Problem 20
Question
The concentration of bacteria \(B\) in millions per milliliter after \(x\) hours is given by $$B(x)=1.33 e^{0.15 x}$$ (a) How many bacteria are there after 2.5 hours? (b) How many bacteria are there after 8 hours? (c) After how many hours will there be 31 million bacteria per milliliter?
Step-by-Step Solution
Verified Answer
(a) 1.935 million, (b) 4.413 million, (c) 20.99 hours
1Step 1: Understanding the Problem
We are given an exponential function that models the concentration of bacteria over time. The objective is to determine the number of bacteria after specific durations and find the time when the bacteria reach a specified concentration.
2Step 2: Calculate Bacteria at 2.5 Hours
Substitute the value \(x = 2.5\) into the function.\[B(2.5) = 1.33 e^{0.15 \times 2.5}\]First, calculate the exponent:\[0.15 \times 2.5 = 0.375\]Then solve:\[B(2.5) = 1.33 e^{0.375}\]Using a calculator, find \(e^{0.375} \approx 1.4543\).\[B(2.5) \approx 1.33 \times 1.4543 \approx 1.935 \]So, there are approximately 1.935 million bacteria per milliliter after 2.5 hours.
3Step 3: Calculate Bacteria at 8 Hours
Substitute the value \(x = 8\) into the function:\[B(8) = 1.33 e^{0.15 \times 8}\]First, calculate the exponent:\[0.15 \times 8 = 1.2\]Then solve:\[B(8) = 1.33 e^{1.2}\]Using a calculator, find \(e^{1.2} \approx 3.3201\).\[B(8) \approx 1.33 \times 3.3201 \approx 4.413\]So, there are approximately 4.413 million bacteria per milliliter after 8 hours.
4Step 4: Determine Time for 31 Million Bacteria
Set the function equal to 31 to find \(x\):\[31 = 1.33 e^{0.15x}\]First, isolate the exponential term:\[e^{0.15x} = \frac{31}{1.33} \approx 23.308\]Take the natural logarithm of both sides:\[\ln(e^{0.15x}) = \ln(23.308)\]Which simplifies to:\[0.15x = \ln(23.308)\]Solving for \(x\):\[x = \frac{\ln(23.308)}{0.15}\]Calculate \(\ln(23.308) \approx 3.148\).\[x \approx \frac{3.148}{0.15} \approx 20.987\]Thus, it will take approximately 20.99 hours for there to be 31 million bacteria per milliliter.
Key Concepts
Bacterial GrowthExponential Growth ModelCalculating Exponents
Bacterial Growth
When studying bacterial growth, we often look at how bacteria multiply over time under certain conditions. Bacteria tend to reproduce by splitting in two, a process known as binary fission. This means a single bacterium divides into two identical daughter cells.
Due to binary fission, bacterial populations can increase rapidly, especially in ideal conditions with plenty of nutrients and a suitable environment. The growth is typically exponential, which means the population doubles at regular intervals.
In laboratory settings, scientists observe and model bacterial growth to understand different behaviors. This helps in areas like medicine, as bacterial growth is key to understanding infections and how to control them.
Due to binary fission, bacterial populations can increase rapidly, especially in ideal conditions with plenty of nutrients and a suitable environment. The growth is typically exponential, which means the population doubles at regular intervals.
In laboratory settings, scientists observe and model bacterial growth to understand different behaviors. This helps in areas like medicine, as bacterial growth is key to understanding infections and how to control them.
Exponential Growth Model
An exponential growth model is a mathematical representation of how a quantity increases over time. It's called 'exponential' because the rate of growth is proportional to the current value, resulting in a faster and faster increase as time progresses.
In our exercise, the exponential growth model is expressed through the equation \(B(x) = 1.33 e^{0.15x}\). Here, \(B(x)\) refers to the concentration of bacteria at any time \(x\) in millions per milliliter. The constant 1.33 represents the initial concentration, and \(e^{0.15x}\) models the exponential growth specifying how fast the growth is taking place.
The Euler's number \(e\) is a crucial part of the formula. It is approximately equal to 2.718 and is widely used in scientific calculations for modeling growth that occurs at a continually compounding rate. Exponential growth models apply not just in biology but in population dynamics, finance, and other fields.
In our exercise, the exponential growth model is expressed through the equation \(B(x) = 1.33 e^{0.15x}\). Here, \(B(x)\) refers to the concentration of bacteria at any time \(x\) in millions per milliliter. The constant 1.33 represents the initial concentration, and \(e^{0.15x}\) models the exponential growth specifying how fast the growth is taking place.
The Euler's number \(e\) is a crucial part of the formula. It is approximately equal to 2.718 and is widely used in scientific calculations for modeling growth that occurs at a continually compounding rate. Exponential growth models apply not just in biology but in population dynamics, finance, and other fields.
Calculating Exponents
Exponents play a significant role in understanding exponential functions. Calculating exponents involves raising a number to the power of another number. The base here is \(e\), and it is raised to the power of \(0.15x\) in the given bacterial growth model.
When calculating exponents like \(e^{0.15x}\), follow these steps:
When calculating exponents like \(e^{0.15x}\), follow these steps:
- First, calculate the multiplier for the exponent (\(0.15 \times x\)).
- Second, use a calculator to find the value of \(e\) raised to this exponent.
- Multiply this result by the initial amount, in this case, 1.33, to determine the bacteria concentration.
Other exercises in this chapter
Problem 19
Decide whether each function is one-to-one. Do not use a calculator. $$y=2 x^{3}+1$$
View solution Problem 19
Solve each equation. Give the exact answer. $$\log _{5} 125=x$$
View solution Problem 20
Find the domain of each logarithmic function analytically. You may wish to support your answer graphically. $$y=\log \left(x^{3}-81 x\right)$$
View solution Problem 20
Decide whether each function is one-to-one. Do not use a calculator. $$y=-2 x^{5}-4$$
View solution