Unformatted text preview: , we use the fact that
after 2 hours, the roast is 125◦ F, which means T (2) = 125. This gives rise to the equation
350 − 310e−2k = 125 which yields k = − 1 ln 45 ≈ 0.1602. The temperature function is
2
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T (t) = 350 − 310e 2 ln( 62 ) ≈ 350 − 310e−0.1602t .
t 45 384 Exponential and Logarithmic Functions 2. To determine when the roast is done, we set T (t) = 165. This gives 350 − 310e−0.1602t = 165
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whose solution is t = − 0.1602 ln 37 ≈ 3.22. It takes roughly 3 hours and 15 minutes to cook
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the roast completely.
If we had taken the time to graph y = T (t) in Example 6.5.4, we would have found the horizontal
asymptote to be y = 350, which corresponds to the temperature of the oven. We can also arrive
at this conclusion by applying a bit of ‘number sense’. As t → ∞, −0.1602t ≈ very big (−) so
that e−0.1602t ≈ very small (+). The larger the value of t, the smaller e−0.1602t becomes so that
T (t) ≈ 350 − very small (+), which indicates the...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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