SOLUTION: Could someone please assist with the question: A pot of boiling soup with an internal temperature of 100� Fahrenheit was taken off the stove to cool in a 69�F room. After fif

Could someone please assist with the question: 
A pot of boiling soup with an internal temperature of 100� Fahrenheit was taken off the stove to cool in a 69�F room.
After fifteen minutes, the internal temperature of the soup was 93�F. 

To the nearest minute, how long will it take the soup to cool to 81�F? 

Here is my work:
69+(100-69) * e= 69+31*e
93=69+31*e(-k*15)
31*e^(-k *15)= 93-69= 24
e^-15k= 24/31= .7742
-15K= ln(.7742)
k= -ln(0.7742)/ 15=0.017

T(t)=  69+31*e (0.017)=81
e(-0.017*t)= 81-69/31=.387
-0.017*t= ln(.387)
t= - ln (.387)/ (0.017)= 55.8 min
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You can use Newton's Law of Cooling.......any of the following 3 formulae should work: 

This author's preference is the 1st formula. In this case, the cooling rate is first needed, and is derived as follows:

       , where:  = time taken to get to a COOLED temperature (15 minutes, in this case)
                                              = TEMPERATURE (T) at a given time (t)___(93oF, in this case) 
                                               = SURROUNDING Temperature (69oF, in this case)
                                               = ORIGINAL/INITIAL temperature (100oF, in this case)
                                                = the CONSTANT or COOLING rate (UNKNOWN, in this case)
                                      
                                       ----- Substituting 15 for t, 69o for , and 100o for 
                                              ----- Substituting 93o for T(15)
                                             
                                        ----- Converting to NATURAL LOGARITHMIC (ln) form
                                              
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To the nearest minute, how long will it take the soup to cool to 81�F?

                                    
                                       ----- Substituting 69o for , 100o for , and .0170622 for k
                                           ----- Substituting 81o for T(t)
                                          
                         ----- Converting to NATURAL LOGARITHMIC (ln) form
Time it takes for the soup to cool to 81oF, or 

As your answer, 55.8 min, or approximately 56 mins, coincides with mine, it is correct!

Great JOB!!