Suppose a bacteria culture begins with 45 bacteria and after 6 hours the population has grown to
68 bacteria.
a) Write a continuous growth function that models the population over time. To do
this you will need to solve for the rate of growth.
Use as your variable and keep at least 4 significant figures on the growth rate.
b) According to your model, what is the population after 9 hours?
bacteria
Round to the nearest whole number.
c) How long will it take for the population to reach 170 bacteria?
hours
Give your answer accurate to at least 2 decimal places.

Ask by Rowe Young. in the United States

Mar 17,2025

Question

Suppose a bacteria culture begins with 45 bacteria and after 6 hours the population has grown to
68 bacteria.
a) Write a continuous growth function that models the population over time. To do
this you will need to solve for the rate of growth.
Use as your variable and keep at least 4 significant figures on the growth rate.
b) According to your model, what is the population after 9 hours?
bacteria
Round to the nearest whole number.
c) How long will it take for the population to reach 170 bacteria?
hours
Give your answer accurate to at least 2 decimal places.

Ask by Rowe Young. in the United States

Mar 17,2025

UpStudy AI · Accepted Answer

**a)** We start with the continuous growth model  
$$
P(t)=P_0 e^{kt}
$$  
Given that initially $$ P(0)=45 $$ and after 6 hours $$ P(6)=68 $$, we have  
$$
45e^{6k}=68.
$$  
Solve for $$ k $$:  
$$
e^{6k}=\frac{68}{45}
$$  
Taking the natural logarithm of both sides,  
$$
6k=\ln\left(\frac{68}{45}\right)
$$  
Thus,  
$$
k=\frac{1}{6}\ln\left(\frac{68}{45}\right) \approx \frac{1}{6}\ln(1.5111) \approx \frac{0.4129}{6} \approx 0.06882.
$$  
So the growth function is  
$$
P(t)=45 e^{0.06882t}.
$$

**b)** To find the population after 9 hours, substitute $$ t=9 $$ into the function:  
$$
P(9)=45 e^{0.06882 \cdot 9}=45 e^{0.61938}.
$$  
Evaluating $$ e^{0.61938}\approx1.857 $$, we have  
$$
P(9)\approx45 \times 1.857\approx83.57.
$$  
Rounded to the nearest whole number,  
$$
P(9)=84 \text{ bacteria}.
$$

**c)** To determine the time $$ t $$ when the population reaches 170 bacteria, set  
$$
170=45 e^{0.06882t}.
$$  
Solve for $$ t $$:  
$$
e^{0.06882t}=\frac{170}{45},
$$  
Take the natural logarithm of both sides:  
$$
0.06882t=\ln\left(\frac{170}{45}\right).
$$  
Thus,  
$$
t=\frac{\ln\left(\frac{170}{45}\right)}{0.06882}.
$$  
Calculating,  
$$
\frac{170}{45}\approx3.7778,\quad \ln(3.7778)\approx1.3303,
$$  
so  
$$
t\approx\frac{1.3303}{0.06882}\approx19.33 \text{ hours}.
$$
**a)** The growth function is $$ P(t) = 45 e^{0.06882t} $$ **b)** After 9 hours, the population is $$ P(9) \approx 84 \text{ bacteria} $$ **c)** It takes approximately $$ t \approx 19.33 \text{ hours} $$ for the population to reach 170 bacteria.

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