Yes, that's good.

OkayNow I plug in y(3) = 16 and A = -99.3

(16^-.01)/-.01 = k(3) + -99.3

Solving for k I get...

[ (16^-.01)/-.01 + 99.3] / 3 = k = .681

Now I plug k and A back into the original equation...

(y^-.01)/-.01 = .681(t) + -99.3

No. y= 1/(-.00681t+ .993)^(1/.01)= 1/(-.00681t+ .993)^(100).The question is asking me when the lim(t -> T) = infinity, right? So that means there is an asymptote at t = T. This means I need to find a t that y(t) does not occupy.

Solving for y...

y^-.01 = (.681(t) + -99.3)*-.01

y^-.01 = -.00681t + .993

y = 1/(-.00681t + .993)^.01

Except for the ".01" power where it should be 100, yes. And that does not affect the final answer, the number of months.So if I solve -.00681t + .993 = 0 for t, then that should be my answer, right? Because that would mean there is a 0 in the denominator.

t = -.993 / -.00681 = (approx) 146 months

Is what I did correct?

Thanks!

P.S. Can anyone tell me the units for k and A?[/QUOTE]