Problem 37
Question
Use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
Step-by-Step Solution
Verified Answer
The population size after 3 hours is 691200.
1Step 1: Understand the Growth Factor
The exercise involves bacteria doubling every 20 minutes. This implies that every 20 minutes, the population of bacteria becomes twice as much. In one hour (60 minutes), the bacteria double 3 times (as 60/20 = 3). For the base of the exponential equation, we identify that each 20-minute interval multiplies the count by 2, so during an hour, the growth factor is given by multiplying three times the doubling rate: \[(2^3 = 8)\] Thus, the growth factor is 8 every hour.
2Step 2: Establish the Equation
Now we need to write an exponential equation for the bacteria growth, starting from an initial population of 1350. With a growth factor of 8 per hour, the equation is \[P(t) = 1350 \times 8^t\] where \(P(t)\) is the population after \(t\) hours, and 8 is the growth factor indicating population growth in each hour.
3Step 3: Calculate Population After 3 Hours
Substitute \(t = 3\) into the exponential equation to find the population after 3 hours: \[P(3) = 1350 \times 8^3\] First calculate \(8^3\): \[8^3 = 8 \times 8 \times 8 = 512\]Then multiply: \[P(3) = 1350 \times 512\]Which results in \[P(3) = 691200\].
4Step 4: Round to the Nearest Whole Number
Since the calculated population is already a whole number, there's no need for further rounding in this instance. Thus, the population size after 3 hours is 691200.
Key Concepts
Growth FactorExponential EquationPopulation DoublingBacteria Growth Model
Growth Factor
The concept of a growth factor is critical in understanding exponential growth, like in the bacteria growth scenario. The growth factor is a number that indicates how many times the population multiplies over a specific interval of time.
In this case, the bacteria double every 20 minutes. This means the population is multiplied by 2 for each 20-minute period.
In one hour, there are 60 minutes, so the bacteria double 3 times (since 60 divided by 20 equals 3). Each doubling event means multiplying the current population by 2. Therefore, the growth factor over one hour is computed as follows:
In this case, the bacteria double every 20 minutes. This means the population is multiplied by 2 for each 20-minute period.
In one hour, there are 60 minutes, so the bacteria double 3 times (since 60 divided by 20 equals 3). Each doubling event means multiplying the current population by 2. Therefore, the growth factor over one hour is computed as follows:
- First doubling: Population Multiplies by 2
- Second doubling: Population Multiplies by 2 again
- Third doubling: Population Multiplies by 2 once more
Exponential Equation
An exponential equation is foundational to model scenarios like bacterial growth, where the increase is rapid and compounds over consistent time intervals. For the bacteria, which start with an initial count of 1350, we use an exponential formula to represent this growth.
We know the growth factor is 8 every hour. Thus, the function expressing this growth is:\[P(t) = 1350 \times 8^t\]Here:
We know the growth factor is 8 every hour. Thus, the function expressing this growth is:\[P(t) = 1350 \times 8^t\]Here:
- \(P(t)\) is the population after \(t\) hours
- 1350 is the initial population
- 8 is the hourly growth factor
Population Doubling
Population doubling is a specific type of exponential growth where a population doubles at constant intervals. When a population doubles, it indicates that the growth is multiplying at a consistent rate, which can be described mathematically.
This scenario is what we observe with the bacteria, doubling every 20 minutes. Over an hour (or any given period), the doubling effect results in significant growth. The doubling time (20 minutes, in this case) allows us to find how many times the population will double in longer time frames, like one hour.
The doubling nature of growth is crucial to understand how rapid exponential growth can be, especially when small intervals cause large-scale increases due to the compounding effect.
This scenario is what we observe with the bacteria, doubling every 20 minutes. Over an hour (or any given period), the doubling effect results in significant growth. The doubling time (20 minutes, in this case) allows us to find how many times the population will double in longer time frames, like one hour.
The doubling nature of growth is crucial to understand how rapid exponential growth can be, especially when small intervals cause large-scale increases due to the compounding effect.
Bacteria Growth Model
The bacteria growth model described here is a classic example of exponential growth, where the bacterial population grows rapidly in recurring cycles.
Such a model is critical in fields like microbiology and environmental science, where understanding bacterial behavior has practical applications.
Initially, the population comprises 1350 bacteria, and given that the population doubles every 20 minutes, we outline the exponential growth using the equation:\[P(t) = 1350 \times 8^t\]After establishing this model, we apply it to determine the population after specific time intervals, like 3 hours. By substituting \(t = 3\) into the equation, we calculate:
Such a model is critical in fields like microbiology and environmental science, where understanding bacterial behavior has practical applications.
Initially, the population comprises 1350 bacteria, and given that the population doubles every 20 minutes, we outline the exponential growth using the equation:\[P(t) = 1350 \times 8^t\]After establishing this model, we apply it to determine the population after specific time intervals, like 3 hours. By substituting \(t = 3\) into the equation, we calculate:
- \(8^3 = 512\), representing the compound growth
- Calculate the population: \(P(3) = 1350 \times 512\)
- The result is a population of 691200 bacteria after 3 hours
Other exercises in this chapter
Problem 36
For the following exercises, use the definition of common and natural logarithms to simplify. $$ \log \left(100^{8}\right) $$
View solution Problem 37
For the following exercises, sketch the graphs of each pair of functions on the same axis.$$f(x)=\log (x) \text { and } g(x)=\log _{\frac{1}{2}}(x)$$
View solution Problem 37
For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to f
View solution Problem 37
Refer to Table. $$ \begin{array}{|c|c|c|c|c|c|c|} \hline \boldsymbol{x} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \boldsymbol{f}(\boldsymbol{x}) & 5.1 & 6.3 & 7.3 & 7.7
View solution