   Chapter 6.5, Problem 4E

Chapter
Section
Textbook Problem

# A bacteria culture grows with constant relative growth rate. The bacteria count was 400 after 2 hours and 25, 600 after 6 hours.(a) What is the relative growth rate? Express your answer as a percentage.(b) What was the initial size of the culture?(c) Find an expression for the number of bacteria after t hours.(d) Find the number of cells after 4.5 hours.(e) Find the rate of growth after 4.5 hours.(f) When will the population reach 50, 000?

To determine

To evaluate:

a) Relative growth rate i.e. k

b) Initial size of the bacteria culture i.e. P(0)

c) Expression for number of cells after t hours

d) Number of bacteria after 4.5 hours

e) Growth rate of bacteria after 4.5 hours

f) When will the population reached 50000 i.e. P(t)=50000

Explanation

We measure time t in hours and let P(t) denotes the number of bacteria after t hours.

Given that:

1. P(2)=400

2. P(6)=25600

Formula used:

(1) Model of population growth: P(t)=P(0)ekt where k denotes the relative growth.

(2) Rate of change of population: dPdt=kP

Calculation:

Consider, the model of growth for the number bacteria

P(t)=P(0)ekt                                            (1)

Part (a):

It is given that P(2)=400. Replace this value in above model, we get400=P(2)     =P(0)e2k                                 From this we get P(0)=400e2k                       (2)Also, P(6)=25600, therfore by replacing this value in the equation (1) we get25600=P(6)          =P(0)e6k          =400e2ke6k                        (By replacing value of P(0) from equation (2))          =400e4k

Divide both sides by 400 we get64=e4kTaking log both side we getln64=lne4k        =4kThis implies that k=ln644=ln(64)14=ln81.039721.04

Conclusion (a): The relative growth rate of bacteria cells is k=1.04. In percentage, relative growth rate is 104%.

Part (b):

Put k=ln8 in equation (2), we getP(0)=400e2×ln8       =400eln8        =4008        =50

Conclusion (b): Thus, the initial size of bacteria is P(0)=50

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