1. ## Exponential Problem

Instant coffee is made by adding boiling water ( $212F$) to coffee mix. If the air temperature is $70F$, Newton's law of cooling says that after $t$ minutes, the temperature of the coffee will be given by a function of the form $f(t) = 70 - Ae^{-kt}$. After cooling for 2 minutes, the coffee is still $15F$ too hot to drink, but 2 minutes later it is just right. What is this ideal temperature for drinking?

$212 = 70 - Ae^{0}$

$A = 212 - 70$

$A = 142$

$15 = 70 - 142e^{-k2}$

$ln{\frac{\frac{(15-70)}{-142}}{-2}} = k$

$0.47 = k$

$f(4) = 70 - 142e^{-0.47\times4}$

$f(4) = 48.697F$

2. Originally Posted by Macleef
Instant coffee is made by adding boiling water ( $212F$) to coffee mix. If the air temperature is $70F$, Newton's law of cooling says that after $t$ minutes, the temperature of the coffee will be given by a function of the form $f(t) = 70 - Ae^{-kt}$. After cooling for 2 minutes, the coffee is still $15F$ too hot to drink, but 2 minutes later it is just right. What is this ideal temperature for drinking?

$212 = 70 - Ae^{0}$

$A = {\color{red}212 - 70}$

$A = 142$

$15 = 70 - 142e^{-k2}$

$ln{\frac{\frac{(15-70)}{-142}}{-2}} = k$

$0.47 = k$

$f(4) = 70 - 142e^{-0.47\times4}$

$f(4) = 48.697F$

Your mistake is in red. It should be $A=70-212=-142$. That would mean then that $f\left(t\right)=70+142e^{-kt}$