A bacteria culture starts with 120 bacteria and grows at a rate proportional to its size. After 5 hours there will be 600 bacteria.

(a) Find the relative rate of growth, [tex]$k$[/tex], of the bacteria. Round to five decimal places.
[tex][tex]$k =$[/tex] \square$[/tex]

(b) Express the population [tex]$P$[/tex] after [tex]$t$[/tex] hours as a function of the form [tex]$P(t) = P_0 e^{kt}[tex]$[/tex].
[tex]$[/tex]P(t) = $ \square$[/tex]

(c) What will be the population after 8 hours?
[tex]\square[/tex] bacteria

(d) How long will it take for the population to reach 1550? Give your answer accurate to at least 2 decimal places.
[tex]\square[/tex] hours



Answer :

Sure, let's work through each part of this question step by step.

### (a) Finding the relative rate of growth, [tex]\( k \)[/tex]

The general form of the exponential growth equation is:
[tex]\[ P(t) = P_0 e^{kt} \][/tex]
where:
- [tex]\( P_0 \)[/tex] is the initial population,
- [tex]\( k \)[/tex] is the relative rate of growth,
- [tex]\( t \)[/tex] is the time,
- [tex]\( P(t) \)[/tex] is the population at time [tex]\( t \)[/tex].

We know:
- [tex]\( P_0 = 120 \)[/tex] (initial population),
- [tex]\( P(5) = 600 \)[/tex] (population after 5 hours).

Plugging in these values, we get:
[tex]\[ 600 = 120 e^{5k} \][/tex]

To solve for [tex]\( k \)[/tex]:
[tex]\[ \frac{600}{120} = e^{5k} \][/tex]
[tex]\[ 5 = e^{5k} \][/tex]
[tex]\[ \ln(5) = 5k \][/tex]
[tex]\[ k = \frac{\ln(5)}{5} \approx 0.32189 \][/tex]

So, the value of [tex]\( k \)[/tex] rounded to five decimal places is:
[tex]\[ k \approx 0.32189 \][/tex]

### (b) Expressing the Population [tex]\( P \)[/tex] After [tex]\( t \)[/tex] Hours

Using the value of [tex]\( k \)[/tex] we found in part (a), the function [tex]\( P(t) \)[/tex] will be:
[tex]\[ P(t) = 120 e^{0.32189 t} \][/tex]

### (c) Population After 8 Hours

To find the population after 8 hours, we use the function [tex]\( P(t) \)[/tex]:
[tex]\[ P(8) = 120 e^{0.32189 \times 8} \][/tex]
[tex]\[ P(8) \approx 1575.92 \][/tex]

So, the population after 8 hours will be approximately 1575.92 bacteria.

### (d) Time for Population to Reach 1550

We want to find the time [tex]\( t \)[/tex] when the population will reach 1550. Using the equation:
[tex]\[ 1550 = 120 e^{0.32189 t} \][/tex]

Solving for [tex]\( t \)[/tex]:
[tex]\[ \frac{1550}{120} = e^{0.32189 t} \][/tex]
[tex]\[ \ln\left(\frac{1550}{120}\right) = 0.32189 t \][/tex]
[tex]\[ t = \frac{\ln\left(\frac{1550}{120}\right)}{0.32189} \approx 7.95 \][/tex]

So, it will take approximately 7.95 hours for the population to reach 1550 bacteria.

### Summary of Results
(a) [tex]\( k \approx 0.32189 \)[/tex]

(b) [tex]\( P(t) = 120 e^{0.32189 t} \)[/tex]

(c) Population after 8 hours: ~1575.92 bacteria

(d) Time to reach a population of 1550: ~7.95 hours