506 CHAPTER 4 Inverse, Exponential, and Logarithmic Functions SOLUTION (a) We must find values for C and k in the given formula. As given, when t = 0, T0 = 20, and the temperature of the coffee is ƒ102 = 100. ƒ1t2 = T0 + Ce-kt Given function 100 = 20 + Ce-0k Let t = 0, ƒ102 = 100, and T 0 = 20. 100 = 20 + C e0 = 1 80 = C Subtract 20. The following function models the data. ƒ1t2 = 20 + 80e-kt Let T 0 = 20 and C = 80. The coffee cools to 60°C after 1 hr, so when t = 1, ƒ112 = 60. ƒ1t2 = 20 + 80e-kt Above function with T 0 = 20 and C = 80 60 = 20 + 80e-1k Let t = 1 and ƒ112 = 60. 40 = 80e-k Subtract 20. 1 2 = e-k Divide by 80. ln 1 2 = ln e-k Take the natural logarithm on each side. ln 1 2 = -k ln e x = x, for all x. k ≈0.693 Multiply by -1, rewrite, and use a calculator. Thus, the model is ƒ1t2 = 20 + 80e-0.693t. (b) To find the temperature after 1 2 hr, let t = 1 2 in the model from part (a). ƒ1t2 = 20 + 80e-0.693t Model from part (a) ƒ a 1 2b = 20 + 80e1-0.693211/22 Let t = 1 2 . ƒ a 1 2b ≈76.6°C Use a calculator. (c) To find how long it will take for the coffee to cool to 50°C, let ƒ1t2 = 50. ƒ1t2 = 20 + 80e-0.693t Model from part (a) 50 = 20 + 80e-0.693t Let ƒ1t2 = 50. 30 = 80e-0.693t Subtract 20. 3 8 = e-0.693t Divide by 80. ln 3 8 = ln e-0.693t Take the natural logarithm on each side. ln 3 8 = -0.693t ln e x = x, for all x. t = ln 3 8 -0.693 Divide by -0.693 and rewrite. t ≈1.415 hr, or about 1 hr, 25 min S Now Try Exercise 27.
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