Help problem with natural log

A pizza pan is removed at 2:00PM from an oven whose temperature is fixed at 450F into a room that is a constant 75F. After 5 minutes, the pizza pan is at 300F

A. At what time is the temperature of the pan 130 F?
B. Determine the time that needs to elapse before the pan is 190 F

By Newton's law of cooling
T(t) = A+(Ti-A)e^-kt
300 = 75+(450-75)e^-5k
225/375 = e^-5k
ln(225/375)=-5k = -0.51108256
k = 0.102165
T(t) = 75+(375)e^-0.102165t
A) 130 = 75+375e^-0.102165t
ln(55/375) = -0.102165t
t = 18.8 minutes
Pan will be at 130F at time 2:19 PM

B) 190 = 75+375e^-0.102165t
ln(115/375) = -0.102165t
t = 11.6 minutes

dT/dt=k(T-75)

dT/(T-75)=kdt

Intergration yields the following:

ln(T-75)=kt+lnlcl

from which we write (T-75)/c=e^(kt)

T=75+ce^(kt), Since T=450 when t=0, it follows that c=375, so that T=75+375e^(kt)

T=75+375e^(kt)
300=75+375e^(k*5)
k=-ln(5/3)/5

k=-.102165

A. At what time is the temperature of the pan 130 F?

130=75+375e^(-.102165*t)

t=18.7891 min


B. Determine the time that needs to elapse before the pan is 190 F

190=75+375e^(-.102165t)

t=11.5695 min