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At the beginning of a study, a certain culture of bacteria has a population of 120. The population grows according to a continuous exponential growth model. After 7 days, there are 228 bacteria.

(a) Let [tex]t[/tex] be the time (in days) since the beginning of the study, and let [tex]y[/tex] be the number of bacteria at time [tex]t[/tex]. Write a formula relating [tex]y[/tex] to [tex]t[/tex]. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.

[tex]y = 120 e^{kt}[/tex]

(b) How many bacteria are there 19 days after the beginning of the study? Do not round any intermediate computations, and round your answer to the nearest whole number.

[tex]\square[/tex] bacteria



Answer :

GinnyAnswer
Certainly! Let's tackle this problem step-by-step.

### Part (a)

We are given:
- Initial population, [tex]\( P_0 = 120 \)[/tex]
- Population after 7 days, [tex]\( P(7) = 228 \)[/tex]

The population grows according to a continuous exponential growth model. The general formula for continuous exponential growth is:

[tex]\[ y = P_0 \cdot e^{kt} \][/tex]

Where:
- [tex]\( y \)[/tex] is the population at time [tex]\( t \)[/tex]
- [tex]\( P_0 \)[/tex] is the initial population
- [tex]\( k \)[/tex] is the growth rate constant
- [tex]\( e \)[/tex] is the base of the natural logarithm
- [tex]\( t \)[/tex] is the time in days

We need to find [tex]\( k \)[/tex]. We can use the data provided for [tex]\( t = 7 \)[/tex] days:

[tex]\[ 228 = 120 \cdot e^{7k} \][/tex]

Solving for [tex]\( k \)[/tex]:

1. Divide both sides by 120:

[tex]\[ \frac{228}{120} = e^{7k} \][/tex]
[tex]\[ 1.9 = e^{7k} \][/tex]

2. Take the natural logarithm on both sides to solve for [tex]\( k \)[/tex]:

[tex]\[ \ln(1.9) = 7k \][/tex]
[tex]\[ k = \frac{\ln(1.9)}{7} \][/tex]

### The formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex]:

[tex]\[ y = 120 \cdot e^{\left(\frac{\ln(1.9)}{7}\right) t} \][/tex]

This is the exact formula for the population growth.

### Part (b)

We need to find the population after 19 days. Using the formula we derived:

[tex]\[ y = 120 \cdot e^{\left(\frac{\ln(1.9)}{7}\right) \cdot 19} \][/tex]

Substituting [tex]\( t = 19 \)[/tex]:

1. Calculate the exponent:

[tex]\[ \left(\frac{\ln(1.9)}{7}\right) \cdot 19 \approx 0.0916934123103421 \times 19 \approx 1.7421748338964999 \][/tex]

2. Calculate the final population:

[tex]\[ y = 120 \cdot e^{1.7421748338964999} \approx 120 \cdot 5.708333333333432 \approx 685 \][/tex]

Therefore, the population after 19 days is approximately 685 bacteria.

### Final Answers

1. Formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex]:

[tex]\[ y = 120 \cdot e^{\left(\frac{\ln(1.9)}{7}\right) t} \][/tex]

2. The number of bacteria 19 days after the beginning of the study:

[tex]\[ \approx 685 \][/tex] bacteria.

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