An infestation of grubs in a farmer's paddock has been increasing since he purchased the property. After 3 years of owning the property, the infestation of grubs reached 60 grubs per m^2. A survey done 5 years later (t=8) indicated there were now 105 grubs per m^2. The equation that models the infestation can be represented by I=Ak^t.

Find the constants A and k. please help

Question

An infestation of grubs in a farmer's paddock has been increasing since he purchased the property. After 3 years of owning the property, the infestation of grubs  reached 60 grubs per m^2. A survey done 5 years later (t=8) indicated there were now 105 grubs per m^2. The equation that models the infestation can be represented by I=Ak^t. 

Find the constants A and k. please help

marigirl · Answer

(3, 60) (8, 105)

daniellelovee · Answer

105/60=Ak8/Ak3 

7/4=k8-3 

7/4=k5 

k=?

marigirl · Answer

my k value is 1.118 (which is the answer in the back of the book). My A value is 42.88
Where as the back of the book says 42.94.. is my answer ok?

danjs · Answer

hmm,

danjs · Answer

$$\large 60=A*k^3~~~~~and~~~~~105=A*k^8$$

daniellelovee · Answer

looks correct to me :)

daniellelovee · Answer

and for A I got A=42.887  which I guess got simplified and that is why your book says 4.92

danjs · Answer

Solve second for A,  A=105/k^8,  and put in first
$$60=\frac{ 105 }{ k^5 }~~~~~k=(\frac{ 105 }{ 60 })^{1/5} \approx 1.118$$

danjs · Answer

It depends on how many decimals you use for the k value

daniellelovee · Answer

yes...but why are you repeating the steps that we have said?

danjs · Answer

exact numbers
$$\large k=\frac{ 7^{1/5}*2^{3/5} }{ 2 }$$
$$\large A=\frac{ 60*7^{2/5}*2^{6/5} }{ 7 }$$

danjs · Answer

That has A=42.88727...  ,  just depends on if you round off k