (in minutes) and \( T \) is the temperature (in degrees Celsius). \( \left(0, 78.0^{\circ}\right) \), \( \left(5 , 66.0^{\circ}\right) \), \( \left(10, 57.5^{\circ}\right) \), \( \left(15 , 51.2^{\circ}\right) \), \( \left(20 , 46.3^{\circ}\right) \), \( \left(25, 42.4^{\circ}\right) \), \( \left(30 , 39.6^{\circ}\right) \) (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \( \left(t , T\right) \) and \( \left(t, T - 21\right) \). (b) An exponential model for the data \( \left(t, T - 21\right) \) is given by \( T - 21 = 54.4\left(0.964\right)^t \) Solve for \( T \) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use a graphing utility to plot the points \( \left(t, In\left(T - 21\right)\right) \) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form \( In\left(T - 21\right) = at + b \). Solve for \( T \), and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \( y \)-coordinates of the revised data points to generate the points \( \dfrac{1}{T - 21} = at + b \). Solve for \( T \), and use a graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

If the annual rate of inflation averages \( 4\% \) over the next \( 10 \) years, the approximate costs \( C \) of goods or services during any year in that decade will be modeled by \( C(t) = P(1.04)^t \), where \( t \) is the time in years and \( P \) is the present cost. The price of an oil change for your car is presently \( \$23.95 \). Estimate the price \( 10 \) years from now.

In Exercises 29 - 44, find the exact value of the logarithmic expression without using a calculator. (If this is not possible,state the reason.) \( \log_6 \sqrt[3]{6} \)

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