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# A population of fruit flies starts with 6 flies. On Day 4, the population has grown to 94 flies. Wrote an exponential growth function to mod

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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

#1
+80935
+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
Sort:

#1
+80935
+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
#2
+91435
+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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