/*! 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 50 The number of bacteria in a cult... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 50

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
The number of bacteria in the culture after 6 hours is approximately equal to \(250e^{6 ln(2) / 9}\). The precise number depends on the value used for \(e\) and the level of precision required.

Step by step solution

01

Utilize the data to Develop Equation for Growth Rate

To find the growth rate, use the information given - the population after 10 hours is double the population after 1 hour. This is expressed as \(P_{10} = 2P_{1}\), which in exponential growth terms becomes \(250e^{10k} = 2(250e^{k})\). This equation simplifies to \(e^{10k} = 2e^{k}\). By taking the natural logarithm on both sides of this equation, we obtain \(10k = k + ln(2)\).
02

Derive the Growth Rate

Solving the equation \(10k = k + ln(2)\) yields \(k = ln(2) / 9\).
03

Calculate the Population after 6 Hours

Now that we have the growth rate, we can use it to find the population after 6 hours through \(P_{6} = P_0 e^{6k} = 250e^{6 ln(2) / 9}\)
04

Final Calculation

Calculate the final number by evaluating the equation above, noting that the value of \(e\) can be approximated to be roughly 2.71828.

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 91影视!

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

Use the following information for determining sound intensity. The level of sound \(\boldsymbol{\beta}\), in decibels, with an intensity of \(I\), is given by \(\boldsymbol{\beta}=10 \log \left(I / I_{0}\right),\) where \(I_{0}\) is an intensity of \(10^{-12}\) watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. In Exercises 65 and 66 , find the level of sound \(\boldsymbol{\beta}\). (a) \(I=10^{-10}\) watt per \(\mathrm{m}^{2}\) (quiet room) (b) \(I=10^{-5}\) watt per \(\mathrm{m}^{2}\) (busy street corner) (c) \(I=10^{-8}\) watt per \(\mathrm{m}^{2}\) (quiet radio) (d) \(I=10^{0}\) watt per \(\mathrm{m}^{2}\) (threshold of pain)

A sport utility vehicle that costs $$\$ 23,300$$ new has a book value of $$\$ 12,500$$ after 2 years. (a) Find the linear model \(V=m t+b\). (b) Find the exponential model \(V=a e^{k t}\) (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.

The demand equation for a limited edition coin set is \(p=1000\left(1-\frac{5}{5+e^{-0.001 x}}\right)\) Find the demand \(x\) for a price of (a) \(p=\$ 139.50\) and (b) \(p=\$ 99.99\).

The amount of time (in hours per week) a student utilizes a math-tutoring center roughly follows the normal distribution \(y=0.7979 e^{-(x-5.4)^{2} / 0.5},\) \(4 \leq x \leq 7,\) where \(x\) is the number of hours. (a) Use a graphing utility to graph the function. (b) From the graph in part (a), estimate the average number of hours per week a student uses the tutoring center.

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$6 \log _{3}(0.5 x)=11$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
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.