Calculus
posted by Bill on .
The temperature of a cup of Starbucks coffee at time t (in minutes) is T(t)= 70 + c e^(kt) . Initially, the temperature of the coffee was 200 degrees F. Three minutes later, it was 180 degrees. When will the temperature of the coffee be 150 degrees F?
The back of the book says it is 8.7 minutes, but I can't figure out how to get that answer.
Thank you for your help.

t(0)=200=70+ce^0
solve for c (130)
then
t(3)=70+130e^k3=180
solve for k
e^k3=11o/130
k3=.167
k=.0557
150=70+130e^(..0557)
.0557t=.486
t=8.7 min 
Thank you so much.