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Sagot :
Certainly! Let's break this down step by step.
### Part (a): Finding the Inverse of the Function
1. Original Function: The given function that models the bacteria population is:
[tex]\[ n = f(t) = 100 \left(2^{t / 3}\right) \][/tex]
where [tex]\( n \)[/tex] is the number of bacteria, and [tex]\( t \)[/tex] is the time in hours.
2. To find the inverse function, we need to express time [tex]\( t \)[/tex] in terms of the population [tex]\( n \)[/tex].
3. Equate to [tex]\( n \)[/tex]:
[tex]\[ n = 100 \left(2^{t / 3}\right) \][/tex]
4. Isolate the exponential term: Divide both sides of the equation by 100.
[tex]\[ \frac{n}{100} = 2^{t / 3} \][/tex]
5. Apply the logarithm: Take the logarithm base 2 on both sides to linearize the exponential.
[tex]\[ \log_2\left(\frac{n}{100}\right) = \frac{t}{3} \][/tex]
6. Solve for [tex]\( t \)[/tex]: Multiply both sides of the equation by 3.
[tex]\[ t = 3 \log_2\left(\frac{n}{100}\right) \][/tex]
Thus, the inverse function, which we can call [tex]\( f^{-1}(n) \)[/tex], is:
[tex]\[ t = f^{-1}(n) = 3 \log_2\left(\frac{n}{100}\right) \][/tex]
#### Meaning of the Inverse Function:
The inverse function [tex]\( f^{-1}(n) \)[/tex] calculates the amount of time [tex]\( t \)[/tex] required to reach a certain population [tex]\( n \)[/tex]. In this context, it answers the question: "Given a population size of [tex]\( n \)[/tex], how many hours have passed since the start when the initial population was 100 bacteria?"
### Part (b): Finding the Time When the Population Reaches 50,000
To determine when the population will reach 50,000 bacteria, we use the inverse function found in part (a).
1. Set the target population [tex]\( n \)[/tex]:
[tex]\[ n = 50,000 \][/tex]
2. Use the inverse function [tex]\( f^{-1}(n) \)[/tex] to find [tex]\( t \)[/tex]:
[tex]\[ t = 3 \log_2\left(\frac{50,000}{100}\right) \][/tex]
3. Simplify the expression inside the logarithm:
[tex]\[ t = 3 \log_2(500) \][/tex]
4. Evaluate [tex]\( \log_2(500) \)[/tex]:
[tex]\[ \log_2(500) \approx 8.96578 \][/tex]
5. Calculate [tex]\( t \)[/tex]:
[tex]\[ t = 3 \times 8.96578 \approx 26.897 \][/tex]
Therefore, the population will reach 50,000 bacteria in approximately [tex]\( 26.897 \)[/tex] hours.
### Part (a): Finding the Inverse of the Function
1. Original Function: The given function that models the bacteria population is:
[tex]\[ n = f(t) = 100 \left(2^{t / 3}\right) \][/tex]
where [tex]\( n \)[/tex] is the number of bacteria, and [tex]\( t \)[/tex] is the time in hours.
2. To find the inverse function, we need to express time [tex]\( t \)[/tex] in terms of the population [tex]\( n \)[/tex].
3. Equate to [tex]\( n \)[/tex]:
[tex]\[ n = 100 \left(2^{t / 3}\right) \][/tex]
4. Isolate the exponential term: Divide both sides of the equation by 100.
[tex]\[ \frac{n}{100} = 2^{t / 3} \][/tex]
5. Apply the logarithm: Take the logarithm base 2 on both sides to linearize the exponential.
[tex]\[ \log_2\left(\frac{n}{100}\right) = \frac{t}{3} \][/tex]
6. Solve for [tex]\( t \)[/tex]: Multiply both sides of the equation by 3.
[tex]\[ t = 3 \log_2\left(\frac{n}{100}\right) \][/tex]
Thus, the inverse function, which we can call [tex]\( f^{-1}(n) \)[/tex], is:
[tex]\[ t = f^{-1}(n) = 3 \log_2\left(\frac{n}{100}\right) \][/tex]
#### Meaning of the Inverse Function:
The inverse function [tex]\( f^{-1}(n) \)[/tex] calculates the amount of time [tex]\( t \)[/tex] required to reach a certain population [tex]\( n \)[/tex]. In this context, it answers the question: "Given a population size of [tex]\( n \)[/tex], how many hours have passed since the start when the initial population was 100 bacteria?"
### Part (b): Finding the Time When the Population Reaches 50,000
To determine when the population will reach 50,000 bacteria, we use the inverse function found in part (a).
1. Set the target population [tex]\( n \)[/tex]:
[tex]\[ n = 50,000 \][/tex]
2. Use the inverse function [tex]\( f^{-1}(n) \)[/tex] to find [tex]\( t \)[/tex]:
[tex]\[ t = 3 \log_2\left(\frac{50,000}{100}\right) \][/tex]
3. Simplify the expression inside the logarithm:
[tex]\[ t = 3 \log_2(500) \][/tex]
4. Evaluate [tex]\( \log_2(500) \)[/tex]:
[tex]\[ \log_2(500) \approx 8.96578 \][/tex]
5. Calculate [tex]\( t \)[/tex]:
[tex]\[ t = 3 \times 8.96578 \approx 26.897 \][/tex]
Therefore, the population will reach 50,000 bacteria in approximately [tex]\( 26.897 \)[/tex] hours.
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