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A population of fruit flies starts with 6 flies. On Day 4, the population has grown to 94 flies. Write an exponential growth function to model the growth of the fly population.

Guest Aug 16, 2015

Best Answer 

 #1
avatar+78618 
+10

We have.....

 

Pt = Pbt-1     where P is the original population ....   Pt is the popoulation after "n" days  ... b is the growth rate... and t is the number of days ...[ t = 1 is the first day.....so, we're looking for the population 3 days after. the first day = the end of the 4th day ]....and we have

 

94 = 6b(4 - 1)   =    6b(3)      divide both sides by 6

 

(94/6)  = b3        take the cube root of both sides

 

b ≈ 2.5022

 

So, our function is  :     Pn =   6(2.5022)(t - 1)

 

 

CPhill  Aug 16, 2015
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2+0 Answers

 #1
avatar+78618 
+10
Best Answer

We have.....

 

Pt = Pbt-1     where P is the original population ....   Pt is the popoulation after "n" days  ... b is the growth rate... and t is the number of days ...[ t = 1 is the first day.....so, we're looking for the population 3 days after. the first day = the end of the 4th day ]....and we have

 

94 = 6b(4 - 1)   =    6b(3)      divide both sides by 6

 

(94/6)  = b3        take the cube root of both sides

 

b ≈ 2.5022

 

So, our function is  :     Pn =   6(2.5022)(t - 1)

 

 

CPhill  Aug 16, 2015
 #2
avatar+91001 
+5

Would it be reasonable for me to do it like this?

 

A population of fruit flies starts with 6 flies. On Day 4, the population has grown to 94 flies. Write an exponential growth function to model the growth of the fly population.

 

$$\\P=6e^{k(t-1)}\\\\
94=6e^{3k}\\\\
15.\dot6 = e^{3k} \\\\
ln(15.\bar6 )= ln(e^{3k}) \\\\
ln(15.\bar6 )= 3k \\\\
k=ln(15.\bar6 )/3 \\\\
k\approx 0.91718\\\\
so\\\\
P=6e^{0.91718(t-1)}\\\\$$

 

Is that ok?

Melody  Aug 17, 2015

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