Pies can only be delivered when their temperature is 120 degrees.When they are taken out of the oven, after two minutes their temperature is

I'm going to use the following function to model this

T = ab^x      where T is the temperature after some "x" minutes have elapsed

So we have, when x = 2

323 = ab²     →   323/b²  = a

And when x = 5   we have

288 = ab⁵  =  (323/b²)b⁵   = 323 b³     divide both sides by 323

288/323 = b³    take the 3rd root of each side

(288/323)^(1/3)  = b = about .9624

So, a = 323/(.9624)² =  about 348.66

So, our function is

T = 348.66(.9624)^x

So, to find out when they will be at 120 degrees, we have

120 = 348.66(.9624)^x        divide both sides by 348.66

120/348.66  =  (.9624)^x    take the log of each side

log ( 120/348.66 )  = log (.9624)^x     and by a log property, we have

log ( 120/348.66 )  =  x log (.9624)    divide both sides by log(.9624)

log ( 120/348.66 ) / log (.9624)= x = about 27.83 minutes

Here's a graph of the function.......https://www.desmos.com/calculator/logb5i0pwa

thanks Chris.  

I don't know Chris,

This might be the answer wanted but

Newton's law of cooling says that 

$$\\\mbox{T is temperature, t is time, and S is the surrounding temperature}\\\\ \frac{dT}{dt}=-k(t-S)\\\\ $I think this means$\\\\ T(t)-S=(T_0-S)e^{-kt}\\\\$$

I don't think that we are given enough information to solve this question.   

http://formulas.tutorvista.com/physics/newton-s-law-of-cooling-formula.html