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**+10 **

We have.....

Pt = Pb^{t-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) = b^{3 } take the cube root of both sides

b ≈ 2.5022

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

CPhill
Aug 16, 2015

#1**+10 **

Best Answer

We have.....

Pt = Pb^{t-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) = b^{3 } 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**+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