Answer:
[tex]k \approx 0.44[/tex]
Step-by-step explanation:
Given function:
[tex]f(t) = (ce)^{-kt}+32[/tex]
As per question statement:
Initial temperature of bottle is 70 [tex]^\circ F[/tex].
i.e. when time = 0 minutes, f(t) = 70 [tex]^\circ F[/tex]
[tex]70 = ce^{-k\times 0}+32\\\Rightarrow 38 = ce^{0}\\\Rightarrow c = 38[/tex]
After t = 3, f(t) = 42[tex]^\circ F[/tex]
[tex]42 = 38 \times e^{-k\times 3}+32\\\Rightarrow 42-32 = 38 \times e^{-3k} \\\Rightarrow 10 = 38 \times e^{-3k} \\\Rightarrow e^{3k} = \dfrac{38}{10}\\\Rightarrow e^{3k} = 3.8\\\\\text{Taking } log_e \text{both the sides:}\\\\\Rightarrow log_e {e^3k} = log_e {3.8}\\\Rightarrow 3k \times log_ee=log_e {3.8} (\because log_pq^r=r \times log_pq)\\\Rightarrow 3k \times 1=log_e {3.8}\\\Rightarrow 3k = 1.34\\\Rightarrow k = \dfrac{1.34}{3}\\\Rightarrow k \approx 0.44[/tex]
Hence, the value is:
[tex]k \approx 0.44[/tex]
