Suppose the number of bacteria in a certain population increases according to a continuous exponential growth model. a sample of 2200 bacteria selected from the population reached the size 2300 bacteria in one and a half hours. Find the hourly growth rate parameter.

Answer :

Answer:

The hourly growth rate parameter of the bacteria is 3%.

Step-by-step explanation:

Continuous exponential growth formula:

[tex]A=Pe^{rt}[/tex]

A= Population of bacteria after t hours

P = Initial population

r= growth rate

t= time in hours

Given that,

A sample of 2200 bacteria selected from the population reached the size 2300 bacteria in one and half hours.

A=2300, P=2200, t [tex]=1\frac12[/tex] hours =[tex]\frac32[/tex] hours,r=?

[tex]\therefore2300=2200e^{\frac32r}[/tex]

[tex]\Rightarrow2200e^{\frac32r}=2300[/tex]

[tex]\Rightarrow e^{\frac32r}=\frac{2300}{2200}[/tex]

[tex]\Rightarrow e^{\frac32t}=\frac{23}{22}[/tex]

Taking ln both sides

[tex]\Rightarrow ln( e^{\frac32r})=ln(\frac{23}{22})[/tex]

[tex]\Rightarrow {\frac32r}=ln(\frac{23}{22})[/tex]

[tex]\Rightarrow r}=\frac{ln|\frac{23}{22}|}{\frac32}[/tex]

[tex]\Rightarrow r}\approx0.030[/tex]

[tex]\Rightarrow r}=3\%[/tex]

The hourly growth rate parameter of the bacteria is 3%.

The hourly growth rate is 2%

An exponential growth is in the form:

y = abˣ;

where y, x are variables, a is the initial value of y and b > 1

Let y represent the number of bacteria after x hours.

There was initially 2200 bacteria, hence a = 2200. The equation becomes:

[tex]y=2200(b)^x\\\\\\After\ 1.5\ hours,y=2300:\\\\\\2300=2200(b)^{1.5}\\\\\\b^{1.5}=1.045\\\\1.5lnb=ln(1.045)\\\\ln(b)=0.0296\\\\b=1.02[/tex]

Therefore the hourly growth rate is 1.02 - 1 = 0.02 = 2%

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