Question 658621
For this problem in Newtons law of cooling, what does the e mean? Why do we use ln and what is ln? 
T (t) = Te + (T0 − Te ) e^-kt
T(t) is he temperature of the object at time "t".
Te is the temperature of the environment.
To is the temperature of the object at some initial time (e.g. time of death)
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The e in e^kt is the irrational number e = 2.718281828...
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160= 69 + (190-69)e^k2
160-69 = 69-69(121)e^k2
91 = (121)ek2
91/121 = (121/121)e^2k
e^2k = 91/121
2k = ln(91/121)
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ln is the "natural log"; it is the power of "e" that gives you some 
particular number.
For example:
ln(e^k2) = k2
ln(5) = 1.609.. because e^(1.609..) = 5
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k = .5(ln(91/121))
k = .5(-.284931)
k = -.1425 
69 + (180 - 69)e^-.1425t = 130
69-69 + 121e-.1425t = 130-69
121e.-1425t = 61
e^-.1425t = 61/121
-.1425t = ln(61/121)
t = (-.684917)/(-.1425)
t = 4.8 minutes 
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Cheers,
Stan H.