Differential equations: Newton's Law of Cooling

Need help on this one. DE Newton's Law of Cooling

At 9 A.M., a thermometer reading 70°F is taken outdoors where the temperature is
l 5°F. At 9: 05 A.M., the thermometer reading is 45°F. At 9: 10 A.M., the thermometer is
taken back indoors where the temperature is fixed at 70°F. Find (a) the reading at
9: 20 A.M. and (b) when the reading, to the nearest degree, will show the correct
(70°F) indoor temperature.

Newton's law of Cooling:
dTdt=k(TTs)

dTTTs=kdt

ln(TTs)=kt+lnC

ln(TTs)=lnekt+lnC

ln(TTs)=lnCekt
 

Hence,
TTs=Cekt
 

At 9:00 am (t = 0 ), the thermometer was taken outdoors where Ts =15°F and the thermometer reading was T = 70°F.
7015=Cek(0)

C=55
 

Therefore,
T15=55ekt
 

At 9:05 am (t = 5), the thermometer reads 45°F:
4515=55ek(5)

k=0.1212271607
 

At 9:10 am (t = 10), the thermometer is taken back indoors.
T15=55e0.1212271607(10)

T=31.364F   ←   thermometer reading just before it was taken back indoors
 

Inside the room, the surrounding temperature Ts is 70°F, and we will reset the t to zero, hence,
TTs=Cekt

31.36470=Ce0.1212271607(0)

C=38.636F
 

Therefore, the equation when the thermometer is inside the room is
T70=38.636e0.1212271607t
 

At 9:20 am, 10 minutes after the thermometer was taken indoors
T70=38.636e0.1212271607(10)

T=58.505F
 

For T = 70°F
7070=38.636e0.1212271607t
 

Since logarithm of zero do not exist, we can approximate the solution. Use T → 70-, say T = 69.99999 then solve for t. It will actually do the job.
 

I leave it to you then.